Text Book Back Questions and Answers - Chapter 6 Optics 12th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Optics
Multiple Choice Questions
Question 1.
The speed of light in an isotropic medium depends on,
(a) its intensity
(b) its wavelength
(c) the nature of propagation
(d) the motion of the source w.r.to medium
Answer:
(b) its wavelength
Question 2.
A rod of length $10 \mathrm{~cm}$ lies along the principal axis of a concav e mirror of focal length $10 \mathrm{~cm}$ in such a way that its end closer to the pole is $20 \mathrm{~cm}$ away from the mirror. The length of the image is, (AIPMT Main 2012)
(a) $2.5 \mathrm{~cm}$
(b) $5 \mathrm{~cm}$
(c) $10 \mathrm{~cm}$
(d) $15 \mathrm{~cm}$
Answer:
(b) $5 \mathrm{~cm}$
Hint:
By mirror formula, image distance of A
$
\begin{aligned}
& \frac{1}{v}+\frac{1}{u}=\frac{1}{f} ; \frac{1}{v_A}+\frac{1}{u}=\frac{1}{f} \\
& \frac{1}{v_A}+\frac{1}{u}+\frac{1}{(-30)}=\frac{1}{(-10)}
\end{aligned}
$

$
\begin{aligned}
& \therefore \mathrm{v}_{\mathrm{A}}=-15 \mathrm{~cm} \\
& \text { Image distance of } \mathrm{C}, \mathrm{v}_{\mathrm{c}}=-20 \mathrm{~cm} \\
& \text { The length of image }=\left|\mathrm{v}_{\mathrm{A}}-\mathrm{v}_{\mathrm{c}}\right| \\
& =|-15+20|=5 \mathrm{~cm}
\end{aligned}
$
Question 3.
An object is placed in front of a convex mirror of focal length of/and the maximum and minimum distance of an object from the mirror such that the image formed is real and magnified. (IEE Main 2009)
(a) $2 f$ and $c$
(b) $c$ and $\infty$
(c) $f$ and $\mathrm{O}$
(d) None of these
Answer:
(d) None of these
Hint:
There is no maximum minimum object distance for convex mirror to form real and inverted image.
Question 4.
For light incident from air onto a slab of refractive index 2. Maximum possible angle of refraction is,
(a) $30^{\circ}$
(b) $45^{\circ}$
(c) $60^{\circ}$
(d) $90^{\circ}$
Answer:
(a) $30^{\circ}$
Hint:
From Snell's law, $\mu=\frac{\sin i}{\sin r}$
Now consider an angle of incident is $90^{\circ}$
$
\begin{aligned}
& \frac{\sin 90^{\circ}}{2} \\
& \sin r=\sin ^{-1}(0.5) \\
& r=30^{\circ}
\end{aligned}
$
Question 5.
If the velocity and wavelength of light in air is $\mathrm{V}_{\mathrm{a}}$ and $\lambda_{\mathrm{a}}$ and that in water is $\mathrm{V}_{\mathrm{a}}$ and $\lambda_{\mathrm{W}}$, then the refractive index of water is,
(a) $\frac{V_w}{V_a}$
(b) $\frac{V_a}{V_w}$

(c) $\frac{\lambda_w}{\lambda_a}$
(d) $\frac{V_a \lambda}{V_w \lambda_w}$
Answer:
(b) $\frac{V_a}{V_w}$
Question 6.
Stars twinkle due to
(a) reflection
(b) total internal reflection
(c) refraction
(d) polarisation
Answer:
(c) refraction
Question 7.
When a biconvex lens of glass having refractive index 1.47 is dipped in a liquid, it acts as a plane sheet of glass. This implies that the liquid must have refractive index,
(a) less than one
(b) less than that of glass
(c) greater than that of glass
(d) equal to that of glass
Answer:
(d) equal to that of glass
Hint:
According to len's maker formula,
$
\frac{1}{f}=\left(\frac{n_2}{n_1}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)
$
Biconvex lens dipped in a liquid, acts as a plane sheet of glass, $f=\infty ; \frac{1}{f}=0$
$
\frac{n_2}{n_1}-1=0 ; \mathrm{n}_2=\mathrm{n}_1
$
Question 8.
The radius of curvature of curved surface at a thin planoconvex lens is $10 \mathrm{~cm}$ and the refractive index is 1.5. If the plane surface is silvered, then the focal length will be,
(a) $5 \mathrm{~cm}$
(b) $10 \mathrm{~cm}$
(c) $15 \mathrm{~cm}$
(d) $20 \mathrm{~cm}$
Answer:
(b) $10 \mathrm{~cm}$
Hint:
According to len's maker formula,

$
\begin{aligned}
& \frac{1}{\mathrm{f}}=(\mathrm{n}-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\
& \mathrm{R}_1=\infty ; \mathrm{R}_2=-6 \\
& \therefore f=\frac{R}{(n-1)}(\mathrm{R}=10 \mathrm{~cm} ; \mathrm{n}=1.5 \\
& f=\frac{10}{(1.5-1)}=20 \mathrm{~cm}
\end{aligned}
$
Question 9.
An air bubble in glass slab of refractive index 1.5 (near normal incidence) is $5 \mathrm{~cm}$ deep when viewed from one surface and $3 \mathrm{~cm}$ deep when viewed from the opposite face. The thickness of the slab is,
(a) $8 \mathrm{~cm}$
(b) $10 \mathrm{~cm}$
(c) $12 \mathrm{~cm}$
(d) $16 \mathrm{~cm}$
Answer:
(c) $12 \mathrm{~cm}$
Hint:
Let $\mathrm{d}_1=5 \mathrm{~cm}$ and $\mathrm{d}_2=3 \mathrm{~cm} ; \mathrm{n}=1.5$
Actual width is the sum of real depth from 2 sides
Thickness of slab $=d_1 \mathrm{n}+d_2 \mathrm{n}$
$
=(5 \times 1.5)+(3 \times 1.5)=12 \mathrm{~cm}
$

Question 10.
A ray of light travelling in a transparent medium of refractive index $\mathrm{n}$ falls, on a surface separating the medium from air at an angle of incidents of $45^{\circ}$. The ray can undergo total internal reflection for the following $\mathrm{n}$,
(a) $\mathrm{n}=1.25$
(b) $\mathrm{n}=1.33$
(c) $\mathrm{n}=1.4$
(d) $\mathrm{n}=1.5$
Answer:
(d) $\mathrm{n}=1.5$
Hint:
For total internal reflection
$\sin i>\sin c \quad$ where, $i$ - angle of incidence
But, $\sin \mathrm{c}=1 / \mathrm{n} \quad \mathrm{c}-$ critical angle
$\sin i>1 / \mathrm{n}$
$\mathrm{n}>\frac{1}{\sin i}$
$\mathrm{n}>\frac{1}{\sin 45} ; \mathrm{n}>\sqrt{ } 2 ; \mathrm{n}>1.414$
Question 11.
A plane glass is placed over a various coloured letters (violet, green, yellow, red) The letter which appears to be raised more is,
(a) red
(b) yellow
(c) green
(d) violet
Answer:
(d) violet
Hint:
Letters appear to be raised depending upon the refractive index of the material. Since violet has a higher refractive index than red (the index increases with frequency), red will be the lowermost.
Question 12.
Two point white dots are $1 \mathrm{~mm}$ apart on a black paper. They are viewed by eye of pupil diameter $3 \mathrm{~mm}$ approximately. The maximum distance at which these dots can be resolved by the eye is, [take 

wavelength of light, $\lambda=500 \mathrm{~nm}$ ]
(a) $1 \mathrm{~m}$
(b) $5 \mathrm{~m}$
(c) $3 \mathrm{~m}$
(d) $6 \mathrm{~m}$
Answer:
(b) $5 \mathrm{~m}$
Hint:
Resolution limit $\sin \theta=\frac{Y}{d}=\frac{1.22 c}{a}$
$
\mathrm{D}=\frac{\mathrm{Y} d}{1.22 \lambda}=\frac{1 \times 10^{-3} \times 3 \times 10^{-3}}{1.22 \times 500 \times 10^{-9}} \approx 5 \mathrm{~m}
$
Question 13.
In a Young's double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to,
(a) $2 \mathrm{D}$
(b) $\frac{D}{2}$
(c) $\sqrt{2 \mathrm{D}}$
(d) $\frac{D}{\sqrt{2}}$
Answer:
(a) $2 \mathrm{D}$
Hint:
Young's double -slite experiment is
$
\beta=\frac{\lambda D}{d} ; \beta^{\prime}=\frac{\lambda D^{\prime}}{d^{\prime}} ; \mathrm{d}^{\prime}=2 \mathrm{~d}
$
Same fringe space, $\beta=\beta^{\prime}$
$
\Rightarrow \frac{\lambda D}{d}=\frac{\lambda D^{\prime}}{d^{\prime}} ; \mathrm{D}^{\prime}=2 \mathrm{D}
$
Question 14.
Two coherent monochromatic light beams of intensities I and 41 are superposed. The maximum and minimum possible intensities in the resulting beam are [IIT-JEE 1988]
(a) $5 I$ and I
(b) $5 \mathrm{I}$ and $3 \mathrm{I}$
(c) $9 I$ and I
(d) $9 I$ and 31
Answer:
(c) 9I and I

Hint:
$
\begin{aligned}
& I_{\max }=(\sqrt{I+\sqrt{4 I}})^2=9 I \\
& I_{\min }=(\sqrt{I-\sqrt{4 I}})^2=I \\
& I_{\max }: I_{\min }=9 I: I
\end{aligned}
$
Question 15.
When light is incident on a soap film of thickness $5 \times 10^{-5}$ $\mathrm{cm}$, the wavelength of light reflected maximum in the visible region is $5320 \mathrm{~A}$. Refractive index of the film will be,
(a) 1.22
(b) 1.33
(c) 1.51
(cl) 1.83
Answer:
(b) 1.33
Hint.
The condition for constructive interference, (for reflection)
$2 \mu \operatorname{tcos} r=(2 n+1) \frac{\lambda}{2}[\therefore \cos r=1]$
$
\mu=\frac{(2 n+1) \lambda}{4 t}
$
For visible region, $\mathrm{n}=2$
$
\mu=\frac{(2(2)+1) \times 5320 \times 10^{-10}}{4 \times 5000 \times 10^{-10}}=1.33
$
Question 16.
First diffraction minimum due to a single slit of width $1.0 \times 10^{-5} \mathrm{~cm}$ is at $30^{\circ}$. Then wavelength of light used is,
(a) $400 \AA$
(b) $500 \AA$
(c) $600 \AA$
Answer:
(b) $500 \AA$
Hint.
For diffraction minima, $d \sin \theta=n \lambda$
$
\begin{aligned}
& \lambda=\frac{d \sin \theta}{n}=\frac{1 \times 10^{-5} \times 10^{-2} \times \sin 30^{\circ}}{1}=0.5 \times 10^{-7} \\
& \lambda=500 \AA
\end{aligned}
$

Question 17.
A ray of light strikes a glass plate at an angle $60^{\circ}$. If the reflected and refracted rays are perpendicular to each other, the refractive index of the glass is,
(a) $\sqrt{3}$
(b) $\frac{3}{2}$
(c) $\sqrt{\frac{3}{2}}$
(d) 2
Answer:
(a) $\sqrt{ } 3$
Hint.
Angle of refraction $r=60^{\circ} ;$ Angle of incident $i=30^{\circ}$
$\sin \mathrm{i}=\mathrm{nx} \sin \mathrm{r}$
$
\mathrm{n}=\frac{\sin 30^{\circ}}{\sin 60^{\circ}}=\sqrt{3}
$
Question 18.
One of the of Young's double slits is covered with a glass plate as shown in figure. The position of central maximum will,

(a) get shifted downwards
(b) get shifted upwards
(c) will remain the same
(d) data insufficient to conclude
Answer:
(b) get shifted upwards
Question 19.
Light transmitted by Nicol prism is,
(a) partiallypolarised
(b) unpolarised
(c) plane polarised
(d) elliptically polarised
Answer:
(c) plane polarised
Question 20.
The transverse nature of light is shown in,
(a) interference
(b) diffraction
(c) scattering
(d) polarisation
Answer:
(d) polarisation
Short Answer Questions
Question 1.
State the laws of reflection.
Answer:
(a) The incident ray, reflected ray and normal to the reflecting surface all are coplanar (ie. lie in the same plane).
(b) The angle of incidence $i$ is equal to the angle of reflection $r$. $i=r$

Question 2.
What is angle of deviation due to reflection?
Answer:
The angle between the incident and deviated light ray is called angle of deviation of the light ray. It is written as, $d=180-(i+r)$
As, $i=r$ in reflection, we can write angle of deviation ' in reflection at plane surface as, $d=180-2 i$
Question 3.
Give the characteristics of image formed by a plane mirror.
Answer:
1. The image formed by a plane mirror is virtual, erect, and laterally inverted.
2. The size of the image is equal to the size of the object.
3. The image distance far behind the mirror is equal to the object distance in front of it.
4. If an object is placed between two plane mirrors inclined at an angle 0 , then the number of images $\mathrm{n}$ formed is as, $\mathrm{n}=\left(\frac{360}{\theta}-1\right)$
Question 4.
Derive the relation between $f$ and $\mathrm{R}$ for a spherical mirror.
Let $C$ be the centre of curvature of the mirror. Consider a light ray parallel to the principal axis is incident on the mirror at $M$ and passes through the principal focus $F$ after reflection. The geometry of reflection of the incident ray is shown in figure. The line $\mathrm{CM}$ is the normal to the mirror at $\mathrm{M}$. Let $\mathrm{i}$ be the angle of incidence and the same will be the angle of reflection. If $\mathrm{MP}$ is the perpendicular from $\mathrm{M}$ on the principal axis, then from the geometry, The angles $\angle \mathrm{MCP}=i$ and $\angle \mathrm{MFP}=2 \mathrm{i}$ From right angle triangles $\triangle \mathrm{MCP}$ and $\triangle \mathrm{MFP}$,

