Additional Questions - Chapter 6 Optics 12th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Questions
Multiple Choice Questions
Question 1.
When a ray of light enters a glass slab from air
(b) its wavelength increases
(c) its frequency increases
(d) neither its wavelength nor its frequency changes
Answer:
(a) its wavelength decreases
Hint:
Wavelength, $\lambda=\frac{\text { Velocity }}{\text { Frequency }}=\frac{u}{v}$
When light travels from air to glass, frequency $\mathrm{v}$ remains unchanged, velocity $\mathrm{u}$ decreases and hence wavelength $\mathrm{X}$ also decreases.
Question 2.
A source emits sound of frequency $600 \mathrm{~Hz}$ inside water. The frequency heard in air (velocity of sound in water $=1500 \mathrm{~m} / \mathrm{s}$, velocity of sound in air $=300 \mathrm{~m} / \mathrm{s}$ ) will be
(a) $300 \mathrm{~Hz}$
(b) $120 \mathrm{~Hz}$
(c) $600 \mathrm{~Hz}$
(d) $6000 \mathrm{~Hz}$
Answer:
(c) $600 \mathrm{~Hz}$
Hint:
Frequency does not change when sound travels from one medium to another
$\therefore$ Frequency of sound in air $=$ Frequency of sound in water $=600 \mathrm{~Hz}$
Question 3.
Two beams of red and violet colours are made to pass separately through a prism (angle of the prism is $60^{\circ}$ ). In the position of minimum deviation, the angle of refraction will be
(a) $30^{\circ}$ for both the colours
(b) greater for the violet colour
(c) greater for the violet colour
(d) equal but not $30^{\circ}$ for both the colours
Answer:
(a) $30^{\circ}$ for both the colours

Hint:
For any prism, $r_1=r_2=A$
In the position of minimum deviation for any wavelength,
$
\mathrm{r}_1=\mathrm{r}_2=\frac{A}{2}=\frac{60^{\circ}}{\theta}=30^{\circ}
$
Question 4.
To get three images of a single object, one should have two plane mirrors at an angle of
(a) $60^{\circ}$
(b) $90^{\circ}$
(c) $120^{\circ}$
(d) $30^{\circ}$
Answer:
(b) $90^{\circ}$
Hint:
The number of images formed,
$
\mathrm{n}=\frac{360^{\circ}}{\theta}-1 \text { or } 3=\frac{360^{\circ}}{\theta}-1 \text { or } \theta=90^{\circ}
$
Question 5.
Which of the following is used in optical fibres?
(a) Total internal reflection
(b) Diffraction
(c) Refraction
(d) Scattering
Answer:
(a) Total internal reflection
Hint:
The working of optical fibres is based on total internal reflection.
Question 6.
Two lenses of power $-15 \mathrm{D}$ and $+15 \mathrm{D}$ are in contact with each other. The focal length of the combination is
(a) $+10 \mathrm{~cm}$
(b) $-20 \mathrm{~cm}$
(c) $-10 \mathrm{~cm}$
(d) $+20 \mathrm{~cm}$
Answer:
(c) $-10 \mathrm{~cm}$
Hint $\mathrm{P}=\mathrm{P}_1+\mathrm{P}_2=-15+5=-10 \mathrm{D}$
$
\mathrm{F}=\frac{1}{P}=\frac{1}{-10} \mathrm{~m}=-10 \mathrm{~cm}
$

Question 7.
The refractive index of glass is 1.520 for red light and 1.525 for blue light. Let $\delta_1$ and $\delta_2$ be angles of minimum deviation for red and blue light respectively in a prism of this glass, then (a) $\delta_1$, can be less than or greater than $\delta_2$ depending upon the values of $\delta_1$ and $\delta_2$
(b) $\delta_1>\delta_2$
(c) $\delta_1<\delta_2$
(d) $\delta_1=\delta_2$
Answer:
(c) $\delta_1<\delta_2$
Hint:
$
\begin{aligned}
& \delta_1=\left(\mu_{\mathrm{R}}-1\right) \mathrm{A}, \delta_2=\left(\mu_{\mathrm{B}}-1\right) \mathrm{A} \\
& \text { As, } \mu_{\mathrm{R}}<\mu_{\mathrm{B}} \therefore \delta_1<\delta_2
\end{aligned}
$
Question 8.
Time image formed by an objective of a compound microscope is
(a) virtual and diminished
(b) real and diminished
(c) real and enlarged
(d) virtual and enlarged
Answer:
(c) real and enlarged
Hint
The image formed by the objective of a compound microscope is real and enlarged.
Question 9.
An astronomical telescope has a large aperture to,
(a) reduce spherical aberration
(b) have high resolution
(c) increase span of observation
(d) have low dispersion
Answer:
(b) have high resolution

Question 10.
Two plane mirrors are inclined to each other at an angle of $60^{\circ}$. A point object is placed in between them. The total number of images produced by both the mirror is
(a) 2
(b) 4
(c) 5
(d) 6
Answer:
(c) 5
Hint:
Number of images formed, $\theta=\frac{360^{\circ}}{\theta}-1=\frac{360}{60}=1=5$.
Question 11.
A boy $1.5 \mathrm{~m}$ tall with his eye level at $1.38 \mathrm{~m}$ stands before a mirror fixed on a wall. The minimum length of mirror required to view the complete image of boy is
(a) $0.75 \mathrm{~m}$
(b) $0.06 \mathrm{~m}$
(c) $0.69 \mathrm{~m}$
(d) $0.12 \mathrm{~m}$
Answer:
(a) $0.75 \mathrm{~m}$
Hint
Minimum length of mirror required $\frac{1}{2} \times$ Height of boy $=\frac{1}{2} \times 1.5=0.75 \mathrm{~m}$.
Question 12.
A pencil of light rays falls on a plane mirror and forms a real image, so the incident rays are
(a) parallel
(b) diverging
(c) converging
(d) statement is false
Answer:
(c) converging
Hint:
When converging rays fall on a plane mirror, they get reflected to a point $d$ in front of the mirror forming a real image.

Question 13.
For a real object, which of the following can produce a real image?
(a) plane mirror
(b) concave lens
(c) convex lens
(d) concave mirror
Answer:
(d) concave mirror
Hint:
Only concave mirror produces real image provided the object is not placed between its focus and pole.
Question 14.
Which mirror is to be used to obtain a parallel beam of light from a small lamp?
(a) Plane mirror
(b) Convex mirror
(c) Concave mirror
(d) None of the above
Answer:
(c) Concave mirror
Hint:
When the small lamp is placed at the focus of the concave mirror, the reflected light is a parallel beam.
Question 15.
When a plane electromagnetic wave enters a glass slab, then which of the following will not change?
(a) Wavelength
(b) Frequency
(c) Speed
(d) Amplitude
Answer:
(b) Frequency
Only the frequency of the electromagnetic wave remains unchanged.
Question 16.
If wavelength of light in air is $2400 \times 10^{-10} \mathrm{~m}$, then what will be the wavelength of light in glass $(\mu=$ $1.5) ?$
(a) $1600 \AA$
(b) $7200 \AA$
(c) $1080 \AA$

(d) None of these
Answer:
(a) $1600 \AA$
Hint:
$
\mu=\frac{\lambda_a}{\lambda_g} \Rightarrow \lambda_\lambda \mathrm{g}=\frac{\lambda_a}{\mu}=\frac{2400 \times 10^{-10} \mathrm{~m}}{1.5}=1600 \AA
$
Question 17.
Why is refractive index in a transparent medium greater than one?
(a) Because the speed of light in vacuum is always less than speed in a transparent medium.
(b) Because the speed of light in vacuum is always greater than the speed in a transparent medium.
(c) Frequency of wave changes when it Gasses medium.
(d) None of the above.
Answer:
(b) Because the speed of light in vacuum is always greater than the speed in a transparent medium. Hint:
$
\mu=\frac{\text { Speed of light in vacuum }}{\text { Speed of light in medium }}=\frac{c}{v}
$
As $\mathrm{c}>\mathrm{v}, \mu>1$
Question 18.
The wavelength of sodium light in air is $5890 \AA$. The velocity of light in air is $3 \times 10^8 \mathrm{~ms}^{-1}$. The wavelength of light in a glass of refractive index 1.6 would be close to
(a) $5890 \AA$
(b) $3681 \AA$
(c) $9424 \AA$
(d) $15078 \AA$
Answer:
(b) $3681 \AA$
Hint:
$
\mu=\frac{\lambda_a}{\lambda_g} \Rightarrow \lambda_{\mathrm{g}}=\frac{\lambda_a}{\mu}=\frac{5890 \AA}{1.6}=3681 \AA
$
Question 19.
A glass slab $(\mu=1.5)$ of thickness $6 \mathrm{~cm}$ is placed over a paper. Qhat is the shift in the letters?
(a) $4 \mathrm{~cm}$
(b) $2 \mathrm{~cm}$
(c) $1 \mathrm{~cm}$
(d) None of these
Answer:
(b) $2 \mathrm{~cm}$
Hint:

