Geometrical Optics
A convex lens of refractive index 1.5 and focal length $f=18 \mathrm{~cm}$ is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is $\alpha \times \mathrm{f}$. The value of $\alpha$ is $\_\_\_\_$ .
(refractive index of water $=4 / 3$ )
Explanation:
The Lens Maker's Formula is
$ \frac{1}{\mathrm{f}}=\left(\frac{\mu_1}{\mu_{\mathrm{m}}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) $
where $\mu_1$ is the refractive index of the lens and $\mu_m$ is the refractive index of the surrounding medium.
Case-1 Focal length in Air $\left(\mathrm{f}_{\mathrm{a}}\right)$
In air, $\mu_{\mathrm{m}}=1$.
$ \frac{1}{\mathrm{f}_{\mathrm{a}}}=\left(\frac{\mu_1}{1}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) $
Let $\mathrm{K}=\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \Rightarrow \frac{1}{\mathrm{f}_{\mathrm{a}}}=\left(\mu_1-1\right) \mathrm{K}$
Given $\mu_1=1.5$ and $\mathrm{f}_{\mathrm{a}}=\mathrm{f}=18 \mathrm{~cm}$ :
$ \frac{1}{\mathrm{f}}=(1.5-1) \mathrm{K}=0.5 \mathrm{~K} \Rightarrow \mathrm{~K}=\frac{2}{\mathrm{f}} $
Case-2 Focal length in Water ( $\mathrm{f}_{\mathrm{w}}$ )
In water, $\mu_{\mathrm{m}}=\mu_{\mathrm{w}}=4 / 3$ :
$\frac{1}{\mathrm{f}_{\mathrm{w}}}=\left(\frac{\mu_{\mathrm{l}}}{\mu_{\mathrm{w}}}-1\right) \mathrm{K}$
$\Rightarrow $ $\frac{1}{\mathrm{f}_{\mathrm{w}}}=\left(\frac{1.5}{4 / 3}-1\right) \mathrm{K}=\left(\frac{4.5}{4}-1\right) \mathrm{K}=\left(\frac{9}{8}-1\right) \mathrm{K}$
$ \frac{1}{\mathrm{f}_{\mathrm{w}}}=\frac{1}{8} \mathrm{~K}=\frac{1}{8} \times(2 \mathrm{f})=\frac{1}{4} \mathrm{f} $
Therefore, $\mathrm{f}_{\mathrm{w}}=4 \mathrm{f}$.
The difference in focal lengths is given as :
$ \Delta \mathrm{f}=\mathrm{f}_{\mathrm{w}}-\mathrm{f}_{\mathrm{a}} $
$\Rightarrow $ $ \Delta \mathrm{f}=4 \mathrm{f}-\mathrm{f}=3 \mathrm{f}=\alpha \times \mathrm{f} \Rightarrow \alpha=3 $
Therefore, the value of $\alpha$ is 3.
Hence, the correct answer is 3.
The size of the images of an object, formed by a thin lens are equal when the object is placed at two different positions 8 cm and 24 cm from the lens. The focal length of the lens is $\_\_\_\_$ cm.
Explanation:
For a thin lens to form images of the same size at two different object positions, one image must be real and the other must be virtual. This typically occurs with a convex lens.
If the object distance is $u$, focal length of lens is $f$ then lens formula is given as,
$ \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v}=\frac{u+f}{u f} $
The magnification is given as $m=\frac{v}{u} \Rightarrow m=\frac{1}{u} \times\left[\frac{u f}{u+f}\right]=\frac{f}{f+u}$
Where:
f is the focal length.
u is the object distance (using sign convention, u is negative).
Since the sizes (magnitudes of magnification) are equal but the nature of the images is different:
$ \left|\mathrm{m}_1\right|=\left|\mathrm{m}_2\right| \Rightarrow \mathrm{m}_1=-\mathrm{m}_2 $
Given:
1. Position $1=\mathrm{u}_1=-8 \mathrm{~cm}$
2. Position $2=\mathrm{u}_2=-24 \mathrm{~cm}$
$\frac{f}{f+(-8)}=-\left(\frac{f}{f+(-24)}\right)$
$\Rightarrow $ $\frac{1}{f-8}=-\frac{1}{f-24}$
$\Rightarrow $ $f-24=-(f-8)$
$\Rightarrow $ $f-24=-f+8$
$\Rightarrow $ $2 \mathrm{f}=32 $
$\Rightarrow $ $ \mathrm{f}=16 \mathrm{~cm}$
The focal length of the lens that produces images of equal size at object distances of 8 cm and 24 cm is 16 cm.