$
\begin{aligned}
& \frac{1}{f}=\left(\frac{n_2}{n_1}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
& \tan \mathrm{i}=\frac{P M}{P C} \text { and } \tan 2 \mathrm{i}=\frac{P M}{P F}
\end{aligned}
$
As the angles are small, $\tan \mathrm{i} \approx \mathrm{i}=\frac{P M}{P C}$ and $\tan 2 \mathrm{i}=\frac{P M}{P F}$
Simplifying further, $2=\frac{P M}{P C}$ and $\tan =\frac{P M}{P F} ; 2 \mathrm{PF}=\mathrm{PC}$ $\mathrm{PF}$ is focal length/and PC is the radius of curvature $\mathrm{R}$.
$
2 f=\mathrm{R} \text { (or) } f=\frac{R}{2}
$
$f=\frac{R}{2}$ is the relation between $f$ and $\mathrm{R}$.
Question 5 .
What are the Cartesian sign conventions for a spherical mirror?
Answer:
1. The Incident light is taken from left to right (i.e. object on the left of mirror).
2. All the distances are measured from the pole of the mirror (pole is taken as origin).
3. The distances measured to the right of pole along the principal axis are taken as positive.
4. The distances measured to the left of pole along the principal axis are taken as negative.
5. Heights measured in the upward perpendicular direction to the principal axis are taken as positive.
6. Heights measured in the downward perpendicular direction to the principal axis, are taken as negative.
Question 6.
What is optical path? Obtain the equation for optical path of a medium of thickness $d$ and refractive index $n$.
Answer:
Optical path of a medium is defined as the distance d' light travels in vacuum in the same time it travels a distance $d$ in the medium.
Let us consider a medium of refractive index $\mathrm{n}$ and thickness $\mathrm{d}$. Light travels with a speed $\mathrm{v}$ through the medium in a time $t$. Then we can write,

$\mathrm{v}=\frac{d}{t} ;$ rewritten as, $\mathrm{t}=\frac{d}{v}$
In the same time, light can cover a greater distance d' in vacuum as it travels with greater speed $\mathrm{c}$ in vacuum.
Then we have, $\mathrm{c}=\mathrm{v}=\frac{d^{\prime}}{t} ;$ rewritten as, $\mathrm{t}=\frac{d^{\prime}}{c}$
As the time taken in both the cases is the same, we can equate the time $t$ as, $\frac{d^{\prime}}{c}=\frac{d}{v}$
rewritten for the optical path $\mathrm{d}^{\prime}$ as $\mathrm{d}^{\prime}=\frac{c}{v} \mathrm{~d}$
As, $\frac{c}{v}=\mathrm{n}$; The optical path $\mathrm{d}^{\prime}$ is, $\mathrm{d}^{\prime}=\mathrm{nd}$
As $\mathrm{n}$ is always greater than 1 , the optical path $\mathrm{d}$ ' of the medium is always greater than $\mathrm{d}$.
Question 7.
State the laws of refraction.
Answer:
Law of refraction is called Snell's law.
Snell's law states that,
(a) The incident ray, refracted ray and normal to the refracting surface are all coplanar (i.e. lie in the same plane).
(b) The ratio of angle of incident $i$ in the first medium to the angle of reflection $\mathrm{r}$ in the second medium is equal to the ratio of refractive index of the second medium $\mathrm{n}_2$ to that of the refractive index of the first medium $\mathrm{n}_1$.
$
\frac{\sin i}{\sin r}=\frac{n_2}{n_1} \text {. }
$
Question 8.
What is angle of deviation due to refraction?
Answer:
The angle between the incident and deviated light is called angle of deviation. When light travels from rarer to denser medium it deviates towards normal. The angle of deviation in this case is, $d=i-r$
Question 9.
What is principle of reversibility?
Answer:
The principle of reversibility states that light will follow exactly the same path if its direction of travel is reversed.

Question 10.
What is relative refractive index?
Answer:
Snell's law, the term $\left(\frac{n_2}{n_1}\right)$ is called relative refractive index of second medium with respect to the first medium which is denoted as $\left(\mathrm{n}_{21}\right) \cdot \mathrm{n}_{21}=\frac{n_2}{n_1}$
Question 11 .
Obtain the equation for apparent depth.
Answer:
Light from the object $\mathrm{O}$ at the bottom of the tank passes from denser medium (water) to rarer medium (air) to reach our eyes. It deviates away from the normal in the rarer medium at the point of incidence B. The refractive index of the denser medium is $\mathrm{n}_1$ and rarer medium is $\mathrm{n}_2$. Here, $\mathrm{n}_1>\mathrm{n}_2$. The angle of incidence in the denser medium is $i$ and the angle of refraction in the rarer medium is $r$. The lines NN' and $\mathrm{OD}$ are parallel. Thus angle $\angle \mathrm{DIB}$ is also $\mathrm{r}$. The angles $\mathrm{i}$ and $\mathrm{r}$ are very small as the diverging light from $\mathrm{O}$ entering the eye is very narrow. The Snell's law in product form for this refraction is, $\mathrm{n}_1 \sin \mathrm{i}=\mathrm{n}_2 \sin \mathrm{r}$
As the angles $i$ and $r$ are small, we can approximate, $\sin i \approx \tan i$;
$\mathrm{n}_1 \tan \mathrm{i}=\mathrm{n}_2 \tan \mathrm{i}$
In triangles $\triangle \mathrm{DOB}$ and $\triangle \mathrm{DIB}$, $\tan (\mathrm{i})=\frac{D B}{D O}$ and $\tan (\mathrm{r})=\frac{D B}{D I}$
$
\mathrm{n}_1=\frac{D B}{D O} \mathrm{n}_2=\frac{D B}{D I}
$
DB is cancelled on both sides, DO is the actual depth d and DI is the apparent depth d'.
Rearranging the above equation for the apparent depth d',
$
\mathrm{d}^{\prime}=\left(\frac{n_2}{n_1}\right) \mathrm{d}
$
As the rarer medium is air and its refractive index $\mathrm{n}_2$ can be taken as $1,\left(\mathrm{n}_2=1\right)$. And the refractive index $\mathrm{n}_1$ of denser medium could then be taken as $\mathrm{n},\left(\mathrm{n}_1=\mathrm{n}\right)$.
In that case, the equation for apparent depth becomes,
$
\mathrm{d}=\frac{d}{n}
$
Question 12.
Why do stars twinkle?
Answer:
The stars actually do not twinkle. They appear twinkling because of the movement of the atmospheric layers with varying refractive indices which is clearly seen in the night sky.

Question 13.
What is critical angle and total internal reflection?
Answer:
The angle of incidence in the denser medium for which the refracted ray graces the boundary is called critical angle $\mathrm{i}_{\mathrm{c}}$.
The entire light is reflected back into the denser medium itself. This phenomenon is called total internal reflection.
Question 14.
Obtain the equation for critical angle.
Answer:
Snell's law in the product form, equation for critical angle incidence becomes,
$\mathrm{n}_1 \operatorname{sini} \mathrm{i}_{\mathrm{c}}=\mathrm{n}_2 \sin 90^{\circ}$
$\mathrm{n}_1 \operatorname{sini} \mathrm{i}_{\mathrm{c}}=\mathrm{n}_2\left(\because \sin 90^{\circ}=1\right)$
$\sin \mathrm{i}_{\mathrm{C}}=\left(\frac{n_2}{n_1}\right)$
Here, $\mathrm{n}_1>\mathrm{n}_2$
If the rarer medium is air, then its refractive index is 1 and can be taken as $n$ itself, i.e. $\left(\mathrm{n}_2=1\right)$ and $\left(\mathrm{n}_1=\right.$ n).
$
\operatorname{sini} i_c=\frac{1}{n} \text { (or) } i_c=\sin ^{-1}\left(\frac{1}{n}\right)
$
Question 15 .
Explain the reason for glittering of diamond.
Answer:
Diamond appears dazzling because the total internal reflection of light happens inside the diamond. The refractive index of only diamond is about 2.417 . It is much larger than that for ordinary glass which is about only 1.5 . The critical angle of diamond is about $24.4^{\circ}$. It is much less than that of glass. A skilled diamond cutter makes use of this larger range of angle of incidence (24.4 ${ }^{\circ}$ to $90^{\circ}$ inside the diamond), to ensure that light entering the diamond is total internally reflected from the many cut faces before getting out. This gives a sparkling effect for diamond.

Question 16.
What are mirage and looming?
Answer:
Mirage: Mirage takes place in hot regions. The light from distant objects appears to be reflected from ground. For mirage to form refractive index goes on increasing as we go up. Looming: Looming takes place in cold regions. The light from distant objects appears to be flying. For looming to form refractive index goes on decreasing.
Question 17.
Write a short notes on the prisms making use of total internal reflection.
Answer:
Prisms can be designed to reflect light by $90^{\circ}$ or by $180^{\circ}$ by making use of total internal reflection. The critical angle ic for the material of the prism must be less than $45^{\circ}$. This is true for both crown glass and flint glass. Prisms are also used to invert images without changing their size.
Question 18.
What is Snell's window?
Answer:
When light entering the water from outside is seen from inside the water, the view is restricted to a particular angle equal to the critical angle ic. The restricted illuminated circular area is called Snell's window.
Question 19.
Write a note on optical fibre.
Answer:
Transmitting signals through optical fibres is possible due to the phenomenon of total internal reflection. Optical fibres consists of inner part called core and outer part called cladding (or) sleeving. The refractive index of the material of the core must be higher than that of the cladding for total internal reflection to happen. Signal in the form of light is made to incident inside the core-cladding boundary at an angle greater than the critical angle. Hence, it undergoes repeated total internal reflections along the length of the fibre without undergoing any refraction.
Question 20.
Explain the working of an endoscope.
Answer:
An endoscope is an instrument used by doctors which has a bundle of optical fibres that are used to see inside a patient's body. Endoscopes work on the phenomenon of total internal reflection. The optical fibres are inserted in to the body through mouth, nose or a special hole made in the body. Even operations could be carried out with the endoscope cable which has the necessary instruments attached at their ends.
Question 21.
What are primary focus and secondary focus of convex lens?
Answer:
Primary focus: The primary focus $F_1$ is defined as a point where an object should be placed to give parallel emergent rays to the principal axis.
Secondary focus: The secondary focus $F_2$ is defined as a point where all the parallel rays travelling close to the principal axis converge to form an image on the principal axis.
Question 22.
What are the sign conventions followed for lenses?
Answer:
The sign conventions for thin lenses differ only in the signs followed for focal lengths.
(a) The sign of focal length is not decided on the direction of measurement of the focal length from the pole of the lens as they have two focal lengths, one to the left and another to the right (primary and secondary focal lengths on either side of the lens).
(b) The focal length of the thin lens is taken as positive for a converging lens and negative for a diverging lens.
Question 23.
Arrive at lens equation from lens maker's formula.
Answer:
From refraction through a double convex lens, the relation between the object distance $\mathrm{u}$, image distance $\mathrm{v}_1$ and radius of curvature $\mathrm{R}_1$ as
$
\frac{\mu_2}{v_1}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R_1}
$
The relation between the object distance image distance $\mathrm{v}_1$ and radius of curvature $R_2$ can be $\frac{\mu_1}{v}-\frac{\mu_2}{v_1}=\frac{\mu_1-\mu_2}{R_2}$
Adding equation (1) and (2)
$
\begin{gathered}
\frac{\mu_1}{v}-\frac{\mu_1}{u}=\left(\mu_2-\mu_1\right)\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right] \\
\frac{1}{v}-\frac{1}{u}=\left(\frac{\mu_2-\mu_1}{\mu_1}\right)\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right]
\end{gathered}
$
If the object is placed at infinity $(\mu=\infty)$, the image will be formed at the focus. i.e.v $=f$
$
\frac{1}{f}=\left(\frac{\mu_2-\mu_1}{\mu_1}\right)\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right]
$
This is len's maker's formula. When the lens is placed in air $\mu_1=1$ and $\mu_2=\mu$.
Equation (4) becomes,

$
\frac{1}{f}=(\mu-1)\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right]
$
From equation (3) and (4), we have $\frac{1}{v}-\frac{1}{u}$
$
=\frac{1}{f}
$
This is the len's equation.
Question 24.
Obtain the equation for lateral magnification for thin lens.
Answer:
Lateral magnification in terms of $u$ and $f$.
The thin lens formula is
$
\frac{1}{v}-\frac{1}{u}=\frac{1}{f}
$
Multiplying on both sides by ' $u$ '
$
\begin{gathered}
\frac{u}{v}-1=\frac{u}{f} \Rightarrow \frac{u}{v}=\frac{f+u}{f} \\
=\frac{v}{u}=\frac{f}{f u}
\end{gathered}
$
In terms of $\mathrm{v}$ and $f$ multiplying by $\mathrm{v}$, we get
$
\begin{aligned}
& 1-\frac{v}{u}=\frac{v}{f} \\
& \frac{v}{u}=1-\frac{v}{f}=\frac{f-v}{f}=m
\end{aligned}
$
Hence, lateral magnification for thin lens,
$
m=\frac{v}{u}=\frac{f}{f+u}=\frac{f-v}{f}
$
Question 25.
What is power of a lens?
Answer:
The power of a lens $P$ is defined as the reciprocal of its focal length.

$
p=\frac{1}{f}
$
The unit of power is diopter D.
Question 26.
Derive the equation for effective focal length for lenses in contact.
Answer:
Consider a two thin lenses in contact. In the absence of second lens $\mathrm{L}_2$, the first lens $\mathrm{L}_1$ will form a real image I'. Using thin lens formula.
$
\frac{1}{f_2}=\frac{1}{v^{\prime}}-\frac{1}{u}
$
The image I' acts as a virtual object $\left(\mathrm{u}=\mathrm{v}^{\prime}\right)$ for the second lens $\mathrm{L}_2$ which finally forms its real image I at distance $\mathrm{v}$. Thus
$
\frac{1}{f_2}=\frac{1}{v}-\frac{1}{v^{\prime}}
$
Adding equation (1) and (2) we get,
$
\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{v}-\frac{1}{u}
$
For the combination of thin lenses in contact, if ' $\mathrm{f}$ ' is the equivalent focal length, then
$
\frac{1}{v}-\frac{1}{u}=\frac{1}{f}
$
From equations (3) and (4), the effective focal length for lenses in contact.
$
\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}
$
Question 27.
What is angle of minimum deviation?
Answer:
The minimum value of the angle of deviation suffered by a ray on passing through a prism is called the angle of minimum deviation.
Question 28.
What is dispersion?
Answer:
The phenomenon of spliting of white light into its component colours on passing through a refracting medium is called dispersion.
Question 29.
How are rainbows formed?
Answer:
Rainbow is formed by dispersion of sunlight into its constituent colours by raindrops which disperse sunlight by refraction and deviate the colours by total internal reflection.