Normal shift, $\mathrm{x}=t\left(1-\frac{1}{\mu}\right)=6\left(1-\frac{1}{1.5}\right) \quad \mathrm{cm}=2 \mathrm{~cm}$
Question 20.
Light traveling from a transparent medium to air undergoes total internal reflection at an angle of incident of $45^{\circ}$. Then refractive index of the medium may be
(a) 1.5
(b) 1.3
(c) 1.1
(d) $\frac{1}{\sqrt{ } 2}$
Answer:
(a) 1.5
Hint:
$
\mu=\frac{1}{\sin i_c}=\frac{1}{\sin 45^{\circ}}=\sqrt{ } 2=1.414 \approx 1.5 \text {. }
$
Question 21.
A point source of light is placed $4 \mathrm{~m}$ below the surface of water of refractive index $5 / 3$. The minimum diameter of a disc which should be placed over the source, on the surface of water to cut-off all light coming out of water is
(a) infinite
(b) $6 \mathrm{~cm}$
(c) $4 \mathrm{~cm}$
(d) $3 \mathrm{~cm}$
Answer:
(b) $6 \mathrm{~cm}$
Hint:
$
\mathrm{r}=\frac{h}{\sqrt{\mu^2-1}}=\frac{4}{\sqrt{\left(\frac{5}{3}\right)^2-1}}=3 \mathrm{~cm}
$
Question 22.
In optical fibres, propagation of light is due to
(a) diffraction
(b) total internal reflection
(c) reflection
(d) refraction
Answer:
(b) total internal reflection
Hint:

In optical fiberes, light propagates due to tatal internal reflection.
Question 23 .
Sparkling of diamond is due to
(a) reflection
(b) dispersion
(c) total internal reflection
(d) high refractive index of diamond
Answer:
(c) total internal reflection
Question 24 .
For a given lens, the magnification was found to be twice as large as when the object was $0.15 \mathrm{~m}$ distant from it as when the distance was $0.2 \mathrm{~m}$. The focal length of the lens is
(a) $1.5 \mathrm{~m}$
(b) $0.20 \mathrm{~m}$
(c) $0.10 \mathrm{~m}$
(d) $0.05 \mathrm{~m}$
Asnwer:
(c) $0.10 \mathrm{~m}$
Hint:
Here $\mathrm{m}_1=2 \mathrm{~m}_2$
$\frac{f}{f-0.15}=2 \frac{f}{f-0.20}$
$2 \mathrm{f}-0.30=\mathrm{f}-0.20 ; \mathrm{f}=0.10 \mathrm{~m}$

Question 25.
Two lenses of focal lengths $f_1$ and $f_2$ are kept in contact coaxially. The resultant power of combination will be
(a) $\frac{f_1 f_2}{f_1-f_2}$
(b) $\frac{f_1+f_2}{f_1 f_2}$
(c) $f_1+f_2$
(d) $\frac{f_1}{f_2}+\frac{f_2}{f_1}$
Answer:
(b) $\frac{f_1+f_2}{f_1 f_2}$
Hint:
$
\mathbf{P}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{f_1+f_2}{f_1 f_2}
$
Question 26.
Two lenses of power 3D and -ID are kept in contact. What is focal length and nature of combined lens?
(a) $50 \mathrm{~cm}$, convex
(b) $200 \mathrm{~cm}$, convex
(c) $50 \mathrm{~cm}$, concave
(d) $200 \mathrm{~cm}$, concave
Answer:
(a) $50 \mathrm{~cm}$, convex
Hint:
$
\begin{aligned}
& \mathrm{P}=\mathrm{P}_1+\mathrm{P}_2=3-1=2 \mathrm{D} \\
& \mathrm{F}=\frac{1}{P}=\frac{1}{2} \mathrm{~m}=50 \mathrm{~cm}
\end{aligned}
$
Question 27.
If two thin lenses are kept coaxially together, then their power is proportional $\left(R_1, R_2\right)$ being the radii of curved surfaces) to
(a) $R_1+R_2$
(b) $\left[\frac{\mathbf{R}_1+\mathbf{R}_2}{\mathbf{R}_1 \mathbf{R}_2}\right]$
(c) $\left[\frac{\mathrm{R}_1 \mathrm{R}_2}{\mathrm{R}_1 \mathrm{R}_2}\right]$
(d) None of these
Answer:
(b) $\left[\frac{\mathbf{R}_1+\mathbf{R}_2}{\mathbf{R}_1 \mathbf{R}_2}\right]$
Hint:

Question 25.
Two lenses of focal lengths $f_1$ and $f_2$ are kept in contact coaxially. The resultant power of combination will be
(a) $\frac{f_1 f_2}{f_1-f_2}$
(b) $\frac{f_1+f_2}{f_1 f_2}$
(c) $f_1+f_2$
(d) $\frac{f_1}{f_2}+\frac{f_2}{f_1}$
Answer:
(b) $\frac{f_1+f_2}{f_1 f_2}$
Hint:
$
\mathbf{P}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{f_1+f_2}{f_1 f_2}
$
Question 26.
Two lenses of power 3D and -ID are kept in contact. What is focal length and nature of combined lens?
(a) $50 \mathrm{~cm}$, convex
(b) $200 \mathrm{~cm}$, convex
(c) $50 \mathrm{~cm}$, concave
(d) $200 \mathrm{~cm}$, concave
Answer:
(a) $50 \mathrm{~cm}$, convex
Hint:
$
\begin{aligned}
& \mathrm{P}=\mathrm{P}_1+\mathrm{P}_2=3-1=2 \mathrm{D} \\
& \mathrm{F}=\frac{1}{P}=\frac{1}{2} \mathrm{~m}=50 \mathrm{~cm}
\end{aligned}
$
Question 27.
If two thin lenses are kept coaxially together, then their power is proportional $\left(R_1, R_2\right)$ being the radii of curved surfaces) to
(a) $R_1+R_2$
(b) $\left[\frac{\mathbf{R}_1+\mathbf{R}_2}{\mathbf{R}_1 \mathbf{R}_2}\right]$
(c) $\left[\frac{\mathrm{R}_1 \mathrm{R}_2}{\mathrm{R}_1 \mathrm{R}_2}\right]$
(d) None of these
Answer:
(b) $\left[\frac{\mathbf{R}_1+\mathbf{R}_2}{\mathbf{R}_1 \mathbf{R}_2}\right]$
Hint:

Question 31.
The sky would appear red instead of blue if
(a) atmospheric particles scatter blue light more than red light
(b) atmospheric particles scatter all colours equally
(c) atmospheric particle scatter red light more than blue light
(d) the sun was much hotter
Answer:
(c) atmospheric particle scatter red light more than blue light
Question 32.
A setting sun appears to be at an altitude higher than it really is. This is because of
(a) absorption of light
(b) reflection of light
(c) refraction of light
(d) dispersion of light
Answer:
(c) refraction of light
Hint:
This is due to refraction of light by the earth's atmosphere.
Question 33.
The reddish appearance of rising and setting sun is due to
(a) reflection of light
(b) diffraction of light
(c) scattering of light
(d) interference of light
Answer:
(c) scattering of light
Hint:
The reddish appearance of the rising and the setting sun is due to scattering of light.
Question 34.
In the formation of a rainbow, the light from the sun on water droplets undergoes
(a) dispersion only
(b) only total internal reflection
(c) dispersion and total internal reflection
(d) none of the above
Answer:
(c) dispersion and total internal reflection