Hence, the correct answer is 16.
A parallel beam of light travelling in air (refractive index 1.0) is incident on a convex spherical glass surface of radius of curvature 50 cm . Refractive index of glass is 1.5 . The rays converge to a point at a distance $x \mathrm{~cm}$ from the centre of the curvature of the spherical surface. The value of $x$ is $\_\_\_\_$ cm .
Explanation:
Imagine a point object $O$ placed on the principal axis of a convex spherical surface that separates two media of refractive indices $\mu_1$ and $\mu_2$.
Here the pole of the surface be P and its centre of curvature be C.
Using the formula for refraction at a spherical surface :
$ \frac{\mu_2}{\mathrm{v}}-\frac{\mu_1}{\mathrm{u}}=\frac{\mu_2-\mu_1}{\mathrm{R}} $
The incident beam is parallel, which means the light is coming from an object at infinity. Therefore, the object distance $u=-\infty$.
The refractive index of the first medium (air) is $\mu_1=1.0$.
The refractive index of the second medium (glass) is $\mu_2=1.5$.
The surface is convex facing the air. This means its centre of curvature lies inside the glass, in the direction of the traveling light. Thus, the radius of curvature is positive: $\mathrm{R}=+50 \mathrm{~cm}$.
Let's plug these into our derived equation to find the image position (v) :
$ \frac{1.5}{v}-\frac{1.0}{-\infty}=\frac{1.5-1.0}{+50} $
$\Rightarrow $ $\frac{1.5}{v}-0=\frac{0.5}{50}$
$\Rightarrow $ $\frac{1.5}{v}=\frac{1}{100}$
$\Rightarrow $ $ \mathrm{v}=150 \mathrm{~cm} $
So, the parallel rays will converge to form an image at a distance of 150 cm to the right of the pole of the spherical surface.
Pole $(P)$ is at 0 cm . and the centre of curvature $(C)$ is at +50 cm .
Image point / Convergence point (I) is at +150 cm .
So, the distance between C and I is
$ x=v-R $
$\Rightarrow $ $ \mathrm{x}=150 \mathrm{~cm}-50 \mathrm{~cm}=100 \mathrm{~cm} $
Therefore, the distance from the centre of curvature is 100 cm .
Hence, the correct answer is $\mathbf{1 0 0}$.
In a microscope the objective is having focal length $f_o=2 \mathrm{~cm}$ and eye-piece is having focal length $f_e=4 \mathrm{~cm}$. The tube length is 32 cm . The magnification produced by this microscope for normal adjustment is $\_\_\_\_$ .
Explanation:
In microscope normal adjustment refers to the condition where the final image is formed at infinity.
The total magnifying power ( M ) of a compound microscope in normal adjustment is the product of the linear magnification of the objective ( $\mathrm{m}_{\mathrm{o}}$ ) and the angular magnification of the eyepiece ( $\mathrm{m}_{\mathrm{e}}$ ):
$ \mathrm{M}=\mathrm{m}_{\mathrm{o}} \times \mathrm{m}_{\mathrm{e}} $
For an image at infinity:
Objective magnification $\mathrm{m}_{\mathrm{o}}=\frac{\mathrm{L}}{\mathrm{f}_{\mathrm{O}}}$, where L is the tube length.
Eyepiece magnification $\mathrm{m}_{\mathrm{e}}=\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}$, where D is the least distance of distinct vision (standard value is 25 cm ).