Question 30 .
What is Rayleigh's scattering?
Answer:
The scattering of light by particles in a medium without a change in wavelength is called as Rayleigh's scattering.
Question 31.
Why does sky appear blue?
Answer:
Blue colour of the sky is due to scattering of sunlight by air molecules. According to Rayleigh's law, intensity of scattered light, $\mathrm{I} \propto \frac{1}{\lambda^4}$ So blue light of shorter wavelength is scattered much more than red light of larger wavelength. The blue component is proportionally more in light coming from different parts of the sky. That is why the sky appears blue.
Question 32 .
What is the reason for reddish appearance of sky during sunset and sunrise?
Answer:
During sunrise or sunset, the sun is near the horizon. Sunlight has to travel a greater distance. So shorter waves of blue region are scattered away by the atmosphere. Red waves of longer wavelength are least scattered and reach the observer. So the sun appears red.
Question 33 .
Why do clouds appear white?
Answer:
Clouds have large particles like dust and water droplets which scatter light of all colours almost equally. Hence clouds generally appear white.
Question 34.
What are the salient features of corpuscular theory of light?
Answer:
- According this theory, light is emitted as tiny, massless (negligibly small mass) and perfectly elastic particles called corpuscles.
- As the corpuscles are very small, the source of light does not suffer appreciable loss of mass even if it emits light for a long time.
- On account of high speed, they are unaffected by the force of gravity and their path is a straight line in a medium of uniform refractive index.

- The energy of light is the kinetic energy of these corpuscles. When these corpuscles impinge on the retina of the eye, the vision is produced.
- The different size of the corpuscles is the reason for different colours of light.
- When the corpuscles approach a surface between two media, they are either attracted or repelled.
- The reflection of light is due to the repulsion of the corpuscles by the medium and refraction of light is due to the attraction of the corpuscles by the medium.
Question 35 .
What is wave theory of light?
Answer:
Light is a disturbance from a source that travels as longitudinal mechanical waves through the either medium that was presumed to pervade all space as mechanical wave requires medium for its propagation. The wave theory could successfully explain phenomena of reflection, refraction, interference and diffraction of light.
Question 36.
What is electromagnetic wave theory of light?
Answer:
Electromagnetic wave theory:
Maxwell (1864) proved that light is an electromagnetic wave which is transverse in nature carrying electromagnetic energy. He could also show that no medium is necessary for the propagation of electromagnetic waves. All the phenomenon of light could be successfully explained by this theory.
Question 37.
Write a short note on quantum theory of light.
Answer:
Albert Einstein (1905), endorsing the views of Max Plank (1900), was able to explain photoelectric effect in which light interacts with matter as photons to eject the electrons. A photon is a discrete packet of energy. Each photon has energy $E$ of, $\mathrm{E}=\mathrm{hv}$
Where, $\mathrm{h}$ is Plank's constant $\left(\mathrm{h}=6.625 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right)$ and $\mathrm{v}$ is frequency of electromagnetic wave. As light has both wave as well as particle nature it is said to have dual nature. Thus, it is concluded that light propagates as a wave and interacts with matter as a particle.
Question 38 .
What is a wave front?
Answer:
A wavefront is the locus of points which are in the same state or phase of vibration.
Question 39.
What is Huygens' principle?
Answer:
According to Huygens principle, each point of the wavefront is the source of secondary wavelets emanating from these points spreading out in all directions with the speed of the wave. These are called as secondary wavelets.

Question 40.
What is interference of light?
Answer:
The phenomenon of addition or superposition of two light waves which produces increase in intensity at some points and decrease in intensity at some other points is called interference of light.
Question 41.
What is phase of a wave?
Answer:
Phase is a particular point in time on the cycle of a waveform, measured as an angle in degrees.
Question 42.
Obtain the relation between phase difference and path difference.
Answer:
Phase difference $(\Phi)$ :
It is the difference expressed in degrees or radians between two waves having same frequency and referenced to same point in time.
Path difference $(\delta)$ :
It is the difference between the lengths of two paths of the two different having same frequency and travelling at same velocity. $\delta=\frac{\lambda}{2 \pi} \Phi$
Question 43.
What are coherent sources?
Answer:
Two light sources are said to be coherent if they produce waves which have same phase or constant phase difference, same frequency or wavelength (monochromatic), same waveform and preferably same amplitude.
Question 44.
What is intensity division?
Answer:
Intensity' or amplitude division: If we allow light to pass through a partially silvered mirror (beam splitter), both reflection and refraction take place simultaneously. As the two light beams are obtained from the same light source, the two divided light beams will be coherent beams. They will be either inphase or at constant phase difference.

Question 45.
How does wavefront division provide coherent sources?
Answer:
Wavefront division is the most commonly used method for producing two coherent sources. A point source produces spherical wavefronts. All the points on the wavefront are at the same phase. If two points are chosen on the wavefront by using a double slit, the two points will act as coherent sources.
Question 46 .
How do source and images behave as coherent sources?
Answer:
Source and images: In this method a source and its image will act as a set of coherent source, because the source and its image will have waves in-phase or constant phase difference. The Instrument, Fresnel's biprism uses two virtual sources as two coherent sources and the instrument, Lloyd's mirror uses a source and its virtual image as two coherent sources.
Question 47.
What is bandwidth of interference pattern?
Answer:
The bandwidth ( $\beta$ ) is defined as the distance between any two consecutive bright or dark fringes.
Question 48.
What is diffraction?
Answer:
Diffraction is bending of waves around sharp edges into the geometrically shadowed region.
Question 49.
Differentiate between Fresnel and Fraunhofer diffraction.
Answer:

Question 50.
Discuss the special cases on first minimum in Fraunhofer diffraction.
Let us consider the condition for first minimum with $(\mathrm{n}=1)$. a $\sin \theta=\lambda$
The first minimum has an angular spread of, $\sin \theta=\frac{\lambda}{a}$. Special cases to discuss on the condition.
1. When $\mathrm{a}<\lambda$, the diffraction is not possible, because $\sin 0$ can never be greater than 1 .
2. When $a \geq \lambda$, the diffraction is possible.
- For $\mathrm{a}=\lambda, \sin \theta=1$ i.e, $\theta=90^{\circ}$. That means the first minimum is at $90^{\circ}$. Hence, the central maximum spreads fully in to the geometrically shadowed region leading to bending of the diffracted light to $90^{\circ}$.
- For $\mathrm{a} \gg>, \sin \theta<1$ i.e, the first minimum will fall within the width of the slit itself. The diffraction will not be noticed at all.
3. When $\mathrm{a}>\lambda$ and also comparable, say $\mathrm{a}=2 \lambda, \sin \theta=\frac{\lambda}{a}=\frac{\lambda}{2 \lambda}=\frac{1}{2}$; then $\theta=30^{\circ}$. These are practical cases where diffraction could be observed effectively.
Question 51.
What is Fresnel's distance? Obtain the equation for Fresnel's distance.
Answer:
Fresnel's distance is the distance up to which the ray optics is valid in terms of rectilinear propagation of light.
Fresnel's distance $z$ as, $z=\frac{a^2}{2 \lambda}$.
Question 52.
Mention the differences between interference and diffraction.
Answer:

Question 53.
What is a diffraction grating?
Answer:
A diffraction grating is an optical component with a periodic structure that splits and diffracts light into several beams travelling in different directions.
Question 54
What are resolution and resolving power?
Answer:
Optical resolution describes the ability of an imaging system to resolve detail in the object that is being imaged. Resolving power is the ability of an optical instrument to resolve or separate the image of two nearby point objects so that they can be distinctly seen. It is equal to the reciprocal of the limit of resolution of the optical instrument.
Question 55.
What is Rayleigh's criterion?
Answer:
The images of two point objects are just resolved when the central maximum of the diffraction pattern of one falls over the first minimum of the diffraction pattern of the other.
Question 56.
What is polarisation?
Answer:
The phenomenon of restricting the vibrations of light (electric or magnetic field vector) to a particular direction perpendicular to the direction of wave propagation motion is called polarization of light.
Question 57.
Differentiate between polarised and unpolarised light.
Answer:

Question 58.
Discuss polarisation by selective absorption.
Answer:
Selective absorption is the property of a material which transmits w'aves whose electric fields vibrate in a plane parallel to a certain direction of orientation and absorbs all other waves. The polaroids or polarisers are thin commercial sheets which make use of the property of selective absorption to produce an intense beam of plane polarised light. Selective absorption is also called as dichroism.
Question 59.
What are polariser and analyser?
Answer:
Polariser:
The Polaroid which plane polarises the unpolarised light passing through it is called a polariser.
Analyser:
The polaroid which is used to examine whether a beam of light is polarised or not is called an analyser.
Question 60.
What are plane polarised, unpolarised and partially polarised light?
Answer:
Plane polarised:
If the vibrations of a wave are present in only one direction in a plane perpendicular to the direction of propagation of the wave is said to be polarised or plane polarised light.
Unpolarised:
A transverse wave which has vibrations in all directions in a plane perpendicular to the direction of propagation is said to be unpolarised light.
Partially polarised light:
If the intensity of light varies between maximum and minimum for every' rotation of $90^{\circ}$ of the analyser, the light is said to be partially polarised light.
Question 61
State and obtain Malus' law.
Answer:
When a beam of plane polarised light of intensity $\mathrm{I}_0$ is incident on an analyser, the light transmitted of intensity I from the analyser varies directly as the square of the cosine of the angle 0 between the transmission axis of polariser and analyser. This is known as Malus' law. $\mathrm{I}=\mathrm{I}_0 \cos ^2 \theta$

Question 62.
List the uses of polaroids.
Answer:
Uses of polaroids:
1. Polaroids are used in goggles and cameras to avoid glare of light.
2. Polaroids are useful in three dimensional motion pictures i.e., in holography.
3. Polaroids are used to improve contrast in old oil paintings.
4. Polaroids are used in optical stress analysis.
5. Polaroids are used as window glasses to control the intensity of incoming light.
Question 63.
State Brewster's law.
Answer:
The law states that the tangent of the polarising angle for a transparent medium is equal to its refractive index, $\tan \mathrm{i}=\mathrm{n}$. This relation is known as Brewster's law.
Question 64.
What is angle of polarisation and obtain the equation for angle of polarisation.
Answer:
The angle of incidence at which a beam of unpolarised light falling on a transparent surface is reflected as a beam of plane polarised light is called polarising angle or Brewster's angle. It is denoted by $i_p$ $i_p=90^{\circ}-\mathrm{R}_{\mathrm{p}}$
Question 65.
Discuss about pile of plates.
Answer:
The phenomenon of polarisation by reflection is used in the construction of pile of plates. It consists of a number of glass plates placed one over the other. The plates are inclined at an angle of $33.7^{\circ}\left(90^{\circ}-\right.$ $\left.56.3^{\circ}\right)$ to the axis of the tube. A beam of unpolarised light is allowed to fall on the pile of plates along the axis of the tube. So, the angle of incidence of light will be at $56.3^{\circ}$ which is the polarising angle for glass.

The vibrations perpendicular to the plane of incidence are reflected at each surface and those parallel to it are transmitted. The larger the number of surfaces, the greater is the intensity of the reflected plane polarised light. The pile of plates is used as a polarizer and also as an analyser.

Question 66.
What is double refraction?
Answer:
When a ray of unpolarised light is incident on a calcite crystal, two refracted rays are produced. Hence, two images of a single object are formed. This phenomenon is called double refraction.
Question 67.
Mention the types of optically active crystals with example.
Answer:
Types of optically active crystals:
Uniaxial crystals:
Crystals like calcite, quartz, tourmaline and ice having only one optic axis are called uniaxial crystals.
Biaxial crystals:
Crystals like mica, topaz, selenite and aragonite having two optic axes are called biaxial crystals.
Question 68.
Discuss about Nicol prism.
Answer:
Nicol prism is an optical device incorporated in optical instruments both for producing and analysing plane polarised light. The construction of a Nicol prism is based on the phenomenon of Double refraction. One of the most common forms of the Nicol prism is made by taking a calcite crystal which is a double refracting crystal with its length three times its breadth.

It is cut into two halves along the diagonal so that their face angles are $72^{\circ}$ and $108^{\circ}$. The two halves are joined $i$ together by a layer of Canada balsam, a transparent cement.
Question 69.
How is polarisation of light obtained by scattering of light?
Answer:
The light from a clear blue portion of the sky shows a rise and fall of intensity when viewed through a polaroid which is rotated. This is because of sunlight, which has changed its I direction (having been scattered) on encountering the molecules of the earth's atmosphere. The electric field of light interact with the electrons present in the air molecules.
Under the influence of the electric field of the incident wave the electrons in the molecules acquire components of motion in both these directions. We have an observer looking at $90^{\circ}$ to the direction of the sun. Clearly, charges accelerating parallel do not radiate energy towards this observer since their acceleration has no transverse component. The radiation scattered by the molecule is therefore polarized perpendicular to the plane.