Hint:
Rainow is formed due to dispersion of sunlight by raindrops which also deviate the colours by total internal reflection.
Question 35.
The angular magnification of a simple microscope can be increased by increasing
(a) focal length of lens
(b) size of object
(c) aperture of lens
(d) power of lens
Answer:
(b) size of object
Question 36.
For compound microscope $\mathrm{f}_0=1 \mathrm{~cm}, \mathrm{f}_{\mathrm{e}}=2.5 \mathrm{~cm}$. An object is placed at distance $1.2 \mathrm{~cm}$ from objective lens. What should be length of microscope for normal adjustment?
(a) $8.5 \mathrm{~cm}$
(b) $8.3 \mathrm{~cm}$
(c) $6.5 \mathrm{~cm}$
(d) $6.3 \mathrm{~cm}$
Answer:
(a) $8.5 \mathrm{~cm}$
Hint:
In the normal adjustment of a compound microscope, $\mathrm{L}=v_0+v_{\mathrm{e}}=\frac{v_0 f_e}{v_0+f_{\mathrm{e}}}+\mathrm{f}_{\mathrm{e}}=\frac{1.2 \times 1}{-1.2+1}+2.5=6+25$ $=8.5 \mathrm{~cm}$
Question 37.
Magnifying power of an astronomical telescope for normal vision with usual notation is
(a) $-\mathrm{f}_0 / \mathrm{f}_{\mathrm{e}}$
(b) $-f_0 \times f_e$
(c) $-f_0 / f_0$
(d) $-\mathrm{f}_0+\mathrm{f}_{\mathrm{e}}$
Answer:
(a) $-\mathrm{f}_0 / \mathrm{f}_{\mathrm{e}}$
Hint:
In normal adjustment of the telescope, $\mathrm{m}=-\mathrm{f}_0 / \mathrm{f}_{\mathrm{e}}$

Question 38.
$F_1$ and $F_2$ are focal length of objective and eyepiece respectively of the telescope. The angular magnification for the given telescope is equal to
(a) $\frac{F_1}{F_2}$
(b) $\frac{F_2}{F_1}$
(c) $\frac{F_1 F_2}{F_1+F_2}$
(d) $\frac{F_1+F_2}{F_1 F_2}$
Answer:
(a) $\frac{F_1}{F_2}$
Hint:
In normal adjustment of the telescope, $|m|=\frac{f_0}{f_e}=\frac{F_1}{F_2}$
Question 39.
Focal length of objective and eyepiece of telescope are $200 \mathrm{~cm}$ and $4 \mathrm{~cm}$ respectively. What is length of telescope for normal adjustment?
(a) $196 \mathrm{~cm}$
(b) $204 \mathrm{~cm}$
(c) $250 \mathrm{~cm}$
(d) $225 \mathrm{~cm}$
Answer:
(b) $204 \mathrm{~cm}$
Hint $\mathrm{L}=-f_0+f_e=200+4=204 \mathrm{~cm}$
Question 40.
For normal vision, what is minimum distance of object from eye?
(a) $30 \mathrm{~cm}$
(b) $25 \mathrm{~cm}$
(c) Infinite
(d) $40 \mathrm{~cm}$
Answer:
(b) $25 \mathrm{~cm}$
Hint:
For normal eye, the least distance of distinct vision is $25 \mathrm{~cm}$.

Question 41 .
The focal length of the objective and eyepiece of a telescope are respectively $100 \mathrm{~cm}$ and $2 \mathrm{~cm}$. The moon subtends angle of $0.5^{\circ}$; the angle subtended by the moon's image will be
(a) $10^{\circ}$
(b) 250
(e) $100^{\circ}$
(d) $75^{\circ}$
Answer:
b) 250
Hint ;
$
\mathrm{m}=\frac{\beta}{\alpha} \beta=\frac{f_0}{f_e} ; \alpha=\frac{100}{2} \times 0.5^{\circ}=25^{\circ}
$
Question 42.
A person cannot clearly see distance more than $40 \mathrm{~cm}$. He is advised to use lens of power,
(a) $-2.5 \mathrm{D}$
(b) $2.5 \mathrm{D}$
(c) $-6.25 \mathrm{D}$
(d) $1.5 \mathrm{D}$
Answer:
(a) $-2.5 \mathrm{D}$
Hint;
For the remedial lens, $\mathrm{u}=\infty$,
$
\begin{aligned}
& \mathrm{v}=-40 \mathrm{~cm}=-0.40 \mathrm{~m} \\
& \therefore \frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{-0.40}-\frac{1}{\infty}=-2.5 \Rightarrow \mathrm{P}=2.5 \mathrm{D}
\end{aligned}
$
Question 43.
The light gathering power of a camera lens depends on
(a) its diameter only
(b) ratio of diameter and focal length
(c) product of focal length and diameter
(d) wavelength of the light used
Answer:
(a) its diameter only
Hint:
The light gathering power of a camera lens is proportional to its area or to the square of its diameter.
Question 44
Amount of light entering into the camera depends upon
(a) focal length of objective lens
(b) product of focal length and diameter of the objective lens

(c) distance of object from camera
(d) aperture setting of the camera
Answer:
(d) aperture setting of the camera
Hint:
The amount of light entering into the camera depends upon the aperture setting of the camera.
Question 45.
Line spectrum can be obtained from
(a) sun
(b) candle
(c) mercury vapour lamp
(d) electric bulb
Answer:
(c) mercury vapour lamp
Question 46.
The Production of band spectra is caused by
(a) atomic nuclei
(b) hot metals
(c) molecules
(d) electrons
Answer:
(c) molecules
Question 47.
Two mirrors are kept at $60^{\circ}$ to each other and a body is placed at middle. The total number of images formed is
(a) $\operatorname{six}$
(b) four
(c) five
(d) three
Answer:
(a) $\operatorname{six}$
Hint:
Number of images formed, $\mathrm{n}=\frac{360}{\theta}-1=\frac{360}{60}-1=5$
Question 48.
A point source kept at a distance of $1000 \mathrm{~m}$ has a illumination I. To change the illumination to 161 , the new distance should become
(a) $250 \mathrm{~m}$
(b) $500 \mathrm{~m}$
(c) $750 \mathrm{~m}$
(d) $800 \mathrm{~m}$

Answer:
(a) $250 \mathrm{~m}$
Hint:
$
\begin{aligned}
& \frac{I_2}{I_1}=\left(\frac{r_1}{r_2}\right)^2 \\
& \frac{16 \mathrm{I}_1}{\mathrm{I}_1}=\frac{(1000)^2}{r_2^2} ; \mathrm{r}_2 \frac{1000}{4}=250 \mathrm{~cm}
\end{aligned}
$
Question 49.
A concave mirror of focal length $15 \mathrm{~cm}$ forms an image having twice the linear dimensions of the object. The position of the object, when the image is virtual will be
(a) $22.5 \mathrm{~cm}$
(b) $7.5 \mathrm{~cm}$
(c) $30 \mathrm{~cm}$
(d) $45 \mathrm{~cm}$
Answer:
(b) $7.5 \mathrm{~cm}$
Hint:
For virtual images, $\mathrm{m}=\frac{-v}{u}=+2$ or $\mathrm{v}=-2 \mathrm{u}$
As $\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$
$
\therefore \frac{1}{u}-\frac{1}{2 u}=\frac{1}{-15} \text { or } \frac{1}{2 u}=\frac{1}{-15} \Rightarrow \mathrm{u}=-7.5 \mathrm{~cm}
$
Question 50.
when a ray of light enters a glass slab, then
(a) its frequency and velocity change
(b) only frequency changes
(c) its frequency and wavelength change
(d) its frequency does not change
Answer:
(d) its frequency does not change
Hint:
When a ray of light enters a glass slab, its velocity and wavelength change while frequency does not change.