Thus, the total magnification formula is:
$ M=\frac{L}{f_o} \times \frac{D}{f_e} $
The focal length of objective is $\mathrm{f}_{\mathrm{o}}=2 \mathrm{~cm}$
The focal length of eyepiece is $\mathrm{f}_{\mathrm{e}}=4 \mathrm{~cm}$
The tube length is $\mathrm{L}=32 \mathrm{~cm}$
Substituting the values,
$ M=\frac{32}{2} \times \frac{25}{4} 100 $
Therefore, the total magnification of given microscope is 100 . Hence, the correct answer is $\mathbf{1 0 0}$.
A collimated beam of light of diameter 2 mm is propagating along $x$-axis. The beam is required to be expanded in a collimated beam of diameter 14 mm using a system of two convex lenses. If first lens has focal length 40 mm , then the focal length of second lens is $\_\_\_\_$ mm.
Explanation:
For a beam to enter collimated (parallel rays) and exit collimated with a larger diameter, the two lenses must be placed such that their focal points coincide.
The first lens focuses the parallel rays at its focal point ( $f_1$ ) and then the second lens is placed so that its focal point $\left(f_2\right)$ is at the same location. It then re-collimates the rays diverging from that point.
From similar triangles in the ray diagram, the ratio of the diameters of the beams is directly proportional to the ratio of the focal lengths of the lenses.
$ \frac{D_2}{D_1}=\frac{f_2}{f_1} $
Where :
$\mathrm{D}_1=$ Initial beam diameter $=2 \mathrm{~mm}$
$\mathrm{D}_2=$ Final beam diameter $=14 \mathrm{~mm}$
$\mathrm{f}_1=$ Focal length of the first lens $=40 \mathrm{~mm}$
$f_2=$ Focal length of the second lens (need to be calculated)
$\Rightarrow $$\mathrm{f}_2=\mathrm{f}_1 \times\left(\frac{\mathrm{D}_2}{\mathrm{D}_1}\right)$
$\Rightarrow $ $\mathrm{f}_2=40 \times\left(\frac{14}{2}\right)$
$\Rightarrow $ $ \mathrm{f}_2=280 \mathrm{~mm} $
Therefore, to expand the beam from a diameter of 2 mm to 14 mm using a first lens of 40 mm focal length, the second lens must have a focal length of 280 mm .
Hence, the correct answer is 280 .
A concave mirror of focal length 10 cm forms an image which is double the size of object when the object is placed at two different positions. The distance between the two positions of the object is
$\_\_\_\_$ cm.
Explanation:
For a mirror, magnification is given as $\mathrm{m}=-\frac{\mathrm{v}}{\mathrm{u}}$, where v is the image distance and u is the object distance.
From the mirror formula $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ :
$ \frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{u-f}{u f} \Rightarrow v=\frac{u f}{u-f} $
Substituting v into the magnification formula:
$ \mathrm{m}=-\frac{\mathrm{f}}{\mathrm{u}-\mathrm{f}}=\frac{\mathrm{f}}{\mathrm{f}-\mathrm{u}} $
A concave mirror ( $\mathrm{f}=-10 \mathrm{~cm}$ ) produces a magnification $|\mathrm{m}|=2$ in two cases:
Case 1 : Real Image ( $\mathrm{m}=-2$ )
$ -2=-\frac{10}{-10-\mathrm{u}_1} $
$\Rightarrow $ $ 20+2 \mathrm{u}_1=-10 \Rightarrow 2 \mathrm{u}_1=-30 \Rightarrow \mathrm{u}_1=-15 \mathrm{~cm} $
Case 2 : Virtual Image ( $\mathrm{m}=+2$ )
$ 2=-\frac{10}{-10-\mathrm{u}_2} $
$\Rightarrow $ $ -20-2 \mathrm{u}_2=-10 \Rightarrow-2 \mathrm{u}_2=10 \Rightarrow \mathrm{u}_2=-5 \mathrm{~cm} $
The distance between the two object positions is:
$ \Delta \mathrm{u}=\left|\mathrm{u}_1-\mathrm{u}_2\right|=|-15-(-5)|=|-10|=10 \mathrm{~cm} $
Therefore, the distance between the two positions is 10 cm .
If sunlight is focused on a paper using convex lens, it starts burning the paper in shortest time when the lens is kept at 30 cm above the paper. If the radius of curvature of the lens is 60 cm then the refractive index of the lens material is $\frac{\alpha}{10}$. The value of $\alpha$ is ________.
