Numerical Problems-1 - Chapter 6 Optics 12th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Numerical Problems
Question 1.
An object is placed at a certain distance from a convex lens of focal length $20 \mathrm{~cm}$. Find the distance of the object if the image obtained is magnified 4 times.
Solution:
$
\begin{aligned}
& f=-20 \mathrm{~cm} \\
& \mathrm{v}=-4 \mathrm{u}
\end{aligned}
$
According to lens formula
$
\begin{aligned}
& \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\
& \frac{1}{(-20)}=\frac{1}{(-4 u)}+\frac{1}{u} \\
& \frac{1}{(-20)}=\left[-\frac{1}{4}+1\right] \\
& =\frac{1}{u} \frac{3}{4} \\
& \mathrm{u}=\frac{3 \times 20}{4}=-15 \mathrm{~cm}
\end{aligned}
$
Question 2.
A compound microscope has a magnification of 30 . The focal length of eye piece is $5 \mathrm{~cm}$. Assuming the final image to be at least distance of distinct vision, find the magnification produced by the objective.
Solution:
Magnification of compound microscope, $M=30$
Focal length, $\mathrm{f}=5 \mathrm{~cm}$
Least distance of distinct vision, $\mathrm{D}=25 \mathrm{~cm}$
Now, $M=\mathrm{M}_0 \times \mathrm{M}_{\mathrm{e}}$
$
\begin{aligned}
& =\mathrm{M}_0 \times\left[1+\frac{\mathrm{D}}{f_e}\right] \\
& 30=\mathrm{M}_0 \times\left[1+\frac{25}{5}\right] \\
& \mathrm{M}_0=\frac{30}{6}=5
\end{aligned}
$

Question 3.
An object is placed in front of a concave mirror of focal length $20 \mathrm{~cm}$. The image formed is three times the size of the object. Calculate two possible distances of the object from the mirror.
Solution:
$\mathrm{m}=+3$ and $\mathrm{m}=-3 ; \mathrm{f}=-20 \mathrm{~cm}$
When, $\mathrm{m}=3$
$
\begin{aligned}
& \mathrm{m}=\frac{v}{u}=3 \\
& \mathrm{v}=-3 \mathrm{u}
\end{aligned}
$
From mirror equation,
$
\begin{aligned}
& \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\
& \frac{1}{-20}=\frac{1}{-3 u}+\frac{1}{u} \\
& \frac{1}{u}\left[-\frac{1}{3}+1\right] \\
& \frac{1}{-20}=\frac{2}{3 u} \\
& \mathrm{u}=-\frac{40}{3} \mathrm{~cm}
\end{aligned}
$
When, $\mathrm{m}=-3 \Rightarrow \mathrm{v}=3 \mathrm{u}$
$
\begin{aligned}
& \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\
& \frac{1}{-20}=\frac{1}{3 u}+\frac{1}{u} \\
& \frac{1}{-20}=\frac{4}{3 u}
\end{aligned}
$
$
\begin{aligned}
& \mathrm{u}=\frac{4 \times 20}{3} \\
& \mathrm{u}=-\frac{80}{3} \mathrm{~cm}
\end{aligned}
$
Question 4.
A small bulb is placed at the bottom of a tank containing water to a depth of $80 \mathrm{~cm}$. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)
Solution:
Actual depth of the bulb in water
$
\mathrm{d}_1=80 \mathrm{~cm}=0.8 \mathrm{~m}
$
Refractive index of water, $\mathrm{p}=1.33$
The given situation is shown in the figure.

Where, $\mathrm{i}=$ Angle of incidence
$\mathrm{r}=$ Angle of refraction $=90^{\circ}$
Since the bulb is a point source, the emergent light can be considered as a circle AT
$
\mathrm{R}=\frac{A C}{2}=\mathrm{AO}=\mathrm{OB}
$
Using snell's law, the relation for the refractive index of water is $\sin$
$
\begin{aligned}
& \mu=\frac{\operatorname{sinr}}{\sin i} \\
& \mathrm{i}=\sin ^{-1}(0.75)=48.75^{\circ}
\end{aligned}
$
Using the given figure, we have the relation
$
\begin{aligned}
& \tan \mathrm{i}=\frac{O C}{B C}=\frac{R}{d_1} \\
& \mathrm{R}=\tan \mathrm{ix} \mathrm{d}=\tan 48.75^{\circ} \times 0.8 \\
& \mathrm{R}=0.91 \mathrm{~m}
\end{aligned}
$
Area of the surface of water $=\pi R^2$
$
=3.14 \times(0.91)^2=2.61 \mathrm{~m}^2
$
Hence, the area of the surface of water through which the light from the bulb can emerge is approximately $2.61 \mathrm{~m}^2$
Question 5.
A thin converging glass lens made of glass with refractive index 1.5 has a power of $+5.0 \mathrm{D}$. When this lens is immersed in a liquid of refractive index $n$, it acts as a divergent lens of focal length $100 \mathrm{~cm}$.
What must be the value of $n$ ?
Solution:
According to Lens maker's formula, the focal length for a convex lens placed in air can be obtained as