Question 70.
Discuss about simple microscope and obtain the equations for magnification for near point focusing and normal focusing.
Answer:
A simple microscope is a single magnifying (converging) lens of small focal length. The idea is to get an erect, magnified and virtual image of the object. For this the object is placed between $F$ and $P$ on one side of the lens and viewed from other side of the lens. There are two magnifications to be discussed for two kinds of focusing.
1. Magnification in near point focusing $\mathrm{m}=1+\frac{D}{f}$
2. Magnification in normal focusing (angular magnification), $\mathrm{m}=\frac{D}{f}$
Question 71.
What are near point and normal focusing?
Answer:
- Near point focusing:
The image is formed at near point, i.e. $25 \mathrm{~cm}$ for normal eye. This distance is also called as least distance $\mathrm{D}$ of distinct vision. In this position, the eye feels comfortable but there is little strain on the eye.
- Normal focusing: The image is formed at infinity. In this position the eye is most relaxed to view the image.
Question 72.
Why is oil immersed objective preferred in a microscope?
Answer:
It is best to use an oil-immersed objective at high magnification as the oil compensates for short focal lengths associated with larger magnifications.
Question 73.
What are the advantages and disadvantages of using a reflecting telescope?
Answer:
Advantages:

- The main advantage is reflector telescope can escape from chromatic aberration because wavelength does not effect reflection.
- The primary mirror is very stable because it is located at the back of the telescope and can be support in the back.
- More cost effective than refractor of similar size.
- Easier to make a high quality mirror than lens because mirror need to only concern with one side of the curvature.
Disadvantages:
- Optical misalignment can occur quite easily.
- Require frequent cleaning because the inside is expose to the atmosphere.
- Secondary mirror can cause diffraction of original incoming light rays causing the "Christmas star effect" where a bright object have spikes.
Question 74
What is the use of an erecting lens in a terrestrial telescope?
Answer:
A terrestrial telescope has an additional erecting lens to make the final image erect.
Question 75.
What is the use of collimator?
Answer:
The collimator is an arrangement to produce a parallel beam of light.
Question 76.
What are the uses of spectrometer?
Answer:
The spectrometer is an optical instrument used to study the spectra of different sources of light and to measure the refractive indices of materials.
Question 77.
What is myopia? What is its remedy?
Answer:
Myopia (or) short sightedness:
It is a vision defect in which a person can see nearby objects clearly but cannot see the distant objects clearly beyond a certain point.
Remedy (correction):
A myopia eye is corrected by using a concave lens of focal length equal to the distance of the far point F from the eye.
Question 78.
What is hypermetropia? What is its remedy?
Answer:
Hypermetropia (or) Long sightedness: It is a vision defect in which a person can see the distant objects clearly but cannot see the nearby objects clearly.
Remedy (correction): A hypermetropic eye is corrected by using a convex lens of suitable focal length.
Question 79.
What is presbyopia?
Answer:
This defect is similar to hypermetropia i.e., a person having this defect cannot see nearby objects distinctly, but can see distant objects without any difficulty. This defect occurs in elderly persons (aged persons).
Question 80.
What is astigmatism?
Answer:
Astigmatism is the defect arising due to different curvatures along different planes in the eye lens. Astigmatic person cannot see all the directions equally well. The defect due to astigmatism is more serious than myopia and hyperopia.
Long Answer Questions
Question 1.
Derive the mirror equation and the equation for lateral magnification.
Answer:
The mirror equation:
1. The mirror equation establishes a relation among object distance $\mathrm{u}$, image distance $\mathrm{v}$ and focal length/for a spherical mirror. An object $A B$ is considered on the principal axis of a concave mirror beyond the center of curvature $\mathrm{C}^{\prime}$.
2. Let us consider three paraxial rays from point $B$ on the object.
3. The first paraxial ray BD travelling parallel to principal axis is incident on the concave mirror at $\mathrm{D}$, close to the pole $\mathrm{P}$. After reflection the ray passes through the focus $\mathrm{F}$. The second paraxial ray $\mathrm{BP}$ incident at the pole $\mathrm{P}$ is reflected along $\mathrm{PBThe}$ third paraxial ray $\mathrm{BC}$ passing through centre of curvature $\mathrm{C}$, falls normally on the mirror at $\mathrm{E}$ is reflected back along the same path.
4. The three reflected rays intersect at the point B'. A perpendicular drawn as A' B' to the principal axis is the real, inverted image of the object $\mathrm{AB}$.

As per law of reflection, the angle of incidence $\angle B P A$ is equal to the angle of reflection $\angle B^{\prime} P^{\prime}$ '. The triangles $\triangle \mathrm{BPA}$ and $\triangle \mathrm{B}^{\prime} \mathrm{PA}^{\prime}$ are similar. Thus, from the rule of similar triangles, $\frac{A^{\prime} B^{\prime}}{A B}=\frac{P A^{\prime}}{P A}$
The other set of similar triangles are, $\mathrm{ADPF}$ and $\mathrm{A} \mathrm{BA}$.' $F$. (PD is almost a straight vertical line) $\frac{A^{\prime} B^{\prime}}{P D}=\frac{A F^{\prime}}{P F}$
As, the distances $\mathrm{PD}=\mathrm{AB}$ the above equation becomes,
$
\frac{A^{\prime} B^{\prime}}{A B}=\frac{A F^{\prime}}{P F}
$
From equations (1) and (2) we can write,
$
\frac{P A^{\prime}}{P A}=\frac{A F^{\prime}}{P F}
$
As, $A^{\prime} \mathrm{F}=\mathrm{PA}^{\prime}-\mathrm{PF}$, the above equation becomes,
$
\frac{P A^{\prime}}{P A}=\frac{P A^{\prime}-P F^{\prime} P}{P F}
$
We can apply the sign conventions for the various distances in the above equation.
$
\mathrm{PA}=-\mathrm{u}, \mathrm{PA}^{\prime}=-\mathrm{v}, \mathrm{PF}=-\mathrm{f}
$
All the three distances are negative as per sign convention, because they are measured to the left of the pole. Now, the equation (3) becomes,
$
\frac{-v}{-u}=\frac{-v(-f)}{-f}
$
On further simplification,
$
\frac{-v}{-u}=\frac{v-f}{f} ; \frac{v}{u}=\frac{v}{f}=1
$
Dividing either side with $\mathrm{v}$,
$
\frac{1}{u}=\frac{1}{f}=\frac{1}{v}
$
After rearranging,
$
\frac{1}{v}+\frac{1}{u}=\frac{1}{f}
$
The above equation (4) is called mirror equation.

Lateral magnification in spherical mirrors:
The lateral or transverse magnification is defined as the ratio of the height of the image to the height of the object. The height of the object and image are measured perpendicular to the principal axis.
$
\begin{aligned}
& \text { magnification }(m)=\frac{\text { height of the image }\left(h^{\prime}\right)}{\text { height of the } \operatorname{object}(h)} \\
& \mathrm{m}=\frac{h^{\prime}}{h} \ldots \ldots .(5) \\
& \text { Applying proper sign conventions for equation (1), } \\
& \frac{A^{\prime} B^{\prime}}{A B}=\frac{P A^{\prime}}{P A} \\
& \mathrm{~A}^{\prime} \mathrm{B}^{\prime}=-\mathrm{h}, \mathrm{AB}=\mathrm{h}, \mathrm{PA}^{\prime}=-\mathrm{v}, \mathrm{PA}=-\mathrm{u} \\
& \frac{-h^{\prime}}{h}=\frac{-v}{-u}
\end{aligned}
$
On simplifying we get,
$
\mathrm{m}=\frac{h^{\prime}}{h}=-\frac{v}{u}
$
Using mirror equation, we can further write the magnification as,
$
\mathrm{m}=\frac{h^{\prime}}{h}-\frac{f-v}{f}=\frac{f}{f-u}
$
Question 2.
Describe the Fizeau's method to determine speed of light.
Answer:
Fizeau's method to determine speed of light:
Apparatus:
The light from the source $S$ was first allowed to fall on a partially silvered glass plate G kept at an angle of $45^{\circ}$ to the incident light from the source. The light then was allowed to pass through a rotating toothed-wheel with $\mathrm{N}$ teeth and $\mathrm{N}$ cuts of equal widths whose speed of rotation could be varied through an external mechanism. The light passing through one cut in the wheel will get reflected by a mirror $\mathrm{M}$ kept at a long distance $\mathrm{d}$, about $8 \mathrm{~km}$ from the toothed w'heel. If the toothed wheel was not rotating, the reflected light from the mirror would again pass through the same cut and reach the eyes of the observer through the partially silvered glass plate.
Working:
The angular speed of rotation of the toothed wheel was increased from zero to a value to until light passing through one cut would completely be blocked by the adjacent tooth. This is ensured by the disappearance of light while looking through the partially silvered glass plate.

Expression for speed of light:
The speed of light in air $\mathrm{v}$ is equal to the ratio of the distance the light travelled from the toothed wheel to the mirror and back $2 \mathrm{~d}$ to the time taken $\mathrm{t}$. $\mathrm{v}=\frac{2 d}{t} \mathrm{c} \ldots .(1)$
The distance $d$ is a known value from the arrangement. The time taken $t$ for the light to travel the distance to and fro is calculated from the angular speed co of the toothed wheel.
The angular speed $\omega$ of the toothed wheel when the light disappeared for the first time is, $\omega=\frac{\theta}{t} \ldots \ldots(2)$
Here, 0 is the angle between the tooth and the slot which is rotated by the toothed wheel within that time $t$.

$
\theta=\frac{\text { total angle of the circle in radian }}{\text { number of teeth }+ \text { number of cuts }}
$
$
\theta=\frac{2 \pi}{2 N}=\frac{\pi}{N}
$
Substituting for 0 in the equation (2) for ,
$
\omega=\frac{\pi / n}{2}=\frac{\pi}{N t}
$
Rewriting the above equation for $t$,
$
\mathrm{t}=\frac{\pi}{N_\omega}
$
Substituting $\mathrm{t}$ from equation (3) in equation (1), $\mathrm{v}=\frac{2 d}{\pi / N \omega}$
After rearranging,
$
\mathrm{v}=\frac{2 d N \omega}{\pi}
$
Fizeau had some difficulty to visually estimate the minimum intensity of the light when blocked by the adjacent tooth, and his value for speed of light was very close to the actual value.
Question 3.
Obtain the equation for radius of illumination (or) Snell's window.
Answer:
The radius of Snell's window can be deduced with the illustration as shown in figure. Light is seen from a point $A$ at a depth $d$. The Snell's law in product form, equation $n_2 \sin i=n_2 \sin \mathrm{r}$ for the refraction happening at the point $\mathrm{B}$ on the boundary between the two media is,
$n_1 \sin i_c=n_2 \sin 90^{\circ}$
$\mathrm{n}_1 \sin \mathrm{i}_{\mathrm{c}}=\mathrm{n}_2 \because \sin 90^{\circ}=1$
$\sin \mathrm{i}_{\mathrm{c}}=\frac{n_2}{n_1}$

From the right angle triangle $\triangle \mathrm{ABC}$,
$\sin \mathrm{i}_{\mathrm{c}}=\frac{R}{\sqrt{d^2+R^2}} \ldots \ldots$
Equating the above two equation (3) and equation (2).
Squaring on both sides, $\frac{R^2}{R^2+d^2}=\left(\frac{n_2}{n_1}\right)^2$
Taking reciprocal, $\frac{R^2+d^2}{R^2}=\left(\frac{n_1}{n_2}\right)^2$
On further simplifying,
$
1+\frac{d^2}{R^2}=\left(\frac{n_1}{n_2}\right)^2 ; \frac{d^2}{R^2}=\left(\frac{n_1}{n_2}\right)^2-1 ; \frac{d^2}{R^2}=\frac{n_1^2}{n_2^2}-1=\frac{n_1^2-n_2^2}{n_2^2}
$
Again taking reciprocal and rearranging,
$
\frac{R^2}{d^2}=\frac{n_2^2}{n_1^2-n_2^2} ; \quad R^2=d^2\left(\frac{n_2^2}{n_1^2-n_2^2}\right)
$
The radius of illumination is,
$
R=d \sqrt{\frac{n_2^2}{n_1^2-n_2^2}}
$
If the rarer medium outside is air, then, $\mathrm{n}_2=1$, and we can take $\mathrm{n}_1=\mathrm{n}$
$
R=d\left(\frac{1}{\sqrt{n^2-1}}\right) \text { (or) } R=\frac{d}{\sqrt{n^2-1}}
$

Question 4.
Derive the equation for acceptance angle and numerical aperture, of optical fiber. Acceptance angle in optical fibre:
Answer:
To ensure the critical angle incidence in the core-cladding boundary inside the optical fibre, the light should be incident at a certain angle at the end of the optical fiber while entering in to it. This angle is called acceptance angle. It depends on the refractive indices of the core $\mathrm{n}_1$, cladding $\mathrm{n}_2$ and the outer medium $n_3$. Assume the light is incident at an angle called acceptance angle $i$ at the outer medium and core boundary at $\mathrm{A}$.
The Snell's law in the product form, equation for this refraction at the point $\mathrm{A}$.

$
\mathrm{n}_3 \sin \mathrm{i}_{\mathrm{a}}=\mathrm{n}_1 \sin \mathrm{r}_{\mathrm{a}} \ldots(1)
$
To have the total internal reflection inside optical fibre, the angle of incidence at the core-cladding interface at B should be atleast critical angle ic. Snell's law in the product form, equation for the refraction at point $B$ is,
$\mathrm{n}_1 \sin \mathrm{i}_{\mathrm{c}}=\mathrm{n}_2 \sin 90^{\circ} \ldots(2)$
$\mathrm{n}_1 \sin \mathrm{i}_{\mathrm{c}}=\mathrm{n}_2 \because \sin 90^{\circ}=1$
$\therefore \sin \mathrm{i}_{\mathrm{c}}=\frac{n_2}{n_1} \ldots$ (3)
From the right angle triangle $\triangle \mathrm{ABC}$,
$
i_{\mathrm{c}}=90^{\circ}-\mathrm{r}_{\mathrm{a}}
$
Now, equation (3) becomes, $\sin \left(90^{\circ}-\mathrm{r}_{\mathrm{a}}\right)=\frac{n_2}{n_1}$
Using trigonometry', $\cos \mathrm{ra}=\mathrm{r}_{\mathrm{a}}=\frac{n_2}{n_1}$
$
\sin r_{\mathrm{a}}=\sqrt{1-\cos ^2 r_a}
$
Substituting for $\cos r_a$