Explanation:
For the paper to burn in the shortest time, the image of the Sun must be formed exactly on the paper.
Since the Sun is very far away, the rays coming from it are nearly parallel. A convex lens brings parallel rays to its principal focus. So, the distance of the paper from the lens is equal to the focal length.
Thus,
$ f = 30 \text{ cm} $
Now use the lens maker’s formula for a thin convex lens in air:
$ \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) $
For a symmetric double convex lens,
$ R_1 = +60 \text{ cm}, \qquad R_2 = -60 \text{ cm} $
So,
$ \frac{1}{30} = (\mu - 1)\left(\frac{1}{60} - \left(-\frac{1}{60}\right)\right) $
$ \frac{1}{30} = (\mu - 1)\left(\frac{2}{60}\right) $
$ \frac{1}{30} = (\mu - 1)\left(\frac{1}{30}\right) $
Hence,
$ \mu - 1 = 1 $
$ \mu = 2 $
Given that refractive index is $\frac{\alpha}{10}$,
$ \frac{\alpha}{10} = 2 $
$ \alpha = 20 $
So, the value of $\alpha$ is
$ \boxed{20} $
A container contains a liquid with refractive index of 1.2 up to a height of 60 cm and another liquid having refractive index 1.6 is added to height H above first liquid. If viewed from above, the apparent shift in the position of bottom of container is 40 cm . The value of H is ________ cm . (Consider liquids are immisible)
Explanation:
$\begin{aligned} & y=\text { apparent depth of bottom } \\ & \frac{y}{1}=\frac{H}{1.6}+\frac{60}{1.2} \\ & \text { Shift }=40 \\ & H+60-y=40 \\ & H+60-\frac{\mathrm{H}}{1.6}-\frac{60}{1.2}=40 \\ & \frac{6}{16} \mathrm{H}=30 \\ & H=80 \mathrm{~cm} \end{aligned}$
Distance between object and its image (magnified by $-\frac{1}{3}$ ) is 30 cm . The focal length of the mirror used is $\left(\frac{x}{4}\right) \mathrm{cm}$, where magnitude of value of $x$ is _________.
Explanation:
$\begin{aligned} &\begin{aligned} & M=-\frac{1}{3} \\ & -\frac{-V}{-U}=\frac{-1}{3} \Rightarrow V=\frac{U}{3} \end{aligned}\\ &\text { Distance b/w object and image : } \end{aligned}$
$\begin{aligned} & U-V=30 \\ & U-\frac{\mu}{3}=30 \\ & \Rightarrow U=45 \quad V=15 \\ & \frac{1}{f}=\frac{1}{V}+\frac{1}{U}=-\frac{1}{15}-\frac{1}{45} \\ & \Rightarrow F=\frac{45}{4} \\ & x=45 \end{aligned}$
Light from a point source in air falls on a spherical glass surface (refractive index, $\mu=1.5$ and radius of curvature $=50 \mathrm{~cm}$ ). The image is formed at a distance of 200 cm from the glass surface inside the glass. The magnitude of distance of the light source from the glass surface is ___________m.
Explanation:
We apply the refraction formula at a spherical surface: μ₂/v − μ₁/u = (μ₂ − μ₁)/R.
Here:
• μ₁ = 1 (air)
• μ₂ = 1.5 (glass)
• R = 50 cm (radius of curvature)
• v = +200 cm (image distance inside glass)
• u = –x (object distance, negative for a real object in front of the surface)
Substitute these values and solve for x:
$\begin{aligned} & \frac{\mu_2}{\mathrm{v}}-\frac{\mu_1}{\mathrm{u}}=\frac{\mu_2-\mu_1}{\mathrm{R}} \\ & \frac{1.5}{200}-\frac{1}{-\mathrm{x}}=\frac{1.5-1}{50} \\ & \frac{1}{\mathrm{x}}=\frac{1}{100}-\frac{3}{400} \\ & \mathrm{x}=400 \mathrm{~cm} \\ & \mathrm{x}=4 \mathrm{~m} \end{aligned}$
So, the light source is 4 m from the glass surface.