$
\begin{aligned}
\frac{1}{f} & =\left[\frac{n_2}{n_a}-1\right]\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right] \\
\mathrm{P}=\frac{1}{f} & =5 \text { and } n_2=1.5, n_a=1 ; f=20 \mathrm{~cm} \\
\frac{1}{20} & =\left[\frac{1.5}{\mathrm{l}}-1\right]\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right] \\
\frac{1}{20} & =0.5\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right] \\
\frac{1}{f_l} & =\left[\frac{n_2}{n_l}-1\right]\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right] \\
\frac{1}{f_l} & =1 ; \frac{1}{f_l}=-100 \mathrm{~cm} ; n_2=1.5 ; \quad n_l=n \\
\frac{1}{-100} & =\left[\frac{1.5}{n}-1\right]\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right]
\end{aligned}
$
Ratio of equation ( 2$) \div(1)$
$
\begin{aligned}
& \frac{1}{-5}=\frac{\left(\frac{1.5}{n}-1\right)}{0.5} \\
& \frac{0.5}{5}=\frac{1.5}{n}-1 \\
& 0.9=\frac{1.5}{n} \\
& \mathrm{n}=\frac{1.5}{0.9}=\frac{5}{3}-1
\end{aligned}
$

Question 6.
If the distance $\mathrm{D}$ between an object and screen is greater than 4 times the focal length of a convex lens, then there are two positions of the lens for which images are formed on the screen. This method is called conjugate foci method. If $\mathrm{d}$ is the distance between the two positions of the lens, obtain the equation for focal length of the convex lens.
Solution:
Let us fix the position of object and place the screen to get the enlarged image first. Also, let us fix the position of screen where we get the enlarged image.
Let $\mathrm{D}$ be the distance between object and screen. Let us mark the position of lens dv Then let us move the lens away from the object to get a diminished image. Let this position of lens be $d_2$. Let $d$ be the distance between the lens position $\mathrm{d}_1$ and $\mathrm{d}_2$. Let $\mathrm{V}$ be the distance $\mathrm{b} / \mathrm{w}$ image and lens. Let ' $\mathrm{u}$ ' be the distance between object and lens.
From mirror equation,

$
\frac{1}{v}+\frac{1}{u}=\frac{1}{f}
$
Let us replace $\mathrm{v}$ by substituting $\mathrm{v}=\mathrm{D}-\mathrm{u}$
$
\frac{1}{D-u}+\frac{1}{u}=\frac{1}{f}
$
we get the equation, $\mathrm{u}^2-\mathrm{Du}+\mathrm{fD}=0$
the quadratic equation for above equation,
$
\mathrm{u}=\frac{\mathrm{D} \pm \sqrt{\mathrm{D}^2-4 f \mathrm{D}}}{2}
$
When $\mathrm{D}=4 \mathrm{f}$, we get only position of lens to get image.
This corresponds to placing the object at $2 \mathrm{f}$ and getting the image at $2 \mathrm{f}$ on the otherside. Hence, for displacement method we need $\mathrm{D}>4 \mathrm{f}$. When this condition is satisfied we get $\mathrm{u}_1=\frac{\mathrm{D}-\sqrt{\mathrm{D}^2-4 f \mathrm{D}}}{2} ;$ corresponding $\mathrm{v}_1-\mathrm{D}-\mathrm{u}_2=\frac{\mathrm{D}+\sqrt{\mathrm{D}^2-4 f \mathrm{D}}}{2}$ after changing the location $\mathrm{u}_1=\frac{\mathrm{D}+\sqrt{\mathrm{D}^2-4 f \mathrm{D}}}{2} ;$ corresponding $\mathrm{v}_2-\mathrm{D}-\mathrm{u}_2=\frac{\mathrm{D}-\sqrt{\mathrm{D}^2-4 f \mathrm{D}}}{2}$
now the displacement $\mathrm{d}=\mathrm{v}_1-\mathrm{u}_1=\mathrm{u}_1-\mathrm{v}_1=\sqrt{D^2-4 f D}$
Hence we get focal length, $f=\frac{D^2-d^2}{4 D}$
Question 7.
A beam of light of wavelength $600 \mathrm{~nm}$ from a distant source falls on a single slit $1 \mathrm{~mm}$ wide and the resulting diffraction pattern is observed on a screen $2 \mathrm{~m}$ away. What is the distance between the first dark fringe on either side of the central bright fringe?
Solution:
For first minimum $(\mathrm{n}=1)$ on either side of central maximum.
$
\sin \theta=\frac{\lambda}{a}
$
Where a is width of the slit Since 0 is very small, $(\sin \theta \approx \theta)$
$
\begin{aligned}
& \theta=\frac{\lambda}{a} \ldots \ldots \text { (1) } \\
& \sin \theta \approx \theta=\frac{x}{2 D}
\end{aligned}
$
Where, $x$ is distance of first dark fringe from central maximum.
$\mathrm{D}$ is distance between slit and screen.
From equation (1) and (2)
$
\begin{aligned}
& \frac{x}{2 D}=\frac{\lambda}{a} \\
& \mathrm{x}=\frac{2 D \lambda}{a}=\frac{2 \times 2 \times 600 \times 10^{-9}}{1 \times 10^{-3}}
\end{aligned}
$