Question 51.
A light wave of frequency $u$ and wavelength A travels from air to glass. Then,
(a) y changes
(b) y does not change, A changes
(c) A does not change
(d) $y$ and A change
Answer:
(b) y does not change, A changes
Hint:
Same reasoning as in the above question.
Question 52.
In refraction, light waves are bent on passing from one medium to the second medium, because in the second medium.
(a) the frequency is different
(b) the coefficient of elasticity is different
(c) the speed is different
(d) the amplitude is smaller
Answer:
(c) the speed is different
Hint:
Speed of light in second medium is different than that in first medium.
Question 53.
A ray of light having wavelength $720 \mathrm{~nm}$ enters in a glass of refractive index 1.5 . The wavelength of the ray within the glass will be
(a) $360 \mathrm{~nm}$
(b) $480 \mathrm{~nm}$
(e) $720 \mathrm{~nm}$
(d) $1080 \mathrm{~nm}$
Answer:
(b) $480 \mathrm{~nm}$
Hint:
$
\lambda_g=\frac{\lambda_0}{\mu}=\frac{790 \mathrm{~nm}}{1.5}=480 \mathrm{~nm}
$

Question 54.
Brilliance of a diamond is due to
(a) shape
(b) cutting
(c) reflection
(d) total internal reflection
Answer:
(d) total internal reflection
Hint:
Brilliance of a diamond is due to total internal reflection of light.
Question 55.
An endoscope is employed by a physician to view the internal parts of a body organ. If is based on the principle of
(a) refraction
(b) reflection
(c) total internal reflection
(d) dispersion
Answer:
(c) total internal reflection
Hint:
An endoscope is made of optical fibres which work on the principle of total internal reflection.
Question 56.
'Mirage' is a phenomenon due to
(a) reflection of light
(b) refraction of light
(c) total internal reflection of light
(d) diffraction of light Hint Mirage occurs due to total internal reflection of light.
Answer:
(c) total internal reflection of light

Question 57.
Two lenses of power $+12 \mathrm{D}$ and $-2 \mathrm{D}$ are combined together. What is their equivalent focal length?
(a) $10 \mathrm{~cm}$
(b) $12.5 \mathrm{~cm}$
(c) $16.6 \mathrm{~cm}$
(d) $8.33 \mathrm{~cm}$
Answer:
(a) $10 \mathrm{~cm}$
Hint:
$
\begin{aligned}
& \mathrm{P}=\mathrm{P}_1+\mathrm{P}_2=+12-2=10 \mathrm{D} \\
& \mathrm{F}=\frac{1}{P}=\frac{1}{10} \mathrm{~m}=10 \mathrm{~cm}
\end{aligned}
$
Question 58 .
If two lenses of power $+1.5 \mathrm{D}$ and $+1.0 \mathrm{D}$ are placed in contact, then the effective power of combination will be
(a) $2.5 \mathrm{D}$
(b) $1.5 \mathrm{D}$
(c) $0.5 \mathrm{D}$
(d) $3.25 \mathrm{D}$
Answer:
(a) $2.5 \mathrm{D}$
Hint:
$
\mathrm{P}=\mathrm{P}_1+\mathrm{P}_2=+1.5+1.0=+2.5 \mathrm{D}
$
Question 59.
The angle of a prism is $6^{\circ}$ and its refractive index for green light is 1.5 . If a green ray passes through it, the deviation will be
(a) $30^{\circ}$
(b) $15^{\circ}$
(c) $3^{\circ}$
(d) $0^{\circ}$
Answer:
(c) $3^{\circ}$
Hint:
$
5=(\mathrm{p}-1) \mathrm{A}=(1.5-1) \times 6=3^{\circ}
$
Question 60 .
Sky appears to be blue in clear atmosphere due to light's
(a) diffraction
(b) dispersion
(c) scattering
(d) polarisation
Answer:
(c) scattering

Hint:
Sky appears blue due to scattering of light by atmospheric modecules.
Question 61.
One can not see through fog, because
(a) fog absorbs the light
(b) light suffers total reflection at droplets
(c) refractive index of the fog is infinity
(d) light is scattered by the droplets
Answer:
(d) light is scattered by the droplets
Question 62.
Fraunhofer lines of the solar system is an example of
(a) emission lines spectrum
(b) emission band spectrum
(c) continuous emission spectrum
(d) line absorption spectrum
Answer:
(d) line absorption spectrum
Hint:
Fraunhofer lines is an example of line absorption spectrum.
Question 63.
A person using a lens as a sample microscope sees an
(a) inverted virtual image
(b) inverted real magnified image
(c) upright virtual image
(d) upright real magnified image
Answer:
(d) upright real magnified image
Hint:
A person sees an upright virtual image in a simple microscope.

Question 64.
Four lenses of focal length $+10 \mathrm{~cm},+50 \mathrm{~cm},+100 \mathrm{~cm}$ and $+200 \mathrm{~cm}$ are available for making an astronomical telescope. To produce the largest magnification, the focal length of the eyepiece should be
(a) $+10 \mathrm{~cm}$
(b) $+50 \mathrm{~cm}$
(c) $+100 \mathrm{~cm}$
(d) $+200 \mathrm{~cm}$
Answer:
(a) $+10 \mathrm{~cm}$
Hint:
To produce the largest magnification, the eyepiece should have minimum focal length.
Question 65.
The camera lens has an aperture of $f$ and the exposure time is $1 / 60 \mathrm{~s}$. What will be the new exposure time if the aperture become $1.4 \mathrm{f}$ ?
(a) $\frac{1}{42} \mathrm{~s}$
(b) $\frac{1}{56} \mathrm{~s}$
(c) $\frac{1}{72} \mathrm{~s}$
(d) $\frac{1}{31} \mathrm{~s}$
Answer:
(d) $\frac{1}{31} \mathrm{~s}$
Hint:
Time of exposure $\propto(\mathrm{f}-\text { number })^2$
$
\frac{t}{\left(\frac{1}{60}\right)}=\left(\frac{1.4}{1}\right)^2 \Rightarrow t=\frac{1.4 \times 1.4}{60} \approx \frac{1}{31} \mathrm{~s}
$
Question 66 .
For a person near point of vision is $100 \mathrm{~cm}$. Then the power of lens he must wear so as have normal vision, should be
(a) $+\mathrm{ID}$
(b) $-\mathrm{ID}$
(c) $+3 \mathrm{D}$
(d) $-3 \mathrm{D}$
Answer:
(c) $+3 \mathrm{D}$

Hint:
$
\mathrm{f}=\frac{y D}{y-D}=\frac{100 \times 25}{100-25}=\frac{100}{3} \mathrm{~cm}=\frac{1}{3} \mathrm{~cm} ; \mathrm{P}=\frac{1}{f}=+3 \mathrm{~d}
$
Question 67.
Ray optics is valid, when characteristic dimension ions are
(a) much smaller than the wavelength of light
(b) much larger than the wavelength of light
(c) of the same order as the wavelength of light
(d) of the order of one millimetre

Answer:

(b) much larger than the wavelength of light
Hint:
Ray optics is valid, when characteristic dimensions are much larger than the wavelength of light.
Question 68.
A tall man of height 6 feet, want to see his full image. Then required minimum length of the mirrorwillbe
(a) 12 feet
(b) 3 feet
(e) 6 feet
(d) any length

Answer:

(b) 3 feet
Hint:
inimum height of mirror required for seeing full image Height of the man $=3$ feet.
Question 69.
The refractive index of water is 1.33 . What will be the speed of light in water?
(a) $3 \times 10^8 \mathrm{~ms}^{-1}$
(b) $2.26 \times 10^8 \mathrm{~ms}^{-1}$
(c) $4 \times 10^8 \mathrm{~ms}^{-1}$
(d) $1.33 \times 10^8 \mathrm{~ms}^{-1}$
Answer:
(b) $2.26 \times 10^8 \mathrm{~ms}^{-1}$
Hint:
As, $\mu=\frac{c}{v} \Rightarrow \mathrm{v}=\frac{c}{\mu}=\frac{3 \times 10^8}{1.33}=2.26 \times 10^8 \mathrm{~ms}^{-1}$

Question 70.
A beam of monochromatic light is refracted from vacuum into a medium of refractive index 1.5. The wavelength of refracted light will be
(a) same
(b) dependent on intensity of refracted light
(c) larger
(d) smaller
Answer:
(d) smaller
Hint:
As light enters the medium, its wavelength decreases and becomes equal to $\frac{\lambda}{\mu}$.
Question 71.
Optical fibers are based on
(a) total internal reflection
(b) less scattering
(c) refraGtion
(d) less absorption coefficient
Answer:
(a) total internal reflection
Question 72.
A convex lens is dipped in a liquid, whose refractive index is equal to the refractive index of the lens.
Then, its focal length will
(a) become zero
(b) becomes infinite
(e) remain unchanged
(d) become small, but non-zero
Answer:
(b) becomes infinite
Hint:
$
\frac{1}{f_1}=\left(\frac{\mu_g}{\mu_e}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)=(1-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)=0 ; \mathrm{f}_1=\infty
$
Question 73.
A convex lens and a concave lens, each having same focal length of $25 \mathrm{~cm}$, are put in contact to form a combination of lenses. The power of the combination (in diopter) is
(a) zero
(b) 25
(c) 50
(d) infinite
Answer:
(a) zero