Explanation:
To find the angle of incidence when a ray of light experiences minimum deviation in a prism, we use the formula for the refractive index $\mu$:
$ \mu = \frac{\sin \left(\frac{A + \delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)} $
Given:
The angle of the prism, $A = 60^{\circ}$
The refractive index of the prism material, $\mu = \sqrt{2}$
Minimum deviation, $\delta_m = 30^{\circ}$
At minimum deviation, the angle of incidence $i$ equals the angle of emergence $e$. Therefore, the relation between the angle of minimum deviation and the angle of the prism is represented as:
$ \delta_m = 2i - A $
Solving for $i$:
$ \delta_m = 2i - A \implies 30^{\circ} = 2i - 60^{\circ} \implies 2i = 90^{\circ} \implies i = 45^{\circ} $
Thus, the angle of incidence is $45^{\circ}$.
Two light beams fall on a transparent material block at point 1 and 2 with angle $\theta_1$ and $\theta_2$, respectively, as shown in figure. After refraction, the beams intersect at point 3 which is exactly on the interface at other end of the block. Given : the distance between 1 and 2, $\mathrm{d}=4 \sqrt{3} \mathrm{~cm}$ and $\theta_1=\theta_2=\cos ^{-1}\left(\frac{n_2}{2 n_1}\right)$. where refractive index of the block $n_2>$ refractive index of the outside medium $\mathrm{n}_1$, then the thickness of the block is ______ cm .
Explanation:
$\begin{aligned} & \mathrm{n}_1 \sin \left(90-\theta_1\right)=\mathrm{n}_2 \sin \theta_3 \\ & \mathrm{n}_1 \cos \theta_1=\mathrm{n}_2 \sin \theta_3 \\ & \mathrm{n}_1 \frac{\mathrm{n}_2}{2 \mathrm{n}_1}=\mathrm{n}_2 \sin \theta_3 \\ & \frac{1}{2}=\sin \theta_3, \theta_3=30 \\ & \tan 30=\frac{\mathrm{d}}{2(\mathrm{t})} \\ & \mathrm{t}=\frac{\mathrm{d} \sqrt{3}}{2}=\frac{4 \sqrt{3} \times \sqrt{3}}{2} \mathrm{~cm}=6 \mathrm{~cm} \end{aligned}$
The driver sitting inside a parked car is watching vehicles approaching from behind with the help of his side view mirror, which is a convex mirror with radius of curvature $\mathrm{R}=2 \mathrm{~m}$. Another car approaches him from behind with a uniform speed of $90 \mathrm{~km} / \mathrm{hr}$. When the car is at a distance of 24 m from him, the magnitude of the acceleration of the image of the car in the side view mirror is ' $a$ '. The value of $100 a$ is __________ $\mathrm{m} / \mathrm{s}^2$.
Explanation:
Given, R = 2m, So, f = 1m (as $f=\frac{R}{2}$)
${V_0} = 90\,km/hr = 90 \times {{1000} \over {3600}} = 25\,m/s$
$v_0$ is uniform, so ${a_0} = 0$
$u = - 24\,m$
Using mirror formula,
${1 \over v} + {1 \over u} = {1 \over f}$
$ \Rightarrow {1 \over v} + {1 \over { - 24}} = {1 \over 1} \Rightarrow {1 \over v} = 1 + {1 \over {24}} = {{25} \over {24}}$
$ \Rightarrow v = {{24} \over {25}}m$
$m = {{ - v} \over u}$
by differentiating mirror formula,
$ - {1 \over {{v^2}}}{{dv} \over {dt}} - {1 \over {{u^2}}}{{du} \over {dt}} = 0$
$ \Rightarrow - {1 \over {{v^2}}}{v_I} - {1 \over {{u^2}}}{v_0} = 0$ ...... (1)
$ \Rightarrow {v_I} = {{ - {v^2}} \over {{u^2}}}{v_0} = - {{{{24}^2}} \over {{{25}^2}}}$
$ \Rightarrow {v_I} = {{ - 1} \over {25}}$ m/s
by differentiating (1) w.r.t. time,