$
\begin{aligned}
& =2400 \times 10^{-6}=2.4 \times 10^{-3} \mathrm{~m} \\
& \mathrm{x}=2.4 \mathrm{~mm}
\end{aligned}
$
Question 8.
In Young's double slit experiment, the slits are $2 \mathrm{~mm}$ apart and are illuminated with a mixture of two wavelength $\lambda_0=750 \mathrm{~nm}$ and $\lambda=900 \mathrm{~nm}$. What is the minimum distance from the common central bright fringe on a screen $2 \mathrm{~m}$ from the slits where a bright fringe from one interference pattern coincides with a bright fringe from the other?
Solution:
Now from the question we can infer that
$
\begin{aligned}
& \mathrm{D}=2 ; \mathrm{d}=2 \\
& \mathrm{n}_1 \lambda_1=\mathrm{n}_2 \lambda_2 \\
& \frac{n_1}{n_2}=\frac{\lambda_2}{\lambda_1}=\frac{900}{750} \frac{1}{2}
\end{aligned}
$
Thus, we have
$
\frac{n_1}{n_1}=\frac{\lambda_2}{\lambda_1}=\frac{6}{5}
$
$5^{\text {th }}$ and $6^{\text {th }}$ fringes will coincide respectively. The minimum distance is given as
$
\begin{aligned}
& \mathrm{X}_{\min }=\frac{n_2 \lambda_2 \mathrm{D}}{d}=\frac{5 \times 900 \times 10^{-9} \times 2}{2 \times 10^{-3}} \\
& =4500 \times 10^{-6}=4.5 \times 10^{-3} \mathrm{~m} \\
& \mathrm{X}_{\min }=4.5 \mathrm{~mm}
\end{aligned}
$
Question 9.
In Young's double slit experiment, 62 fringes are seen in visible region for sodium light of wavelength $5893 \AA$. If violet light of wavelength $4359 \AA$ is used in place of sodium light, then what is the number of fringes seen?
Solution:
From young's double slit experiment,
$
\begin{aligned}
& \lambda_1=5893 \AA ; \lambda_2=4359 \AA \\
& \frac{n_1 \lambda_1 D}{d}=\frac{n_2 \lambda_2 \mathrm{D}}{d}
\end{aligned}
$
The above condition is total extent of fringes is constant for both wavelengths.
$
\begin{aligned}
& \frac{62 \times 5893 \times 10^{-10} \times D}{d}=\frac{n_2 \times 4359 \times 10^{-10} \times D}{d} \\
& \mathrm{n}_2=\frac{62 \times 5893}{4359}=\frac{365366}{4359}=83.8 \\
& \mathrm{n}_2=84
\end{aligned}
$

Question 10.
A compound microscope has a magnifying power of 100 when the image is formed at infinity. The objective has a focal length of $0.5 \mathrm{~cm}$ and the tube length is $6.5 \mathrm{~cm}$. What is the focal length of the eyepiece.
Solution:
Magnifying Power, $\mathrm{m}=100$
Focal length of the objective, $\mathrm{f}_0=0.5 \mathrm{~cm}$
Tube length, $1=6.5 \mathrm{~cm}$
Since the image is formed at infinity, the real image produced by the objective lens should lie on the focus of the eyepiece. $\mathrm{v}_0+f_{\mathrm{e}}=6.5 \mathrm{~cm} \ldots . .(1)$
The magnifying power for normal adjustment is given by
$
\begin{aligned}
& \mathrm{m}=\left(\frac{v_o}{u_o}\right) \times \frac{\mathrm{D}}{f_e} \\
& =-\left[1-\frac{v_o}{f_o}\right] \frac{\mathrm{D}}{f_e} \\
& 100=-\left[1-\frac{v_o}{0.5}\right] \times \frac{25}{f_e} \\
& 2 \mathrm{v}_0-4 f_{\mathrm{e}}=1
\end{aligned}
$
On solving equations (1) and (2), we get $\mathrm{v}_0=4.5 \mathrm{~cm}$ and $f_{\mathrm{e}}=2 \mathrm{~cm}$
Thus, the focal length of the eyepiece is $2 \mathrm{~cm}$.