$
\sin r_a=\sqrt{1-\left(\frac{n_2}{n_1}\right)^2}=\sqrt{\frac{n_1^2-n_2^2}{n_1^2}}
$
Substituting this in equation (1)
$
n_3 \sin i_a=n_1 \sqrt{\frac{n_1^2-n_2^2}{n_1^2}}=\sqrt{n_1^2-n_2^2}
$
On further simplification,
$
\begin{aligned}
& \sin i_a=\sqrt{\frac{n_1^2-n_2^2}{n_3}} \text { (or) } \sin i_a=\sqrt{\frac{n_1^2-n_2^2}{n_3^2}} \\
& i_a=\sin ^{-1}\left(\sqrt{\frac{n_1^2-n_2^2}{n_3^2}}\right)
\end{aligned}
$
If outer medium is air, then $\mathrm{n}_3=1$. The acceptance angle $i_{\mathrm{a}}$ becomes,
$
\mathrm{r}_{\mathrm{a}}=\sin ^{-1}\left(\sqrt{n_1^2-n_2^2}\right)
$
Light can have any angle of incidence from 0 to ia with the normal at the end of the optical fibre forming a conical shape called acceptance cone. In the equation (6), the term $\left(\mathrm{n}_3 \sin \mathrm{i}_{\mathrm{a}}\right)$ is called numerical aperture NA of the optical fibre.
$
\mathrm{NA}=\mathrm{n}_3 \sin \mathrm{i}_{\mathrm{a}}\left(\sqrt{n_1^2-n_2^2}\right)
$
If outer medium is air, then $\mathrm{n}_3=1$. The numerical aperture NA becomes,
$
\mathrm{NA}=\sin \mathrm{i}_{\mathrm{a}}=\left(\sqrt{n_1^2-n_2^2}\right)
$
Question 5.
Obtain the equation for lateral displacement of light passing through a glass slab.
Answer:
Consider a glass slab of thickness $t$ and refractive index $n$ is kept in air medium. The path of the light is $A B C D$ and the refractions occur at two points $B$ and $C$ in the glass slab. The angles of incidence $i$ and refraction $\mathrm{r}$ are measured with respect to the normal $\mathrm{N}_1$ and $\mathrm{N}_2$ at the two points $\mathrm{B}$ and $\mathrm{C}$ respectively.
The lateral displacement $\mathrm{L}$ is the perpendicular distance CE drawn between the path of light and the undeviated path of light at point $\mathrm{C}$. In the right angle triangle $\triangle \mathrm{BCE}$,

$
\sin (\mathrm{i}-\mathrm{r})=\frac{L}{B C} ; \mathrm{BC}=\frac{L}{\sin (i-r)}
$
In the right angle triangle $\triangle \mathrm{BCF}$,
$
\cos (\mathrm{r})=\frac{t}{B C} ; \mathrm{BC}=\frac{t}{\cos (r)}
$
Equating equations (1) and (2)
$
\frac{L}{\sin (i-r)}=\frac{t}{\cos (r)}
$
After rearranging,
$
\mathrm{L}=\mathrm{t}\left(\frac{\sin (i-r)}{\cos (r)}\right)
$
Lateral displacement depends upon the thickness of the slab. Thicker the slab, greater will be the lateral displacement. Greater the angle of incident, larger will be the lateral displacement.
Question 6.
Derive the equation for refraction at single spherical surface.
Answer:
Equation for refraction at single spherical surface:
Let us consider two transparent media having refractive indices $\mathrm{n}_1$ and $\mathrm{n}_2$ are separated by a spherical surface. Let $C$ be the centre of curvature of the spherical surface. Let a point object $O$ be in the medium $\mathrm{n}_1$. The line OC cuts the spherical surface at the pole $\mathrm{P}$ of the surface. As the rays considered are paraxial rays, the perpendicular dropped for the point of incidence to the principal axis is very close to the pole or passes through the pole itself.

Light from $\mathrm{O}$ falls on the refracting surface at $\mathrm{N}$. The normal drawn at the point of incidence passes through the centre of curvature C. As $\mathrm{n}_2>\mathrm{n}_1$, light in the denser medium deviates towards the normal and meets the principal axis at I where the image is formed.
Snell's law in product form for the refraction at the point $N$ could be written as, $\mathrm{n}_1 \sin \mathrm{i}=\mathrm{n}_2 \sin \mathrm{r} \ldots(1)$
As the angles are small, sin of the angle could be approximated to the angle itself.
$\mathrm{n}_1 \mathrm{i}=\mathrm{n}_2 \mathrm{r}$
Let the angles,
$
\begin{aligned}
& \angle \mathrm{NOP}=\alpha, \angle \mathrm{NCP}=\beta, \angle \mathrm{NIP}=\gamma \\
& \tan \alpha=\frac{P N}{P O} ; \tan \beta=\frac{P N}{P C} ; \tan \gamma=\frac{P N}{P I}
\end{aligned}
$
As these angles are small, tan of the angle could be approximated to the angle itself.
$
\alpha=\frac{P N}{P O} ; \beta=\frac{P N}{P C} ; \gamma=\frac{P N}{P I}
$
For the triangle, $\triangle \mathrm{ONC}$,
$
I=\alpha+\beta
$
For the triangle, $\Delta \mathrm{INC}$,
$
\beta=r+\gamma \text { (or) } r=\beta-\gamma \ldots \ldots \text { (5) }
$
Substituting for I and $\mathrm{r}$ from equations (4) and (5) in the equation (2).
$\mathrm{n}_1(\alpha+\beta)=\mathrm{n}_2(\beta-\gamma)$
Rearranging,
$\mathrm{n}_1 \alpha+\mathrm{n}_2 \gamma=\left(\mathrm{n}_2-\mathrm{n}_1\right) \beta$
Substituting for $\alpha, \beta$ and $\gamma$ from equation (3)
$
\mathrm{n}_1=\left(\frac{P N}{P O}\right)+\mathrm{n}_2=\left(\frac{P N}{P I}\right)=\left(\mathrm{n}_2-\mathrm{n}_1\right)\left(\frac{P N}{P C}\right)
$
Further simplifying by cancelling PN,
$
\frac{n_1}{P O}+\frac{n_2}{P I}=\frac{n_2-n_1}{P C}
$
Following sign conventions, $P O=-u, P I=+\mathrm{v}$ and $\mathrm{PC}=+\mathrm{R}$ in equation (6),
$
\frac{n_2}{-v}-\frac{n_1}{u}=\frac{\left(n_2-n_1\right)}{R}
$
After rearranging, finally we get,
$
\frac{n_2}{v}-\frac{n_1}{u}=\frac{\left(n_2-n_1\right)}{R}
$
Equation (7) gives the relation among the object distance $\mathrm{n}$, image distance $\mathrm{v}$, refractive indices of the two media $\left(\mathrm{n}_1\right.$ and $\mathrm{n}_2$ ) and the radius of curvature $\mathrm{R}$ of the spherical surface. It holds for any spherical surface.
If the first medium is air then, $\mathrm{n}_1=1$ and the second medium is taken just as $\mathrm{n}_2=\mathrm{n}$, then the equation is reduced to,
$
\frac{n}{v}-\frac{1}{u}=\frac{(n-1)}{R}
$
Question 7.
Obtain lens maker's formula and mention its significance.
Answer:
Lens maker's formula and lens equation:
Let us consider a thin lens made up of a medium of refractive index $\mathrm{n}_2$, is placed in a medium of refractive index $n_1$. Let $R_1$ and $R_2$ be the radii of curvature of two spherical surfaces (1) and (2)
respectively and $P$ be the pole. Consider a point object $O$ on the principal axis. The ray which falls very close to $\mathrm{P}$, after refraction at the surface (1) forms image at I'. Before it does so, it is again refracted by the surface (2). Therefore the final image is formed at I. The general equation for the refraction at a spherical surface is given by
$
\frac{n_2}{v}-\frac{n_1}{u}=\frac{\left(n_2-n_1\right)}{R}
$
For the refracting surface (1), the light goes from $\mathrm{n}_1$ ton $\mathrm{n}_2$.
$
\frac{n_2}{v^{\prime}}-\frac{n_1}{u}=\frac{\left(n_2-n_1\right)}{R}
$
For the refracting surface (2), the light goes from medium $\mathrm{n}_2$ ton ${ }_1$.
$
\frac{n_1}{v}-\frac{n_2}{v^{\prime}}=\frac{\left(n_1-n_2\right)}{R} 2 \ldots \ldots .(2)
$
Adding the above two equation (1) and (2)

$
\frac{n_1}{v}-\frac{n_1}{u}=\left(\mathrm{n}_2-\mathrm{n}_1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)
$
Furhter simplifying and rearranging,
$
\begin{aligned}
& \frac{1}{v}-\frac{1}{u}=\left(\frac{n_2-n_1}{n_1}\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
& \frac{1}{v}-\frac{1}{u}=\left(\frac{n_2}{n_1}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)
\end{aligned}
$
If the object is at infinity, the image is formed at the focus of the lens. Thus, for $u=\infty, v=f$. Then the equation becomes.
$
\begin{aligned}
& \frac{1}{f}-\frac{1}{\infty}=\left(\frac{n_2}{n_1}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
& \frac{1}{f}=\left(\frac{n_2}{n_1}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \quad \cdots .
\end{aligned}
$
If the refractive index of the lens is and it is placed in air, then $\mathrm{n}_2=\mathrm{n}$ and $\mathrm{n}_1=1$ equation (4) becomes, $\frac{1}{f}=(\mathrm{n}-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)$
The above equation is called the lens maker's formula, because it tells the lens manufactures what curvature is needed to make a lens of desired focal length with a material of particular refractive index. This formula holds good also for a concave lens. By comparing the equations (3) and (4) we can write, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
This equation is known as lens equation which relates the object distance $\mathrm{u}$ and image distance $\mathrm{v}$ with the focal length $\mathrm{f}$ of the lens. This formula holds good for a any type of lens.
Question 8.
Derive the equation for thin lens and obtain its magnification.
Answer:
Lateral magnification in thin lens:
Let us consider an object $00^{\prime}$ of height $\mathrm{h}_1$ placed on the principal axis with its height perpendicular to the principal axis. The ray Op passing through the pole of the lens goes undeviated. The inverted real image II' formed has a height $h_2$.

The lateral or transverse magnification $m$ is defined as the ratio of the height of the image to that of the object.
$
\mathrm{m}=\frac{I I^{\prime}}{O O^{\prime}}
$
From the two similar triangles $\Delta \mathrm{POO}^{\prime}$ and $\Delta \mathrm{PII}$ ', we can write,
$
\frac{H^{\prime}}{O O^{\prime}}=\frac{P I}{P O}
$
Applying sign convention,
$
\frac{-h_2}{h_1}=\frac{v}{-u}
$
Substituting this in the equation (1) for magnification,
$
\mathrm{m}=\frac{-h_2}{h_1}=\frac{v}{-u}
$
After rearranging,
$
\mathrm{m}=\frac{h_2}{h_1}=\frac{v}{u}
$
The magnification is negative for real image and positive for virtual image. In the case of a concave lens, the magnification is always positive and less than one. We can also have the equations for magnification by combining the lens equation with the formula for magnification as,
$
\mathrm{m}=\frac{h_2}{h_1}=\frac{f}{f+u} \text { (or) } \mathrm{m}=\frac{h_2}{h_1}=\frac{f-v}{f}
$
Question 9.
Derive the equation for effective focal length for lenses in out of contact.
Answer:
Consider a two lenses of focal length $\mathrm{f}_1$ and $\mathrm{f}_2$ arranged coaxially but separated by a distance $\mathrm{d}$ can be considered. For a parallel ray that falls on the arrangement, the two lenses produce deviations $\delta_1$ and $\delta_2$ respectively and The net deviation $\delta$ is,
$
\delta=\delta_1+\delta_2
$
From Angle of deviation in lens equation, $\delta=\frac{h}{f}$
$
\delta_1=\frac{h_1}{f_1} ; \delta_2=\frac{h_2}{f_2} \delta=\frac{h_1}{f}
$
The equation (1) becomes,

$
\frac{h_1}{f}=\frac{h_1}{f_1}+\frac{h_2}{f_2}
$
From the geometry,
$
\begin{aligned}
& \mathrm{h}_2-\mathrm{h}_1=\mathrm{P}_2 \mathrm{C}=\mathrm{CG} \\
& \mathrm{h}_2-\mathrm{h}_1=\mathrm{BG} \tan \delta_1 \approx \mathrm{BG} \delta_1 \\
& \mathrm{~h}_2-\mathrm{h}_1=\mathrm{d} \frac{h_1}{f_1} \\
& \mathrm{~h}_2=\mathrm{h}_1+\mathrm{d} \frac{h_1}{f_1} \ldots . . . \text { (4) }
\end{aligned}
$
Substituting the above equation in Equation (3)
$
\frac{h_1}{f}=\frac{h_1}{f_1}+\frac{h_1}{f_2}+\frac{h_1 d}{f_1 f_2}
$
On further simplification,
$
\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{d}{f_1 f_2}
$
The above equation could be used to find the equivalent focal length.
Question 10 .
Derive the equation for angle of deviation produced by a prism and thus obtain the equation for refractive index of material of the prism.
Answer:
Angie of deviation produced by prism:
Let light ray $P Q$ is incident on one of the refracting faces of the prism. The angles of incidence and refraction at the first face $A B$ are $i_1$ and $r_1$. The path of the light inside the prism is $Q R$. The angle of incidence and refraction at the second face $\mathrm{AC}$ is $\mathrm{r}_2$ and $i_2$ respectively. RS is the ray emerging from $\mathrm{p}$ the second face. Angle $i_2$ is also called angle of emergence. The angle between the direction of the incident ray $P Q$ and the emergent ray $R S$ is called the angle of deviation $d$.