Hint:
$
\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{+25}+\frac{1}{-25}=0 ; \mathbf{P}=\frac{1}{F}=0
$
Question 74
The focal length of a converging lens is measured for violet, green and red colours. If is $f_V, f_G$ and $f_R$ respectively. We will get
(a) $f_V=f_G$
(b) $f_G=f_R$
(c) $f_{\mathrm{V}} (d) f $_{\mathrm{V}}>\mathrm{f}_{\mathrm{R}}$
Answer:
(c) $\mathrm{f}_{\mathrm{V}}<\mathrm{f}_{\mathrm{R}}$
Hint:
$
\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{\mathrm{R}_2}\right) \text { i.e., } \mathrm{f} \propto \frac{1}{\mu-1}
$
As $\mu_{\mathrm{V}}>\mu_{\mathrm{R}}$, so, $\mathrm{f}_{\mathrm{V}}<\mathrm{f}_{\mathrm{R}}$
Question 75.
Rainbow is formed due to combination of
(a) refraction and scattering
(b) refraction and absorption
(c) dispersion and total internal reflection
(d dispersion and focusing
Answer:
(c) dispersion and total internal reflection
Hint:
Rainbow is formed due to dispersion and total internal reflection of sunlight by raindrops.
Question 76.
The blue colour of the sky is due to the phenomenon of
(a) scattering
(b) dispersion
(c) reflection
(d) refraction
Answer:
(a) scattering
Hint:
The blue colour of the sky is due to the scattering of sunlight by atmospheric molecules.

Question 77.
An astronomical telescope often fold angular magnification has a length of $44 \mathrm{~cm}$. The focal length of the object is
(a) $4 \mathrm{~cm}$
(b) $40 \mathrm{~cm}$
(c) $44 \mathrm{~cm}$
(d) $440 \mathrm{~cm}$
Answer:
(b) $40 \mathrm{~cm}$
Hint:
Here, $\mathrm{f}_0+\mathrm{f}_{\mathrm{e}}=44 \mathrm{~cm}$
$
\mathrm{m}=\frac{f_0}{f_e} \text { (or) } \mathrm{f}_0=10 \mathrm{f}_{\mathrm{e}}
$
$\therefore 10 \mathrm{f}_{\mathrm{e}}+10 \mathrm{f}_{\mathrm{e}}=44 \mathrm{~cm}$ (or) $\mathrm{f}_{\mathrm{e}}=4 \mathrm{~cm}$
Hence, $\mathrm{f}_0=10 \times 4$ (or) $\mathrm{f}_0=40 \mathrm{~cm}$
Question 78.
Exposure time of a camera lens at the $\frac{f}{2.8}$ Setting is $\frac{1}{200}$ second. The correct time of exposure at $\frac{f}{5.6}$ is
(a) 0.20 second
(b) 0.40 second
(c) 0.02 second
(d) 0.04 second
Answer:
(c) 0.02 second
Hint:
Time of exposure, $t \propto(f-\text { number })^2$
$
\therefore \frac{t}{\left(\frac{1}{200}\right)}=\left(\frac{5.6}{2.8}\right)^2=4 \Rightarrow \mathrm{t}=0.02 \mathrm{~s}
$
Question 79.
Which of the following is not due to total internal reflection?
(a) Working of optical fibre
(b) Difference between apparent and real depth of a pond
(c) Mirage on hot summer day
(d) Brilliance of diamond
Answer:
(b) Difference between apparent and real depth of a pond

Hint:

Difference between apparent and real depth of a pond is due to refraction of light and the other three phenomena involve total internal reflection.
Question 80.
An object is placed at a distance of $0.5 \mathrm{~m}$ infront of a plane mirror. The distance between object and image will be
(a) $0.25 \mathrm{~m}$
(b) $0.5 \mathrm{~m}$
(c) $1.0 \mathrm{~m}$
(d) $2.0 \mathrm{~m}$
Answer:
(c) $1.0 \mathrm{~m}$
Hint:
Distance between object and image $=0.5+0.5=1.0 \mathrm{~m}$.
Question 81.
An observer moves towards a stationary plane mirror at a speed of $4 \mathrm{~ms}^{-1}$ with what speed - will his image move towards him?
(a) $2 \mathrm{~ms}^{-1}$
(b) $4 \mathrm{~ms}^{-1}$
(c) $8 \mathrm{~ms}^{-1}$
(d) the image will stay at rest
Answer:
(c) $8 \mathrm{~ms}^{-1}$
Hint:
Speed of the image towards the observer $=2 \times 4=8 \mathrm{~ms}^{-1}$
Question 82.
If two mirrors are kept at $60^{\circ}$ to each other and a body is placed at the middle, then total number of images formed is
(a) $\operatorname{six}$
(b) four
(c) five
(d) three
Answer:
(c) five
Hint:
Number of images formed $=\frac{360}{\theta}-1=\frac{360}{60}-1=5$

Question 83.
If an object is placed at $10 \mathrm{~cm}$ infront of a concave mirror of focal length $15 \mathrm{~cm}$. The magnification of image is
(a) -1.5
(b) 1.5
(c) -3
(d) 3
Answer:
d) 3
Hint:
$
\mathrm{m}=\frac{f}{f-u}=\frac{-15}{-15(-10)}=+3
$
Question 84.
An object of length $2.5 \mathrm{~cm}$ is placed at the principal axis of a concave mirror at a distance $1.5 \mathrm{f}$. The image height is
(a) $+5 \mathrm{~m}$
(b) $-5 \mathrm{~cm}$
(c) $-10 \mathrm{~cm}$
(d) $+1 \mathrm{~cm}$
Answer:
(b) $-5 \mathrm{~cm}$
Hint:
$
\mathrm{m}=\frac{f}{f-u}=\frac{f}{f-1.5 f}=-2 \mathrm{~cm}
$
Height image $=\mathrm{m} \times$ height of object $-2 \times 2.5=-5 \mathrm{~cm}$
Question 85.
Which of the following mirror is used by a dentist to examine a small cavity?
(a) Concave mirror
(b) Convex mirror
(c) Combination of (a) and (b)
(d) None of these
Answer:
(a) Concave mirror
Hint:
A concave mirror, because it forms erect and enlarged image when held closer to the cavity.

Question 86.
When a ray of light enters from one medium to another, then which of the following does not change?
(a) Frequency
(b) Wavelength
(c) Speed
(d) Amplitude
Answer:
(a) Frequency
Hint:
Only frequency remains unchanged.
Question 87.
When light travels from one medium to the other medium of which the refractive index is different, then which of the following will change?
(a) Frequency, wavelength and velocity
(6) Frequency and wavelength
(c) Frequency and velocity
(d) Wavelength and velocity
Answer:
(d) Wavelength and velocity
Hint:
Wavelength and velocity will change while frequency remains unchanged.
Question 88.
The time taken by the light to cross a glass of thickness $4 \mathrm{~mm}$ and refractive index $(\mu=3)$, will be
(a) $4 \times 10^{-11} \mathrm{sec}$
(b) $16 \times 10^{-11} \mathrm{sec}$
(c) $8 \times 10^{-11} \mathrm{sec}$
(d) $24 \times 10^{-10} \mathrm{sec}$
Answer:
(a) $4 \times 10^{-11} \mathrm{sec}$
Hint:
$
\begin{aligned}
& \text { Time, } t=\frac{\text { Thickness of glass }}{\text { Speed of light in glass }} \\
& \frac{d}{c / a}=\frac{\mu d}{c}=\frac{3 \times 4 \times 10^{-3}}{3 \times 10^8}=4 \times 10^{-11} \mathrm{~s}
\end{aligned}
$
Question 89.
The critical angle of a medium with respect to air is $45^{\circ}$. The refractive index of medium is
(a) 1.41
(b) 1.2
(c) 1.5
(d) 2