${{ - 1} \over {{v^2}}}{a_i} + {2 \over {{v^3}}}v_I^2 - {1 \over {{u^2}}}{a_0} + {2 \over {{u^3}}}v_0^2 = 0$
${a_I} = {2 \over v}v_I^2 + {{2{v^2}} \over {{u^3}}}v_0^2$
$ = {2 \over {24}}{1 \over {{{25}^2}}} + {2 \over { - {{24}^3}}}{{{{24}^2}} \over {{{25}^2}}}{25^2}$
$ = {1 \over {12}}\left[ {{1 \over {25}} - 1} \right] = {1 \over {12}} \times {{ - 24} \over {25}} = - {2 \over {25}}$
$ \Rightarrow 100\,{a_I} = {{ - 2} \over {25}} \times 100 = 8\,m/{s^2}$
A solid glass sphere of refractive index $n=\sqrt{3}$ and radius $R$ contains a spherical air cavity of radius $\frac{R}{2}$, as shown in the figure. A very thin glass layer is present at the point 0 so that the air cavity (refractive index $n=1$ ) remains inside the glass sphere. An unpolarized, unidirectional and monochromatic light source $S$ emits a light ray from a point inside the glass sphere towards the periphery of the glass sphere. If the light is reflected from the point 0 and is fully polarized, then the angle of incidence at the inner surface of the glass sphere is $\theta$. The value of $\sin \theta$ is ________.
Explanation:
The reflected light at O is "fully polarized," which means the incident angle at 0 is the Brewster angle. Brewster's angle occurs when the reflected and refracted rays are at $90^{\circ}$ to each other.
so Brewster angle is defined by
$ \tan \theta_B=\frac{n_2}{n_1} $
So $\tan \beta=\frac{\sqrt{3}}{1}$ (as at the point 0, light travels from air to glass so
$ \begin{aligned} & \text { light, travels } \left.n_2=\sqrt{3}, n_1=1\right) \\ & \Rightarrow \beta=60^{\circ} \end{aligned} $
$ \text { So } \angle D N O=60^{\circ}\left(\begin{array}{c} \text { As in } \triangle D N O, DO = DN=\frac{R}{2} \\ \text { so } \angle D O N=\angle D N O) \end{array}\right. $
Now, using snell's law at point $N$,
$ \begin{aligned} & n_1 \sin \theta_1=n_2 \sin \theta_2 \\ \Rightarrow & \sqrt{3} \sin \alpha=1 \times \sin 60^{\circ} \\ \Rightarrow & \sqrt{3} \sin \alpha=\frac{\sqrt{3}}{2} \\ \Rightarrow & \sin \alpha=\frac{1}{2} \Rightarrow \alpha=30^{\circ} \end{aligned} $
In $\triangle D O N, \angle D\ O N+\angle O N D+\angle N D O=180^{\circ}$
$ \begin{aligned} & 60^{\circ}+60^{\circ}+\angle N D O=180^{\circ} \\ & \Rightarrow \angle N D O=60^{\circ} \end{aligned} $
So $\angle C D N_{}=180^{\circ}-60^{\circ}=120^{\circ}$
$ C N=2 C P $
we can find $C P$ using $\triangle D P C$,
$ \begin{aligned} \quad \cos 30^{\circ} & =\frac{C P}{C D} \\ \Rightarrow \quad \frac{\sqrt{3}}{2} & =\frac{C P}{R / 2} \\ \Rightarrow \quad C P & =\frac{\sqrt{3} R}{4} \\ \text { So } \quad C N & =2\left(\frac{\sqrt{3} R}{4}\right)=\frac{\sqrt{3} R}{2} \end{aligned} $
Now, using sine rule in $\triangle C N M$,
$ \begin{aligned} & \frac{\sin \theta}{C N}=\frac{\sin 120^{\circ}}{R} \\ \Rightarrow \quad & \sin \theta=\frac{\sqrt{3}}{2} B\left(\frac{\sqrt{3}, R}{2}\right) \\ \Rightarrow \quad & \sin \theta=\frac{3}{4}=0.75 \quad \text { Ans. } \end{aligned} $
The refractive index of prism is $\mu=\sqrt{3}$ and the ratio of the angle of minimum deviation to the angle of prism is one. The value of angle of prism is _________$^\circ$.