The two normals drawn at the point of incidence $\mathrm{Q}$ and emergence $\mathrm{R}$ are $\mathrm{QN}$ and $\mathrm{RN}$. They meet at point $\mathrm{N}$. The incident ray and the emergent ray meet at a point $\mathrm{M}$. The deviation $d_1$ at the surface $A B$ is, angle $\angle \mathrm{RQM}=\mathrm{d}_1=\mathrm{i}_1-\mathrm{r}_1 \ldots$ (1) The deviation $d_2$ at the surface $A C$ is, angle $\angle \mathrm{QRM}=\mathrm{d}_2=\mathrm{i}_2-\mathrm{r}_2 \ldots$ (2) Total angle of deviation d produced is, $\mathrm{d}=\mathrm{d}_1+\mathrm{d}_2 \ldots(3)$
Substituting for $\mathrm{d}_1$ and $\mathrm{d}_2$, $\mathrm{d}=\left(\mathrm{i}_1-\mathrm{r}_1\right)+\left(\mathrm{i}_2-\mathrm{r}_2\right)$
After rearranging, $\mathrm{d}=\left(\mathrm{i}_1-\mathrm{r}_1\right)+\left(\mathrm{i}_2-\mathrm{r}_2\right) \ldots(4)$
In the quadrilateral $A Q N R$, two of the angles (at the vertices $\mathrm{Q}$ and $\mathrm{R}$ ) are right angles. Therefore, the sum of the other angles of the quadrilateral is $180^{\circ}$. $\angle \mathrm{A}+\angle \mathrm{QNR}=180^{\circ} \ldots(5)$ From the triangle $\triangle \mathrm{QNR}$, $\mathrm{r}_1+\mathrm{r}_2 \angle \mathrm{QNR}=180^{\circ} \ldots(6)$ Comparing these two equations (5) and (6) we get, $\mathrm{r}_1+\mathrm{r}_2=\mathrm{A}$
Substituting this in equation (4) for angle of deviation, $\mathrm{d}=\mathrm{i}_1+\mathrm{i}_2-\mathrm{A}$...(8)
Thus, the angle of deviation depends on the angle of incidence angle of emergence and the angle for the prism.
Refractive index of the material of the prism:
At minimum deviation, $i_1=i_2=i$ and $r_1=r_2=r$
Now, the equation (8) becomes,
$\mathrm{D}=\mathrm{i}_1+\mathrm{i}_2$-A (or) $\mathrm{i}=\frac{(A+D)}{2}$
The equation (7) becomes,
$\mathrm{r}_1+\mathrm{r}_2=\mathrm{A} \Rightarrow 2 \mathrm{r}=\mathrm{A}$ (or) $\mathrm{r}=\frac{A}{2}$
Substituting $i$ and $r$ in Snell's law, $\mathrm{n}=\frac{\sin i}{\sin r}$]

$
\mathrm{n}=\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \left(\frac{A}{2}\right)}
$
The above equation is used to find the refractive index of the material of the prism.
Question 11.
What is dispersion? Obtain the equation for dispersive power of a medium.
Answer:
Dispersion. Dispersion is splitting of white light into its constituent colours.
Dispersive Power:
Consider a beam of white light passes through a prism; It gets dispersed into its constituent colours. Let $\delta_{\mathrm{V}}, \delta_{\mathrm{R}}$ are the angles of deviation for violet and red light. Let $\mathrm{n}_{\mathrm{V}}$ and $\mathrm{n}_{\mathrm{R}}$ are the refractive indices for the violet and red light respectively.

The refractive index of the material of a prism is given by the equation
$
\mathrm{n}=\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \left(\frac{A}{2}\right)}
$
Here $\mathrm{A}$ is the angle of the prism and $\mathrm{D}$ is the angle of minimum deviation. If the angle of prism is small of the order of $10^{\circ}$, the prism is said to be a small angle prism. When rays of light pass through such prisms, the angle of deviation also becomes small. If $\mathrm{A}$ be the angle of a small angle prism and 5 the angle of deviation then the prism formula becomes.
$
\mathrm{n}=\frac{\sin \left(\frac{A+\delta}{2}\right)}{\sin \left(\frac{A}{2}\right)}
$
For small angles of $\mathrm{A}$ and $\delta_{\mathrm{m}}$
$
\begin{aligned}
& \sin \left(\frac{\mathrm{A}+\delta}{2}\right) \approx\left(\frac{\mathrm{A}+\delta}{2}\right) \ldots \ldots(2) \\
& \sin \left(\frac{\mathrm{A}}{2}\right) \approx\left(\frac{\mathrm{A}}{2}\right) \quad \ldots .(3) \\
& \therefore \mathrm{n}=\frac{\left(\frac{A+\delta}{2}\right)}{\left(\frac{A}{2}\right)}=\frac{A+\delta}{A}=1+\frac{\delta}{A}
\end{aligned}
$
Further simplifying, $\frac{\delta}{A}=\mathrm{n}-1$
$
\delta=(\mathrm{n}-1) \mathrm{A}
$
When white light enters the prism, the deviation is different for different colours. Thus, the refractive index is also different for different colours.
For Violet light, $\delta_{\mathrm{V}}=\left(\mathrm{n}_{\mathrm{V}}-1\right) \mathrm{A}$...(5)
For Red light, $\delta_R=\left(n_R-1\right) \ldots(6)$
As, angle of deviation for violet colour $\delta_V$ is greater than angle of deviation for red colour $\delta_R$, the

refractive index for violet colour $\mathrm{n}_{\mathrm{V}}$ is greater than the refractive index for red colour $\mathrm{n}_{\mathrm{R}}$. Subtracting $\delta_V$ from $\delta_R$ we get,
$$
\delta_{\mathrm{V}}-\delta_{\mathrm{R}}=\left(\mathrm{n}_{\mathrm{V}}-\mathrm{n}_{\mathrm{R}}\right) \mathrm{A}
$$
The term $\left(\delta_V-\delta_{\mathrm{R}}\right)$ is the angular separation between the two extreme colours (violet and red) in the spectrum is called the angular dispersion. Clearly, the angular dispersion produced by a prism depends upon.
1. Angle of the prism
2. Nature of the material of the prism.
If we take 8 is the angle of deviation for any middly ray (green or yellow) and $\mathrm{n}$ the corresponding refractive index. Then, $8=(\mathrm{n}-1) \mathrm{A} \ldots(8)$
Dispersive power $(\omega)$ is the ability of the material of the prism to cause dispersion. It is defined as the ratio of the angular dispersion for the extreme colours to the deviation for any mean colour. Dispersive power $(\omega)$,
$$
\omega=\frac{\text { angular dispersion }}{\text { mean deviation }}=\frac{\delta_{\mathrm{V}}-\delta_{\mathrm{R}}}{\delta}
$$
Substituting for $\left(\delta_{\mathrm{V}}-\delta_{\mathrm{R}}\right)$ and $(\delta)$,
$$
\omega=\frac{\left(n_{\mathrm{V}}-n_{\mathrm{R}}\right)}{(n-1)}
$$
Dispersive power is a dimensionless quality. It has no unit. Dispersive power is always positive. The dispersive power of a prism depends only on the nature of material of the prism and it is independent of the angle of the prism.
Question 12.
Prove laws of reflection using Huygens' principle.
Answer:
Proof for laws of reflection using Huygens' Principle:
Let us consider a parallel beam of light, incident on a reflecting plane surface such as a plane mirror $\mathrm{XY}$. The incident wavefront is $\mathrm{AB}$ and the reflected wavefront is $\mathrm{A}^{\prime} \mathrm{B}^{\prime}$ ' in the same medium. These wavefronts are perpendicular to the incident rays $\mathrm{L}, \mathrm{M}$ and reflected rays $\mathrm{L}$ ', $\mathrm{M}$ ' respectively.
By the time point A of the incident wavefront touches the reflecting surface, the point $\mathrm{B}$ is yet to travel a distance BB' to touch the reflecting surface a B'. When the point B falls on the reflecting surface at $\mathrm{B}^{\text {' }}$, the point $\mathrm{A}$ would have reached $\mathrm{A}^{\prime}$. This is applicable to all the points on the wavefront.
Thus, the reflected wavefront A'B' emanates as a plane wavefront. The two normals $\mathrm{N}$ and N' are considered at the points where the rays $L$ and $M$ fall on the reflecting surface. As reflection happens in the same medium, the speed of light is same before and after the reflection. Hence, the time taken for the ray to travel from $B$ to $B^{\prime}$ is the same as the time taken for the ray to travel from $A$ to $A^{\prime}$. Thus, the distance $\mathrm{BB}^{\prime}$ is equal to the distance $\mathrm{AA}^{\prime} ;\left(\mathrm{AA}^{\prime}=\mathrm{BB}^{\prime}\right)$.

(i) The incident rays, the reflected rays and the normal are in the same plane.
(ii) Angle of incidence, $\angle \mathrm{i}=\angle \mathrm{NAL}=90^{\circ}-\angle \mathrm{NAB}=\angle \mathrm{BAB}$
Angle of reflection, $\angle \mathrm{r}=\angle \mathrm{N}^{\prime} \mathrm{B}^{\prime} \mathrm{M}^{\prime}=90^{\circ}-\angle \mathrm{N}^{\prime} \mathrm{B}^{\prime} \mathrm{A}^{\prime}=\angle \mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{A}$
For the two right angle triangles, $\mathrm{A} A \mathrm{AB}^{\prime}$ ' and $\mathrm{A} \mathrm{B}^{\prime} \mathrm{A}^{\prime} \mathrm{A}$, the right angles, $\angle \mathrm{B}$ and $\angle \mathrm{A}^{\prime}$ are equal, ( $\angle \mathrm{B}$ and $\left.\angle \mathrm{A}^{\prime}=90^{\circ}\right)$; the two sides, $\mathrm{AA}^{\prime}$ and $\mathrm{BB}^{\prime}$ are equal, $\left(\mathrm{AA}^{\prime}=\mathrm{BB}^{\prime}\right)$; the side $\mathrm{AB}^{\prime}$ is the common. Thus, the two triangles are congruent. As per the property of congruency, the two angles, $\angle \mathrm{BAB}^{\prime}$ and $\angle \mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{A}$ must also be equal. $i=r$ Hence, the laws of reflection are proved.
Question 13.
Prove laws of refraction using Huygens' principle.
Answer:
Let us consider a parallel beam of light is incident on a refracting plane surface XY such as a glass surface. The incident wavefront $\mathrm{AB}$ is in rarer medium (1) and the refracted wavefront $\mathrm{A}^{\text {' }} \mathrm{B}^{\text {' }}$ is in denser medium (2). These wavefronts are perpendicular to the incident rays $L, M$ and refracted rays $L^{\prime}, M^{\prime}$ respectively. By the time the point $\mathrm{A}$ of the incident wavefront touches the refracting surface, the point $\mathrm{B}$ is yet to travel a distance $\mathrm{BB}^{\prime}$ to touch the refracting surface at $\mathrm{B}^{\prime}$

When the point B falls on the refracting surface at B', the point A' would have reached A in the other medium. This is applicable to all the points on the wavefront. Thus, the refracted wavefront A'B' emanates as a plane wavefront.
The two normals $\mathrm{N}$ and $\mathrm{N}^{\prime}$ are considered at the points where the rays $\mathrm{L}$ and $\mathrm{M}$ fall on the refracting surface. As refraction happens from rarer medium (1) to denser medium (2), the speed of light is $v_1$ and $\mathrm{v}_2$ before and after refraction and $\mathrm{v}_1$ is greater than $\mathrm{v}_2\left(\mathrm{v}_1>\mathrm{v}_2\right)$. But, the time taken $t$ for the ray to
travel from B to B' is the same as the time taken for the ray to travel from $\mathrm{A}$ to $\mathrm{A}^{\prime}$ '.
$
\mathrm{t}=\frac{B B^{\prime}}{v_1}=\frac{A A^{\prime}}{v_2} \text { (or) } \frac{B B^{\prime}}{A A^{\prime}}=\frac{v_1}{v_2}
$
(i) The incident rays, the refracted rays and the normal are in the same plane.
(ii) Angle of incidence,
$
\text { i }=\angle \mathrm{NAL}=90^{\circ}-\angle \mathrm{NAB}=\angle \mathrm{BAB} \text { ' }
$
Angle of refraction,
$
\mathrm{r}=\angle \mathrm{N}^{\prime} \mathrm{B}^{\prime} \mathrm{M}^{\prime}=90^{\circ}-\angle \mathrm{N}^{\prime} \mathrm{B}^{\prime} \mathrm{A}^{\prime}=\angle \mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{A}^{\prime}
$
For the two right angle triangles $\triangle \mathrm{ABB}^{\prime}$ and $\triangle \mathrm{B}^{\prime}$ 'A'A,
$
\frac{\operatorname{sini}}{\operatorname{sinr}}=\frac{B B^{\prime} / A B^{\prime}}{A A^{\prime} / A B^{\prime}}=\frac{B B^{\prime}}{A A^{\prime}}=\frac{v_1}{v_2}=\frac{c / v_2}{c / v_1}
$
Here, $\mathrm{c}$ is speed of light in vacuum. The ratio $\mathrm{c} / \mathrm{v}$ is the constant, called refractive index of the medium. The refractive index of medium (1) is, $c / v_1=n_1$ and that of medium (2) is, $c / v_2=n_2$.
$
\frac{\sin i}{\operatorname{sinr}}=\frac{n_2}{n_1}
$
In product form,
$\mathrm{n}_1 \sin \mathrm{i}=\mathrm{n}_2 \sin \mathrm{r}$
Hence, the laws of refraction are proved.
Question 14.
Obtain the equation for resultant intensify' due to interference of light.
Answer:
Let us consider two light waves from the two sources $S_1$ and $S_2$ meeting at a point $P$. The wave from $S_1$ at an instant $t$ at $P$ is, $\mathrm{y}_1=\mathrm{a}_1 \sin \omega \mathrm{t} \ldots(1)$
The wave form $S_2$ at an instant $t$ at $P$ is,

$
\mathrm{y}_2=\mathrm{a}_2 \sin (\omega t+\Phi) \ldots(2)
$
The two waves have different amplitudes $a_1$ and $a_2$, same angular frequency $\omega$, and a phase difference of $\Phi$ between them. The resultant displacement will be given by,
$
\mathrm{y}=\mathrm{y}_1+\mathrm{y}_2=\mathrm{a}_1 \sin \omega \mathrm{t}+\mathrm{a}_1 \sin _2(\omega \mathrm{t}+\Phi) \ldots \text { (3) }
$
The simplification of the above equation by using trigonometric identities gives the equation, $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}+\theta) \ldots(4)$
Where, $\mathrm{A}=\sqrt{a_1^2+a_2^2+2 a_1 a_2 \cos \phi}$
The resultant amplitude is maximum,
$
\mathrm{A}_{\max }=\sqrt{\left(a_1+a_2\right)^2} ; \text { when } \Phi=0, \pm 2 \pi, \pm 4 \pi \ldots, \ldots \ldots
$
The resultant amplitude is minimum,
$\mathrm{A}_{\max }=\sqrt{\left(a_1-a_2\right)^2}$; when $\Phi= \pm \pi, \pm 3, \pi \pm 5 \pi \ldots \ldots, \ldots$ (8)
The intensity of light is proportional to square of amplitude,
$I \propto \mathrm{A}^2 \ldots(9)$
Now, equation (5) becomes,
$\mathrm{I} \propto \mathrm{I}_1+\mathrm{I}_2+2 \sqrt{I_1 I_2} \cos \Phi$
In equation (10) if the phase difference, $\Phi=0, \pm 2 \pi, \pm 4 \pi \ldots$. it corresponds to the condition for maximum intensity of light called as constructive interference. The resultant maximum intensity is, $\mathrm{I}_{\max } \propto\left(\mathrm{a}_1+\mathrm{a}_2\right)^2 \propto \mathrm{I}_1+\mathrm{I}_2+2 \sqrt{I_1 I_2}$
In equation (10) if the phase difference, $\Phi= \pm \pi, \pm 3, \pi \pm 5 \pi \ldots$, it corresponds to the condition for minimum intensity of light called destructive interference. The resultant minimum intensity is, $\mathrm{I}_{\max } \propto\left(\mathrm{a}_1-\mathrm{a}_2\right)^2 \propto \mathrm{I}_1+\mathrm{I}_2-2 \sqrt{I_1 I_2}$
Question 15.
Explain the Young's double slit experimental setup and obtain the equation for path difference.
Answer:
I Experimental setup:
1. Wavefronts from $S_1$ and $S_2$ spread out and overlapping takes place to the right side of double slit. When a screen is placed at a distance of about 1 meter s from the slits, alternate bright and dark fringes which are equally spaced appear on the screen. These are called interference fringes or bands.