Answer:
(a) 1.41
Hint:
$
\mu=\frac{1}{\sin i_c}=\frac{1}{\sin 45^{\circ}}=\frac{1}{1 / \sqrt{ } 2} \approx 1.41
$
Question 90.
If the critical angle for total internal reflection from a medium to vacuum is $30^{\circ}$, then velocity of light in the medium is
(a) $6 \times 10^8 \mathrm{~m} / \mathrm{sec}$
(b) $2 \times 10^8 \mathrm{~m} / \mathrm{sec}$
(c) $3 \times 10^8 \mathrm{~m} / \mathrm{sec}$
(d) $1.5 \times 10^8 \mathrm{~m} / \mathrm{sec}$
Answer:
(c) $3 \times 10^8 \mathrm{~m} / \mathrm{sec}$
Hint:
$
\mu=\frac{1}{\sin i_c}=\frac{c}{v}
$
Question 91.
When a ray of light enter from one medium to another, its velocity is doubled. The critical angle for the ray for two internal reflection will be
(a) $30^{\circ}$
(b) $60^{\circ}$
(c) $90^{\circ}$
(d) Information is incomplete
Answer:
(a) $30^{\circ}$
Hint:
$
\mu=\frac{1}{\sin i_{\mathrm{c}}}=\frac{v_1}{v_2}=\frac{2 v_2}{v_2}=2 ; \sin \mathrm{i}_{\mathrm{c}}=\frac{1}{2} \therefore \mathrm{i}_{\mathrm{c}}=30^{\circ}
$
Question 92.
A driver at a depth $12 \mathrm{~m}$ inside water $(\mu=4 / 3)$ see the sky in a cone of semi-vertical angle is
(a) $\sin ^{-1}\left(\frac{4}{3}\right)$
(b) $\tan ^{-1}\left(\frac{4}{3}\right)$
(c) $\sin ^{-1}\left(\frac{3}{4}\right)$
(d) $90^{\circ}$

Answer:
(c) $\sin ^{-1}\left(\frac{3}{4}\right)$
Hint:
Required semi vertical angle $=$ Critical angle $i_{\mathrm{C}}=\sin ^{-1} \frac{1}{\mu}=\sin ^{-1}\left(\frac{3}{4}\right)$
Question 93.
The principle behind optical fibres is
(a) total internal reflection
(b) total external reflection
(c) both (a) and (b)
(d) diffraction
Answer:
(a) total internal reflection
Hint:
Optical fibres work on the principle of total internal reflection.
Question 94.
Air bubble in water behaves as
(a) some times concave, sometimes convex lens
(b) concave lens
(c) convex lens
(d) always refracting surface
Answer:
(b) concave lens
Question 95.
A convex lens of $40 \mathrm{~cm}$ focal length is combined with a concave lens of focal length $25 \mathrm{~cm}$. The power of combination is
(a) $-1.5 \mathrm{D}$
(b) $-6.5 \mathrm{D}$
(c) $+6.6 \mathrm{D}$
(d) $+6.5 \mathrm{D}$
Answer:
(a) $-1.5 \mathrm{D}$
Hint:
$
\mathbf{P}=\mathrm{P}_1+\mathrm{P}_2=\frac{1}{f_1}+\frac{1}{f_2}=\frac{100}{40}+\frac{100}{-25}=-1.5 \mathrm{~d}
$

Question 96.
Two thin lenses, one of focal length $+60 \mathrm{~cm}$ and the other of focal length $-20 \mathrm{~cm}$ are put in contact, the combined focal length is,
(a) $15 \mathrm{~cm}$
(b) $-15 \mathrm{~cm}$
(c) $-30 \mathrm{~cm}$
(d) $30 \mathrm{~cm}$
Answer:
(c) $-30 \mathrm{~cm}$
Hint:
$
\mathrm{F}=\frac{f_1 f_2}{f_1+f_2}=\frac{60 \times(-20)}{60-20}=30 \mathrm{~cm}
$
Question 97.
How does refractive index $(\mu)$ of a material vary with respect to wavelength ( $\lambda$ ). (A and B are constants).
(a) $\mu=\mathrm{A}+\frac{B}{\lambda^2}$
(b) $\mu=\mathrm{A}+\mathrm{B} \lambda^2$
(c) $\mu=\mathrm{A}+\frac{B}{\lambda}$
(d) $\mu=\mathrm{A}+\mathrm{B} \lambda$
Answer:
(a) $\mu=\mathrm{A}+\frac{B}{\lambda^2}$
Hint:
According to cauchy's relation, $\mu=\mathrm{A}+\frac{B}{\lambda^2}$
Question 98.
a prism of a refracting angle $60^{\circ}$ is made with a material of refractive index $p$. For a certain wavelength of light, the angle of minimum deviation is $30^{\circ}$. For this wavelength, the value of $p$ of material is
(a) 1.820
(b) 1.414
(c) 1.503
(d) 1.231
Answer:
(b) 1.414
Hint:
$
\mu=\frac{\sin \left(\frac{60^{\circ}+30^{\circ}}{2}\right)}{\sin 30^{\circ}}=\frac{\sin 45^{\circ}}{\sin 30^{\circ}}=\frac{1}{\sqrt{2}} \times \frac{2}{1} \sqrt{ } 2=1.414
$

Question 99.
Refractive index of red and violet light are 1.52 and 1.54 respectively. If the angle of prism is $10^{\circ}$, the angular dispersion will be
(a) $0.02^{\circ}$
(b) $0.20^{\circ}$
(c) $3.06^{\circ}$
(d) $30.6^{\circ}$
Answer:
(b) $0.20^{\circ}$
Hint:
Angular dispersion $=\delta_{\mathrm{V}}-\delta_{\mathrm{R}}=\mathrm{A}\left(\mu_{\mathrm{V}}-\mu_{\mathrm{V}}\right)=10^{\circ}(1.54-1.52)=0.20^{\circ}$
Question 100.
In a simple microscope, if the final image is located at $25 \mathrm{~cm}$ from the eye placed close to the lens, then magnifying power is
(a) $\frac{25}{f}$
(b) $1+\frac{25}{f}$
(c) $\frac{f}{25}$
(d) $\frac{f}{25}+1$
Answer:
(b) $1+\frac{25}{f}$
Hint:
When the final image is formed at the least distance of distinct vision in a simple microscope, $\mathrm{m}=1+\frac{25}{f}$
Question 101.
Magnification at least distance of distinct vision of a simple microscope of focal length $5 \mathrm{~cm}$ is
(a) 2
(b) 5
(c) 4
(d) 6
Answer:
(d) 6
Hint:
$
\mathrm{m}=1+\frac{25}{f}=1+\frac{25}{5}=6
$

Question 102.
Magnification of a compound microscope is 30 . Focal length of eyepiece is $5 \mathrm{~cm}$ and the image is formed at a distance of distinct vision of $25 \mathrm{~cm}$. The magnification of the objective lens is
(a) 6
(b) 5
(c) 7.5
(d) 10
Answer:
(b) 5
Hint:
$
\mathrm{m}_{\mathrm{e}}=\frac{v}{u}=\frac{D}{u_c}=1+\frac{D}{f_e}=1+\frac{25}{5}=6
$
For the compound microscope, $\mathrm{m}=\mathrm{m}_0 \times \mathrm{m}_{\mathrm{e}} \Rightarrow 30=\mathrm{m}_0 \times 6$ (or) $\mathrm{m}_0=5$
Question 103 .
The astronomical microscope consists of objective and eyepiece. The focal length of the objective is
(a) equal to that of the eyepiece
(b) shorter than that of the eyepiece
(c) greater than that of the eyepiece
(d) five times shorter than that of eyepiece
Answer:
(c) greater than that of the eyepiece
Hint:
For producing large magnification, $f_0>f_e$
Question 104.
The number of lenses in terrestrial telescope is
(a) 2
(b) 4
(c) 3
(d) 6
Answer:
(c) 3
Hint:
A terrestrial telescope consists of three lenses: objective, erecting lens and eyepiece.
Question 105 .
An achromatic combination of lenses is formed by joining
(a) 2 convex lens
(b) 1 convex, 1 concave lens
(c) 2 concave lenses
(d) 1 convex and 1 plane mirror
Answer:

(b) 1 convex, 1 concave lens
Hint:
An achromatic doublet should satisfy the condition $\frac{w_1}{f_1}+\frac{w_2}{f_2}=0$.
Question 106.
The amount of light received by a camera depends upon
(a) diameter only
(b) ratio of focal length and diameter
(c) product of focal length and diameter
(d) only one of the focal length
Answer:
(b) ratio of focal length and diameter
Hint:
The amount of light received by a camera depends on the ratio of the focal length and diameter of the aperture.
Question 107.
Myopia is corrected by using a
(a) cylindrical lens
(b) bifocal lens
(c) convex lens
(d) concave lens
Answer:
(d) concave lens
Question 108.
The critical angle for total internal reflection in diamond is $24.5^{\circ}$. The angle refractive index of diamond is
(a) 2.41
(b) 1.41
(c) 2.59
(d) 1.59
Answer:
(a) 2.41
Hint:
$
\mu=\frac{1}{\sin i_c}=\frac{1}{\sin 24.5^{\circ}}=2.41
$
Question 109.
When a glass lens with $\mu=1.47$ is immersed in a trough of liquid, it looks to be disappeared. The liquid in the trough could be
(a) water
(b) kerosene
(c) glycerine
(d) alcohol

Answer:
(c) glycerine
Hint:
Glass lens will disappear, if $\mu_1=\mu_{\mathrm{g}}$.
Question 110.
In optical fibres, the refractive index of the core is
(a) greater than that of the cladding
(b) equal to that of the cladding
(c) smaller than that of the cladding
(d) independent of that of the cladding
Answer:
(a) greater than that of the cladding
Hint:
In optical fibres, refractive index of core material > refractive index of the cladding.
Question 111.
For a wavelength of light ' $\lambda$ ' and scattering object of size ' $a$ ', all wavelength are scattered nearly equally, if
(a) $\mathrm{a}=\lambda$
(b) $\mathrm{a} \gg \lambda$
(c) $\mathrm{a}<<\lambda$
(d) $\mathrm{a} \geq \lambda$
Answer:

(b) $\mathrm{a} \gg \lambda$
Hint:
For a $\gg \lambda$, the scattering power is not selective.
Question 112 .
Two coherent monochromatic light beams of intensities I and $4 \mathrm{I}$ are supperposed. The maximum and minimum possible intensities in the resulting beams are
(a) $5 \mathrm{I}$ and I
(b) $9 I$ and I
(c) $5 I$ and $3 I$
(d) $9 I$ and $3 \mathrm{I}$
Answer:
(b) 9I and I
Hint:

$
\begin{aligned}
& \mathrm{I}_{\max }=\left(\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}\right)^2=(\sqrt{41}+\sqrt{1})^2=9 \mathrm{I} \\
& \mathrm{I}_{\max }=\left(\sqrt{\mathrm{I}_1}-\sqrt{\mathrm{I}_2}\right)^2=(\sqrt{41}-\sqrt{1})^2=\mathrm{I}
\end{aligned}
$
Question 113.

screen is doubled. The fringe width is
(a) unchanged
(b) halved
(c) doubled
(d) quadrupled
Answer:
(d) quadrupled
Hint:
$
\beta=\frac{\lambda D}{d} ; \beta=\frac{\lambda \times 2 D}{D / 2}=4 \beta
$
Question 114.
In a young's double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen, when light of wavelength $600 \mathrm{~nm}$ is used. If the wavelength of light is changed to $400 \mathrm{~nm}$,
number of fringes observed in the same segment of the screen is given by
(a) 12
(b) 18
(c) 24
(d) 30
Answer:
(b) 18
Hint:
$
\mathrm{n}_1 \lambda_1=\mathrm{n}_2 \lambda_2 \Rightarrow 12 \times 600=\mathrm{n}_2 \times 400 \text { or } \mathrm{n}_2=18
$

Question 115 .
Consider ffaunhoffer diffraction pattern obtained with a single slit illuminated at normal incident. At the angular position of the first diffraction minimum the phase difference between the wavelets from the opposite edges of the slits is
(a) $\frac{\pi}{4}$
(b) $\frac{\pi}{2}$
(c) $2 \pi$
(d) $\pi$
Answer:
(c) $2 \pi$
Hint:
At the angular position of first minimum wavelets from opposite edges of the slits have a path difference of $\mathrm{X}$ and a phase difference of $2 \pi$ radian.
Question 116.
A beam of light of wavelength $600 \mathrm{~nm}$ from a distant source falls on a single slit $1.00 \mathrm{~mm}$ wide and the resulting diffraction pattern is observed on a screen $2 \mathrm{~m}$ away. The distance between the first dark fringes on either side of the central bright fringe is
(a) $1.2 \mathrm{~cm}$
(b) $1.2 \mathrm{~mm}$
(c) $2.4 \mathrm{~cm}$
(d) $2.4 \mathrm{~mm}$
Answer:
(c) $2.4 \mathrm{~cm}$
Hint:
Distance between the first dark fringes on either side $=$ width of central maximum
$
=\frac{2 D \lambda}{d}=\frac{2 \times 2 \times 600 \times 10^{-9}}{1.00 \times 10^{-3}} \mathrm{~m}=2.4 \times 10^{-3} \mathrm{~m}=2.4 \mathrm{~mm}
$
Question 117.
A young's double slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen is.
(a) hyperbola
(b) circle
(c) straight line
(d) parabola
Answer:
(a) hyperbola
Hint:

In young's double slit experiment, the fringes obtained are hyperbolic in shape. But in a small interference pattern, the fringes appear straight.
Question 118.
The initial shape of the wavefront of the beam is
(a) planar
(b) convex
(c) concave
(d) convex near the axis and concave near the periphery
Answer:
(a) planar
Hint:
As the beam is initially parallel, the shape of wavefront is planar.
Question 119.
The angle of incident at which reflected light is totally polarised for reflection from air to glass (refractive index $\mu$ ) is
(a) $\sin ^{-1}(\mu)$
(b) $\sin ^{-1}\left(\frac{1}{\mu}\right)$
(c) $\tan ^{-1}\left(\frac{1}{\mu}\right)$
(d) $\tan ^{-1}(\mu)$
Answer:
(d) $\tan ^{-1}(\mu)$
According to Brewster's law,
$
\mu=\tan i_p \therefore i_p=\tan ^{-1}(\mu)
$
Question 120.
According to Huygen's principle, light is a form of
(a) particle
(b) rays
(c) wave
(d) none of the above
Answer:
(c) wave
Hint:
According to Huygen's principle, light travels in the form of a longitudinal wave.
Question 121.
Which one of the following phenomena is not explained by Huygen's construction of wavefront?
(a) refraction
(b) reflection
(c) diffraction

(d) origin of spectra
Answer:
(d) origin of spectra
Hint:
Huygen's construction of wavefront cannot explain origin of spectra which can be explained on the basis of quantum theory.
Short Answer:
Question 1.
Light from a point source in air falls on a convex spherical glass surface $(\mathrm{n}=1.5$, radius of curvature $=$ $20 \mathrm{~cm}$ ). The distance of light source from the glass surface is $100 \mathrm{~cm}$. At What position is the image formed?
Solution:
$
\begin{aligned}
& \mathrm{n}_1=1 ; \mathrm{n}_2=1.5 \\
& \mathrm{u}=100 \mathrm{~cm} ; \mathrm{R}=+20 \mathrm{~cm}
\end{aligned}
$
( $\mathrm{R}$ is $+\mathrm{Ve}$ for a convex refracting surfce)
As $\frac{1}{2}=\frac{1}{2}$
$
\begin{aligned}
& \frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R} \\
& \frac{1.5}{v}+\frac{1}{100}=\frac{1.5-1}{20}=\frac{1}{40} \\
& \frac{3}{2 v}=\frac{1}{40}-\frac{1}{100}=\frac{5-2}{200}=\frac{3}{200} \\
& \frac{1}{v}=\frac{1}{100} \\
& \mathrm{v}=100 \mathrm{~cm}
\end{aligned}
$
Thus the image is formed at a distance of $100 \mathrm{~cm}$ from the glass surface in the direction of incident light.
Question 2 .
Find the value of critical angle for a material of refractive index $\sqrt{3}$.
Solution:
Here, $\mathrm{n}=\sqrt{3}$
$
\begin{aligned}
& \sin \mathrm{i}_{\mathrm{c}}=\frac{1}{n}=\frac{1}{\sqrt{ } 3}=\frac{\sqrt{ } 3}{3} \\
& \therefore \text { critical angle, } \mathrm{i}_{\mathrm{c}}=35.3^{\circ}
\end{aligned}
$

Question 3.
The radius of curvature of each face of biconcave lens, made of glass of refractive index 1.5 is $30 \mathrm{~cm}$. Calculate the focal length of the lens in air.
Solution:
Here $\mathrm{n}=1.5 ; \mathrm{R}_1=-30 \mathrm{~cm} ; \mathrm{R}_2=30 \mathrm{~cm}$
Using len's maker's formula,
$
\begin{aligned}
& \frac{1}{f}=(\mathrm{n}-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\
& =(1.5-1)\left[\frac{1}{-30}-\frac{1}{30}\right]=0.5 \times\left(\frac{-2}{30}\right) \\
& \frac{1}{f}=\frac{1}{30} \\
& \mathrm{f}=-30
\end{aligned}
$
Question 4.
The radii of curvature of the faces of a double convex lens are $10 \mathrm{~cm}$ and $15 \mathrm{~cm}$. If focal length is 12 $\mathrm{cm}$. What is the refractive index of glass?
Solution:
$
\begin{aligned}
& \mathrm{f}=+12 \mathrm{~cm} ; \mathrm{R}_1=10 \mathrm{~cm} ; \mathrm{R}_2=-15 \mathrm{~cm} ; \mathrm{n}=? \\
& \text { As, } \frac{1}{f}=(\mathrm{n}-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\
& \left.\frac{1}{12}=(\mathrm{n}-1)\left(\frac{1}{10}+\frac{1}{15}\right)=(\mathrm{n}-1) \mathrm{x} \text { latex }\right] \text { frac }\{5\}\{30\}[/ \text { latex }]
\end{aligned}
$
$
\begin{aligned}
& (\mathrm{n}-1)=\frac{6}{12}=0.5 \\
& \mathrm{n}=0.5+1 \\
& \mathrm{n}=1.5
\end{aligned}
$
Question 5.
A double convex lens made of glass of refractive index 1.5 has its both surfaces of equal radii of curvature of $20 \mathrm{~cm}$ each. An object of $5 \mathrm{~cm}$ height is placed at a distance of $10 \mathrm{~cm}$ from the lens. Find the position, nature and size of the image.
Solution:
Here $\mathrm{n}=1.5 ; \mathrm{R}_1=+20 \mathrm{~cm} ; \mathrm{R}_2=-20 \mathrm{~cm}$
Using lens maker's formula,
$
\begin{aligned}
& \frac{1}{f}=(\mathrm{n}-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\
& (1.5-1)=\left[\frac{1}{20}-\frac{1}{-20}\right] 0.5 \times\left(\frac{2}{20}\right) \\
& \frac{1}{f}=\frac{1}{20} \\
& \mathrm{f}=20 \mathrm{~cm}
\end{aligned}
$

Now, $\mathrm{u}=-10 \mathrm{~cm}$ andf $=+20 \mathrm{~cm}$
From thin lens formula, $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{20}-\frac{1}{10}=\frac{1}{20}$
$\therefore \mathrm{v}=-20 \mathrm{~cm}$
Magnification, $\mathrm{m}=\frac{h_2}{h_1}=\frac{v}{u}$
$\frac{h_2}{5 c m}=\frac{-20}{-10}$
$\mathrm{h}_2$
Hence a virtual and erect image of height $10 \mathrm{~cm}$ is formed at a distance of $20 \mathrm{~cm}$ from the lens on the same side as the the object.
Question 6.
The image obtained with a convex lens is erect and its length is four times the length of the object. If the focal length of the lens is $20 \mathrm{~cm}$, calculate the object and image distances.
Solution.
Heref $=20 \mathrm{~cm}, \mathrm{~m}=+4$ for a virtual image.
To calculate $\mathrm{u}$, we have
$
\mathrm{m}=\frac{f}{u+f}
$
Question 7.
The radius of curvature of each surface of a convex lens of refractive index 1.5 is $40 \mathrm{~cm}$. Calculate its power.
Solution:
Here, $\mathrm{n}=105 ; \mathrm{R}_1=+40 \mathrm{~cm}=0.40 \mathrm{~m}$
$
\mathrm{R}_2=-40 \mathrm{~cm}=-0.40 \mathrm{~m}
$
Power $(\mathrm{p})=\frac{1}{f}=(\mathrm{n}-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$
\begin{aligned}
& =(1.5-1)\left[\frac{1}{0.40}-\frac{1}{(-0.40)}\right]=0.5 \times \frac{2}{0.40} \\
& \mathrm{p}=2.5 \mathrm{D}
\end{aligned}
$

Question 8.
A ray of light incident on an equilateral glass prism shows minimum deviation of $30^{\circ}$. Calculate the speed of light through the prism.
Solution:
Here, $\mathrm{A}=60^{\circ} ; \mathrm{D}=30^{\circ}$
Refractive index, $\mathrm{n}=\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \left(\frac{A}{2}\right)}=\frac{\sin \left(\frac{60+30}{2}\right)}{\sin \left(\frac{60}{2}\right)}$
$
\mathrm{n}=\frac{\sin 45^{\circ}}{\sin 30^{\circ}}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}=\sqrt{ } 2
$
$
\mathrm{n}=1.414
$
Velocity or light in glass, $\mathrm{v}=\frac{c}{n}=\frac{3 \times 10^8}{1.414}$
$
\mathrm{v}=2.12 \times 10^8 \mathrm{~ms}^{-1}
$
Question 9.
Two sources of intensity I and 41 are used in an interference experiment. Find the intensity at points where the waves from two sources superimpose with a phase
(i) Zero
(ii) $\frac{\pi}{2}$
(iii) $\pi$
Solution:
The resultant intensity at a point where phase difference is $\Phi$ is
$\mathrm{I}_{\mathrm{R}}+\mathrm{I}_1+\mathrm{I}_2 2+\sqrt{\mathrm{I}_1 \mathrm{I}_2} \cos \Phi$
As $\mathrm{I}_1+\mathrm{I}$ and $\mathrm{I}_2=4 \mathrm{I}$, therefore,
$
\begin{aligned}
& \mathrm{I}_{\mathrm{R}}=\mathrm{I}+4 \mathrm{I}+2 \sqrt{I .4 I} \cos \Phi \\
& \mathrm{I}_{\mathrm{R}}=5 \mathrm{I}+4 \mathrm{I} \cos \Phi
\end{aligned}
$
(i) When $\Phi=0 ; I_R=5 I+4 I \cos 0=9 I$
(ii) When $\Phi=\frac{\pi}{2} ; \mathrm{I}_{\mathrm{R}}=5 \mathrm{I}+4 \mathrm{I} \cos \frac{\pi}{2}=5 \mathrm{I}$
(iii) When $\Phi=\pi ; I_R=5 I+4 I \cos \pi=51-41=I$
$\Phi=0 ; \mathrm{I}_{\mathrm{R}}=9 \mathrm{I}$
$\Phi=\frac{\pi}{2} ; \mathrm{I}_{\mathrm{R}}=5 \mathrm{I}$
$\Phi=\pi ; \mathrm{I}_{\mathrm{R}}=\mathrm{I}$

Question 10 .
Assume that light of wavelength $600 \mathrm{~A}$ is coming from a star. What is the limit of resolution of a telescope whose objective has a diameter of 100 inch?
Solution:
The limit of resolution of a telescope, $\mathrm{d} \theta=\frac{1.22 \lambda}{D}$
$
\begin{aligned}
& \mathrm{D}=100 \text { inch }=254 \mathrm{~cm} \\
& {[\therefore 1 \text { inch }=2.54 \mathrm{~cm}]} \\
& \lambda=6000 \AA=6000 \times 10^{-10} \mathrm{~m} / \mathrm{s} \\
& \mathrm{d} \theta=\frac{1.22 \times 6000 \times 10^{-10}}{254 \times 10^{-2}}=2.9 \times 10^{-7} \\
& \mathrm{~d} \theta=2.9 \times 10^{-7} \mathrm{rad} .
\end{aligned}
$
Question 11.
Two polarising sheet have their polarising directions parallel so that the intensity of the transmitted light is maximum. Through what angle must the either sheet be turned if the intensity is to drop by one-half?
Solution:
Here, $\mathrm{I}=\frac{I_o}{2}$
Using Malus law,
$
\begin{aligned}
& \mathrm{I}=\mathrm{I}_0 \cos ^2 \theta \\
& \frac{I_o}{2}=I_0 \cos ^2 \theta \\
& \cos \theta \pm \frac{1}{\sqrt{ } 2} \\
& \theta= \pm 45^{\circ}, \pm 135^{\circ}
\end{aligned}
$