2. Using an eyepiece the fringes can be seen directly. At the center point $O$ on the screen, waves from $S_1$ and $S_2$ travel equal distances and arrive in-phase. These two waves constructively interfere and bright fringe is observed at $\mathrm{O}$. This is called central bright fringe.
3. The fringes disappear and there is uniform illumination on the screen when one of the slits is covered. This shows clearly that the bands are due to interference.
II Equation for path difference:
1. Let $d$ be the distance between the double slits $S_1$ and $S_2$ which act as coherent sources of wavelength $\lambda$. A screen is placed parallel to the double slit at a distance $\mathrm{D}$ from it. The mid-point of $\mathrm{S}_1$ and $\mathrm{S}_2$ is $\mathrm{C}$ and the midpoint of the screen $\mathrm{O}$ is equidistant from $\mathrm{S}_1$ and $\mathrm{S}_2$. P is any point at a distance y from $\mathrm{O}$.

2. The waves from $S_1$ and $S_2$ meet at $P$ either inphase or out-of-phase depending upon the path difference between the two waves.
3. The path difference $\delta$ between the light waves from $S_1$ and $S_2$ to the point $P$ is, $\delta=\mathrm{S}_2 \mathrm{P}-\mathrm{S}_1$
4. A perpendicular is dropped from the point $S_1$ to the line $S_2 P$ at $M$ to find the path difference more precisely.
$
\delta=\mathrm{S}_2 \mathrm{P}-\mathrm{MP}=\mathrm{S}_2 \mathrm{M}
$
The angular position of the point $P$ from $C$ is $\theta . \angle O C P=\theta$.
From the geometry, the angles $\angle O C P$ and $\angle S_2 S_1$ M are equal. $\angle O C P=\angle S_2 S_1 M=\theta$
In right angle triangle $\Delta S_1 S_2 M$, the path difference, $S_2 M=d \sin \theta$
$
\delta=\mathrm{d} \sin \theta
$
If the angle 0 is small, $\sin \theta \approx \tan \theta \approx \theta$. From the right angle triangle $\triangle O C P, \tan \theta=\frac{y}{D}$
The path difference, $\delta=\frac{d y}{D}$
Question 16.
Obtain the equation for bandwidth in Young's double slit experiment.
Answer:
Condition for bright fringe (or) maxima
The condition for the constructive interference or the point $P$ to be have a bright fringe is, Path difference, $\delta=\mathrm{n} \lambda$
where, $\mathrm{n}=0,1,2, \ldots$
$
\begin{aligned}
& \therefore \frac{d y}{D}=\mathrm{n} \lambda \\
& \mathrm{y}=\mathrm{n} \frac{\lambda D}{d} \text { (or) } \mathrm{y}_{\mathrm{n}}=\mathrm{n} \frac{\lambda D}{d}
\end{aligned}
$
This is the condition for the point $P$ to be a bright fringe. The distance is the distance of the nth bright fringe from the point $\mathrm{O}$.
Condition for dark fringe (or) minima
The condition for the destructive interference or the point $P$ to be have a dark fringe is, Path difference, $\delta=(2 n-1) \frac{\lambda}{2}$
where, $\mathrm{n}=1,2,3 \ldots$

$
\begin{aligned}
& \therefore \frac{d y}{D}=(2 \mathrm{n}-1) \frac{\lambda}{2} \\
& \mathrm{y}=\frac{(2 n-1)}{2} \frac{\lambda D}{d} \text { (or) } \mathrm{y}_{\mathrm{n}}=\frac{(2 n-1)}{2} \frac{\lambda D}{d}
\end{aligned}
$
Equation for bandwidth
The bandwidth $(\beta)$ is defined as the distance between any two consecutive bright or dark fringes.
The distance between $(\mathrm{n}+1)^{\text {th }}$ and $\mathrm{n}^{\text {th }}$ consecutive bright fringes from $O$ is given by,
$
\begin{aligned}
& \beta=\mathrm{y}_{(\mathrm{n}+1)}-\mathrm{y}_{\mathrm{n}}=\left((n+1) \frac{\lambda \mathrm{D}}{d}\right)-\left(n \frac{\lambda \mathrm{D}}{d}\right) \\
& \beta=\frac{\lambda D}{d}
\end{aligned}
$
Similarly, the distance between $(\mathrm{n}+1)^{\text {th }}$ and $\mathrm{n}^{\text {th }}$ consecutive dark fringes from $\mathrm{O}$ is given by,
$
\begin{aligned}
& \beta=\mathrm{y}_{(\mathrm{n}+1)}-\mathrm{y}_{\mathrm{n}}=\left(\frac{(2(n+1)-1)}{2} \frac{\lambda D}{d}\right)-\left(\frac{(2 n-1)}{2} \frac{\lambda D}{d}\right) \\
& \beta=\frac{\lambda D}{d}
\end{aligned}
$
The above equation show that the bright and dark fringes are of same width equally spaced on either side of central bright fringe.
Question 17.
Obtain the equations for constructive and destructive interference for transmitted and reflected waves in thin films.
Answer:
Interference in thin films:
Let us consider a thin film of transparent material of refractive index $p$ (not to confuse with order of fringe $\mathrm{n}$ ) and thickness $\mathrm{d}$. A parallel beam of light is incident on the film at an angle $i$. The wave is divided into two parts at the upper surface, one is reflected and the other is refracted.
The refracted part, which enters into the film, again gets divided at the lower surface into two parts; one is transmitted out of the film and the other is reflected back in to the film. Reflected as well as refracted waves are sent by the film as multiple reflections take place inside the film. The interference is produced by both the reflected and transmitted light.

For transmitted light:
The light transmitted may interfere to produce a resultant intensity. Let us consider the path difference between the two light waves transmitted from $\mathrm{B}$ and $\mathrm{D}$. The two waves moved together and remained in phase up to $B$ where splitting occured. The extra path travelled by the wave transmitted from $\mathrm{D}$ is the path inside the film, $\mathrm{BC}+\mathrm{CD}$.
If we approximate the incidence to be nearly normal $(i=0)$, then the points $B$ and $D$ are very close to each other. The extra distance travelled by the wave is approximately twice thickness of the film, $\mathrm{BC}+$ $\mathrm{CD}=2 \mathrm{~d}$. As this extra path is traversed in a medium of refractive index $\mu$, the optical path difference is, $\delta=2 \mu \mathrm{d}$
The condition for constructive interference in transmitted ray is, $2 \mu \mathrm{d}=\mathrm{n} \lambda$
Similarly, the condition for destructive interference in transmitted ray is, $2 \mu \mathrm{d}=(2 \mathrm{n}-1) \frac{\lambda}{2}$
For reflected light:
It is experimentally and theoretically proved that a wave while travelling in a rarer medium and getting reflected by a denser medium, undergoes a phase change of $\pi$. Hence, an additional path difference of $\frac{\lambda}{2}$ should be considered.

Let us consider the path difference between the light waves reflected by the upper surface at $\mathrm{A}$ and the other wave coming out at $\mathrm{C}$ after passing through the film. The additional path travelled by wave coming out from $\mathrm{C}$ is the path inside the film, $\mathrm{AB}+\mathrm{BC}$. For nearly normal incidence this distance could be approximated as, $A B+B C=2 \mathrm{~d}$. As this extra path is travelled in the medium of refractive index $p$, the optical path difference is, $\delta=2 \mu \mathrm{d}$.
The condition for constructive interference for reflected ray is, $2 \mu \mathrm{d}+\frac{\lambda}{2}=\mathrm{n} \lambda$ (or) $2 \mu \mathrm{d}=(2 \mathrm{n}-1) \frac{\lambda}{2}$
The additional path difference $\frac{\lambda}{2}$ is due to the phase change of $\mathrm{n}$ in rarer to denser reflection taking place at A. The condition for destructive interference for reflected ray is, $2 \mu \mathrm{d}+\frac{\lambda}{2}=(2 \mathrm{n}+1) \frac{\lambda}{2}$ (or) $2 \mu \mathrm{d}=\mathrm{n} \lambda$

Question 18.
Discuss diffraction at single slit and obtain the condition for $\mathrm{n}^{\text {th }}$ minimum.
Answer:
Diffraction at single slit:
Let a parallel beam of light fall normally on a single slit $A B$ of width. The diffracted beam falls on a screen kept at a distance. The center of the slit is C. A straight line through C perpendicular to the plane of slit meets the center of the screen at $O$. We would like to find the intensity at any point $P$ on the screen. The lines joining $P$ to the different points on the slit can be treated as parallel lines, making an angle 9 with the normal CO.

All the waves start parallel to each other from different points of the slit and interfere at point $P$ and other points to give the resultant intensities. The point $P$ is in the geometrically shadowed region, up to which the central maximum is spread due to diffraction. We need to give the condition for the point $P$ to be of various minima.
The basic idea is to divide the slit into much smaller even number of parts. Then, add their contributions at $\mathrm{P}$ with the proper path difference to show that destructive interference takes place at that point to make it minimum. To explain maximum, the slit is divided into odd number of parts.

Condition for $\mathrm{P}$ to be first minimum:
Let us divide the slit $\mathrm{AB}$ into two half's $\mathrm{AC}$ and $\mathrm{CB}$. Now the width of $\mathrm{AC}$ is $(\mathrm{a} / 2)$. We have different points on the slit which are separated by the same width (here $\mathrm{a} / 2$ ) called corresponding points.

The path difference of light waves from different corresponding points meeting at point $\mathrm{P}$ and interfere destructively to make it first minimum. The path difference 8 between waves from these corresponding points is, $\delta=\frac{a}{2} \sin \theta$
The condition for $P$ to be first minimum, $\frac{a}{2} \sin \theta=\frac{\lambda}{2}$ a $\sin \theta=\lambda$ (first minimum) ..... (1) Condition for $P$ to be first minimum:
Let us divide the slit $\mathrm{AB}$ into two half's $\mathrm{AC}$ and $\mathrm{CB}$. Now the width of $\mathrm{AC}$ is $(\mathrm{a} / 2)$. We have different points on the slit which are separated by the same width (here $\mathrm{a} / 2$ ) called corresponding points. The path difference of light waves from different corresponding points meeting at point $\mathrm{P}$ and interfere destructively to make it first minimum.
The path difference $\delta$ between waves from these corresponding points $\delta=\frac{a}{4} \sin \theta$.
The condition for $\mathrm{P}$ to be first minimum, $\frac{a}{4} \sin \theta=\frac{\lambda}{2}$ a $\sin \theta=2 \lambda$ (second minimum) ... (2) Condition for $P$ to be third order minimum:
The same way the slit is divided into six equal parts to explain the condition for $\mathrm{P}$ to be third minimum is, $\frac{a}{6} \sin \theta=\frac{\lambda}{2}$
a $\sin \theta=3 \lambda$ (third minimum) ...(3)
Condition for $P$ to be $\mathrm{n}^{\text {th }}$ order minimum:
Dividing the slit into $2 \mathrm{n}$ number of (even number of) equal parts makes the light produced by one of the corresponding points to be cancelled by its counterpart. Thus, the condition for $\mathrm{n}^{\text {th }}$ order minimum is, $\frac{a}{2 n} \sin \theta=\frac{\lambda}{2}$
$\mathrm{a} \sin \theta=\mathrm{n} \lambda\left(\mathrm{n}^{\text {th }}\right.$ minimum)
Question 19.
Discuss the diffraction at a grating and obtain the condition for the $\mathrm{m}^{\text {th }}$ maximum.
Answer:
A plane transmission grating is represented by $\mathrm{AB}$. Let a plane wavefront of monochromatic light with wavelength $\lambda$ be incident normally on the grating. As the slits size is comparable to that of wavelength, the incident light diffracts at the grating.

A diffraction pattern is obtained on the screen when the diffracted waves are focused on a screen using a convex lens. Let us consider a point $P$ at an angle 0 with the normal drawn from the center of the grating to the screen. The path difference 5 between the diffracted waves from one pair of corresponding points is, $\delta=(\mathrm{a}+\mathrm{b}) \sin \theta$
This path difference is the same for any pair of corresponding points. The point $P$ will be bright, when $\delta=\mathrm{m} \lambda$ where $\mathrm{m}=0,1,2,3$
Combining the above two equations, we get, $(\mathrm{a}+\mathrm{b}) \sin \theta=\mathrm{m} \lambda$
Here, $m$ is called order of diffraction.
Condition for zero order maximum, $\mathrm{m}=0$
For $(a+b) \sin \theta=0$, the position, $\theta=0 . \sin \theta=0$ and $m=0$. This is called zero order diffraction or central maximum.
Condition for first order maximum, $\mathrm{m}=1$
If $(a+b) \sin \theta_1=\lambda$, the diffracted light meet at an angle $\theta_1$ to the incident direction and the first order maximum is obtained.
Condition for second order maximum, $\mathrm{m}=2$
Similarly, $(a+b) \sin \theta_2=2 \lambda$, forms the second order maximum at the angular position $\theta_2$.
Condition for higher order maximum
On either side of central maxima different higher orders of diffraction maxima are formed at different angular positions.
If we take, $\mathrm{N}=\frac{1}{a+b}$
Then, $N$ gives the number of grating elements or rulings drawn per unit width of the grating. Normally, this number $\mathrm{N}$ is specified on the grating itself. Now, the equation becomes, $\frac{1}{N} \sin \theta=\mathrm{m} \lambda$, (or) $\sin \theta=\mathrm{Nm} \lambda$

Question 20.
Discuss the experiment to determine the wavelength of monochromatic light using diffraction grating.

Answer:
Experiment to determine the wavelength of monochromatic light:
The wavelength of a spectral line can be very accurately determined with the help of a diffraction grating and a spectrometer. Initially all the preliminary adjustments of the spectrometer are made. The slit of collimator is illuminated by a monochromatic light, whose wavelength is to be determined. The telescope is brought in line with collimator to view the image of the slit. The given plane transmission grating is then mounted on the prism table with its plane perpendicular to the incident beam of light coming from the collimator.

The telescope is turned to one side until the first order diffraction image of the slit coincides with the vertical cross wire of the eye piece. The reading of the position of the telescope is noted. Similarly the first order diffraction image on the other side is made to coincide with the vertical cross wire and corresponding reading is noted. The difference between two positions gives $2 \theta$. Half of its value gives $\theta$, the diffraction angle for first order maximum. The wavelength of light is calculated from the equation, $\lambda=\frac{\sin \theta}{N m}$
Here, $N$ is the number of rulings per metre in the grating and $m$ is the order of the diffraction image.
Question 21.
Discuss the experiment to determine the wavelength of different colours using diffraction grating.
Answer:
Determination of wavelength of different colours:
When white light is used, the diffraction pattern consists of a white central maximum and on both sides continuous coloured diffraction patterns are formed. The central maximum is white as all the colours meet here constructively with no path difference.

As $\theta$ increases, the path difference, $(a+b) \sin \theta$, passes through condition for maxima of diffraction of different orders for all colours from violet to red. It produces a spectrum of diffraction pattern from violet to red on either side of central maximum. By measuring the angle at which these colours appear for various orders of diffraction, the wavelength of different colours could be calculated using the formula $\lambda=\frac{\sin \theta}{N m}$
Here, $N$ is the number of rulings per metre in the grating and $m$ is the order of the diffraction image.
Question 22.
Obtain the equation for resolving power of optical instrument.
Answer:
Resolution:

The effect of diffraction has an adverse impact in the image formation by the optical instruments such as microscope and telescope. For a single rectangular slit, the half angle $\theta$ subtended by the spread of central maximum (or position of first minimum) is given by the "relation, a $\sin \theta=\lambda \ldots(1)$
Similar to a rectangular slit, when a circular aperture or opening (like a lens or the iris of our eye) forms an image of a point object, the image formed will not be a point but a diffraction pattern of concentric circles that becomes fainter while moving away from the center. These are known as Airy's discs. The circle of central maximum has the half angular spread given by the equation, a $\sin \theta=1.22 \lambda \ldots(2)$
Here, the numerical value 1.22 comes for central maximum formed by circular apertures. This involves higher level mathematics which is avoided in this discussion.

Airy's discs
For small angles, $\sin \theta \approx \theta$
$\mathrm{a} \theta=1.22 \lambda \ldots(3)$
Rewriting further, $\theta=\frac{1.22 \lambda}{a}$ and $\frac{r_0}{f}=\frac{1.22 \lambda}{a}$
$
\mathrm{r}_0=\frac{1.22 \lambda f}{a}
$
When two point sources close to each another form image on the screen, the diffraction pattern of one point source can overlap with another and produce a blurred image. To obtain a good image of the two sources, the two point sources must be resolved i.e., the point sources must be imaged in such a way that their images are sufficiently far apart that their diffraction patterns do not overlap.
According to Rayleigh's criterion, for two point objects to be just resolved, the minimum distance between their diffraction images must be in such a way that the central maximum of one coincides with the first minimum of the other and vice versa. Such an image is said to be just resolved image of the object. The Rayleigh's criterion is said to be limit of resolution.

According to Rayleigh's criterion the two point sources are said to be just resolved when the distance between the two maxima is at least $\mathrm{r}_0$. The angular resolution has a unit in radian (rad) and it is given by the equation.
$
\theta=\frac{1.22 \lambda f}{a}
$
It shows that the first order diffraction angle must be as small as possible for greater resolution. This further shows that for better resolution, the wavelength of light used must be as small as possible and the size of the aperture of the instrument used must be as large as possible. The equation (4) is used to calculate spacial resolution. The inverse of resolution is called resolving power. This implies, smaller the resolution, greater is the resolving power of the instrument. The ability of an optical instrument to • separate or distinguish small or closely adjacent objects through the image formation is said to be resolving power of the instrument.
Question 23.
Discuss about simple microscope and obtain the equations for magnification for near point focusing and normal focusing.
Answer:
Simple microscope:
A simple microscope is a single magnifying (converging) lens of small focal length. The idea is to get an erect, magnified and virtual image of the object. For this the object is placed between $F$ and $P$ on one side of the lens and viewed from other side of the lens. There are two magnifications to be discussed for two kinds of focusing.
- Near point focusing :
The image is formed at near point, i.e. $25 \mathrm{~cm}$ for normal eye. This distance is also called as least distance $\mathrm{D}$ of distinct vision. In this position, the eye feels comfortable but there is little strain on the eye.
- Normal focusing :
The image is formed at infinity. In this position the eye is most relaxed to view the image.
Magnification in near point focusing:
Object distance $u$ is less than $f$. The image distance is the near point $D$. The magnification $\mathrm{m}$ is given by the relation,
$
\mathrm{m}=\frac{v}{u}
$
With the help of lens equation, $\frac{1}{v}-\frac{1}{u} \frac{1}{f}$ the magnification can further be writen as, $\mathrm{m}=1-\frac{v}{f}$
Substituting for $\mathrm{v}$ with sign convention, $\mathrm{v}=-$ Derivem $\mathrm{m}=1+\frac{D}{f}$
This is the magnification for near point focusing

Magnification in normal focusing (angular magnification):
We will now find the magnification for the image formed at infinity. If we take the ratio of height of image to height of object $\left(m=\frac{h^{\prime}}{h}\right)$ to find the magnification, we will not get a practical relation as the image will also be of infinite size when the image is formed at infinity. Hence, we can practically use the angular magnification. The angular magnification is defined as the ratio of angle $0 \mathrm{j}$ subtended by the image with aided eye to the angle 90 subtended by the object with unaided eye.

$
\mathrm{m}=\frac{\theta_i}{\theta_0}
$
For unaided eye, $\tan \theta_0 \approx \theta_0=\frac{h}{D}$
For aided eye, $\tan \theta_{\mathrm{i}} \approx \theta_{\mathrm{i}}=\frac{h}{f}$
The angular magnification is, $\mathrm{m}=\frac{\theta_i}{\theta_0}=\frac{h / f}{h / d}$ $\mathrm{m}=\frac{d}{f}$
This is the magnification for normal focusing. The magnification for normal focusing is one less than that for near point focusing.
Question 24.
Explain about compound microscope and obtain the equation for magnification. Compound microscope:
Answer:
The lens near the object, called the objective, forms a real, inverted, magnified image of the object. This serves as the object for the second lens which is the eyepiece. Eyepiece serves as a simple microscope that produces finally an enlarged and virtual image. The first inverted image formed by the objective is to be adjusted close to, but within the focal plane of the eyepiece so that the final image is formed nearly at infinity or at the near point. The final image is inverted with respect to the original object. We can obtain the magnification for a compound microscope.
Magnification of compound microscope :
From the ray diagram, the linear magnification due to the objective is,

$
\mathrm{m}_0=\frac{h^{\prime}}{f}
$
From the figure, $\tan \beta=\frac{h}{f_0}=\frac{h^{\prime}}{L^{\prime}}$, then
$
\frac{h^{\prime}}{h}=\frac{L}{f_0} ; \mathrm{m}_0=\frac{L}{f_0}
$
Here, the distance $L$ is between the first focal point of the eyepiece to the second focal point of the objective. This is called the tube length
$L$ of the microscope as $f_0$ and $f_e$ are comparatively smaller than $L$. If the final image is formed at $P$ (near point focusing), the magnification $\mathrm{m}_{\mathrm{e}}$ of the eyepiece is,
$
\mathrm{m}_{\mathrm{e}}=1+\frac{D}{f_e}
$
The total magnification $m$ in near point focusing is,
$
\mathrm{m}=\mathrm{m}_0 \mathrm{~m}_{\mathrm{e}}=\left(\frac{L}{f_o}\right)\left(1+\frac{D}{f_e}\right)
$
If the final image is formed at infinity (normal focusing), the magnification me of the eyepiece is, $\mathrm{m}_{\mathrm{e}}=\frac{D}{f_e}$
The total magnification $\mathrm{m}$ in normal focusing is,
$
\mathrm{m}=\mathrm{m}_0 \mathrm{~m}_{\mathrm{e}}=\left(\frac{L}{f_o}\right)\left(\frac{D}{f_e}\right)
$
Question 25.
Obtain the equation for resolving power of microscope.
Answer:
Resolving power of microscope:
The diagram related to the calculation of resolution of microscope. A microscope is used to see the details of the object under observation. The ability of microscope depends not only in magnifying the object but also in resolving two points on the object separated by a small distance $\mathrm{d}_{\text {min }}$.
Smaller the value of $d_{\min }$ better will be the resolving power of the microscope.

The radius of central maxima is, $\mathrm{r}_0=\frac{1.22 \lambda v}{a}$
In the place of focal length $f$ we have the image distance $v$. If the difference between the two points on the object to be resolved is $\mathrm{d}_{\min }$, then the magnification $\mathrm{m}$ is, $\mathrm{m}=\frac{r_o}{d_{\min }}$ $\mathrm{d}_{\min }=\frac{r_o}{m}=\frac{1.22 \lambda v}{a m}=\frac{1.22 \lambda v}{a(v / u)}=\frac{1.22 \lambda u}{a}[\because \mathrm{m}=\mathrm{v} / \mathrm{u}][\because[\mathrm{u} \approx f]$
On the object side, $2 \tan \beta \approx 2 \sin \beta=\frac{a}{f}$
$
\mathrm{d}_{\min }=\frac{1.22 \lambda}{2 \sin \beta} \therefore[\mathrm{a}=f 2 \sin \beta]
$
To further reduce the value of dmin the optical path of the light is increased by immersing the objective of the microscope in to a bath containing oil of refractive index $n$.
$
\mathrm{d}_{\min }=\frac{1.22 \lambda}{2 n \sin \beta}
$
Such an objective is called oil immersed objective. The term $\mathrm{n} \sin \mathrm{p}$ is called numerical aperture NA. $\mathrm{d}_{\min }=\frac{1.22 \lambda}{2(N A)}$
Question 26.
Discuss about astronomical telescope.
Answer:
Astronomical telescope:
An astronomical telescope is used to get the magnification of distant astronomical objects like stars, planets, moon etc. the image formed by astronomical telescope will be inverted. It has an objective of long focal length and a much larger aperture than the eyepiece. Light from a distant object enters the objective and a real image is formed in the tube at its second focal point. The eyepiece magnifies this image producing a final inverted image.

Magnification of astronomical telescope:
The magnification $\mathrm{m}$ is the ratio of the angle $\beta$ subtended at the eye by the final image to the angle a which the object subtends at the lens or the eye
$
\mathrm{m}=\frac{\beta}{\alpha}
$
From the diagram, $\mathrm{m}=\frac{h / f_c}{h / f_0}$
$
\mathrm{m}=\frac{f_0}{f_e}
$
The length of the telescope is approximately, $L=f_0+f_e$
Question 27.
Mention different parts of spectrometer and explain the preliminary adjustments. Spectrometer:
Answer:
The spectrometer is an optical instrument used to study the spectra of different sources of light and to measure the refractive indices of materials. It consists of basically three parts. They are (i) collimator, (ii) prism table and (iii) Telescope.
Adjustments of the spectrometer:
The following adjustments must be made before doing the experiment using spectrometer.
(a) Adjustment of the eyepiece:
The telescope is turned towards an illuminated surface and the eyepiece is moved to and fro until the cross wires are clearly seen.
(b) Adjustment of the telescope:
The telescope is adjusted to receive parallel rays by turning it towards a distant object and adjusting the distance between the objective lens and the eyepiece to get a clear image on the cross wire.
(c) Adjustment of the collimator:
The telescope is brought along the axial line with the collimator. The slit of the collimator is illuminated by a source of light. The distance between the slit and the lens of the collimator is adjusted until a clear image of the slit is seen at the cross wire of the telescope. Since the telescope is already adjusted for
parallel rays, a well-defined image of the slit can be formed, only when the light rays emerging from the collimator are parallel.
(d) Levelling the prism table:
The prism table is adjusted or levelled to be in horizontal position by means of levelling screws and a spirit level.

Question 28.
Explain the experimental determination of material of the prism using spectrometer. Determination of refractive index of material of the prism:
Answer:
The preliminary adjustments of the telescope, collimator and the prism table of the spectrometer are made. The refractive index of the prism can be determined by knowing the angle of the prism and the angle of minimum deviation.
(i) Angle of the prism (A):
The prism is placed on the prism table with its refracting edge facing the collimator. The slit is illuminated by a sodium light (monochromotic light). The parallel rays coming from the collimator fall on the two faces $\mathrm{AB}$ and $\mathrm{AC}$. The telescope is rotated to the position $\mathrm{T}_1$ until the image of the slit formed by the reflection at the face $A B$ is made to coincide with the vertical cross wire of the telescope.

The readings of the verniers are noted. The telescope is then rotated to the position $T_2$ where the image of the slit formed by the reflection at the face $\mathrm{AC}$ coincides with the vertical cross wire. The readings are again noted.
The difference between these two readings gives the angle rotated by the telescope, which is twice the angle of the prism. Half of this value gives the angle of the prism A.
(ii) Angle of minimum deviation (D):
The prism is placed on the prism table so that the light from the collimator falls on a refracting face, and the refracted image is observed through the telescope. The prism table is now rotated so that the angle of deviation decreases. A stage comes when the image stops for a moment and if we rotate the prism table further in the same direction, the image is seen to recede and the angle of deviation increases. The vertical cross wire of the telescope is made to coincide with the image of the slit where it turns back. This gives the minimum deviation position.

The readings of the verniers are noted. Now, the prism is removed and the telescope is turned to receive the direct ray and the vertical cross wire is made to coincide with the image. The readings of the verniers are noted. The difference between the two readings gives the angle of minimum deviation $\mathrm{D}$. The refractive index of the material of the prism $\mathrm{n}$ is calculated using the formula, $\mathrm{n}=\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
The refractive index of a liquid may be determined in the same way using a hollow glass prism filled with the given liquid.