Explanation:
To find the value of the angle of the prism given the refractive index of the prism $\mu = \sqrt{3}$ and the ratio of the angle of minimum deviation ($\delta_m$) to the angle of the prism ($A$) is 1 (i.e., $\frac{\delta_m}{A} = 1 \Rightarrow \delta_m = A$), we can use the prism formula relating these variables.
The formula that relates the angle of deviation $\delta$, refractive index $\mu$, angle of prism $A$, and angle of minimum deviation $\delta_m$ is given by:
$\mu = \frac{\sin \left(\frac{\delta_m + A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$.
Given that $\delta_m = A$, our formula becomes:
$\sqrt{3} = \frac{\sin \left(\frac{A + A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
Simplifying this, we have:
$\sqrt{3} = \frac{\sin(A)}{\sin \left(\frac{A}{2}\right)}$
Now, recall the trigonometric identity:
$\sin(2\theta) = 2\sin(\theta)\cos(\theta)$
By setting $\theta = \frac{A}{2}$, we get:
$\sin(A) = 2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)$
Substituting back into our equation:
$\sqrt{3} = \frac{2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$
$\sqrt{3} = 2\cos\left(\frac{A}{2}\right)$
Dividing by 2 and solving for $\cos\left(\frac{A}{2}\right)$:
$\cos\left(\frac{A}{2}\right) = \frac{\sqrt{3}}{2}$
This corresponds to an angle $\frac{A}{2}$ of $30^\circ$ since the cosine of 30 degrees is $\frac{\sqrt{3}}{2}$. Thus:
$\frac{A}{2} = 30^\circ$
So, the angle of the prism $A$ is:
$A = 2 \times 30^\circ = 60^\circ$.
Therefore, the value of the angle of the prism is $60^\circ$.
A light ray is incident on a glass slab of thickness $4 \sqrt{3} \mathrm{~cm}$ and refractive index $\sqrt{2}$ The angle of incidence is equal to the critical angle for the glass slab with air. The lateral displacement of ray after passing through glass slab is ______ $\mathrm{cm}$.
(Given $\sin 15^{\circ}=0.25$)
Explanation:
$\begin{aligned} & \mu=\sqrt{2} \\ & \sin \theta_C=\frac{1}{\sqrt{2}} \\ & Q_C=45^{\circ} \\ & i=Q_C=45^{\circ} \end{aligned}$
$\begin{aligned} & \text { ( } \phi \text { ) lateral displacement }=\frac{t \sin (i-r)}{\cos r} \\ & \sin 45^{\circ}=\sqrt{2} \sin r \\ & \Rightarrow \quad r=30^{\circ} \\ & \therefore \quad d=\frac{4 \sqrt{3} \sin \left(45^{\circ}-30^{\circ}\right)}{\cos 30^{\circ}} \\ & =\frac{4 \sqrt{3} \times \frac{1}{4}}{\frac{\sqrt{3}}{2}}=2 \end{aligned}$
Explanation:
$\begin{aligned} & \mathrm{v}=3 \mathrm{u} \\\\ & \mathrm{v}-\mathrm{u}=20 \mathrm{~cm} \\\\ & 2 \mathrm{u}=20 \mathrm{~cm} \\\\ & \mathrm{u}=10 \mathrm{~cm}\end{aligned}$
$\begin{aligned} & \frac{1}{(-30)}-\frac{1}{(-10)}=\frac{1}{f} \\\\ & f=15 \mathrm{~cm}\end{aligned}$
Light from a point source in air falls on a convex curved surface of radius $20 \mathrm{~cm}$ and refractive index 1.5. If the source is located at $100 \mathrm{~cm}$ from the convex surface, the image will be formed at ________ $\mathrm{cm}$ from the object.
Explanation:
In the problem, you're dealing with refraction at a convex surface. The light is coming from a point source located in air (with a refractive index, $ \mu_1 = 1.0 $) and entering a medium with a refractive index of $ \mu_2 = 1.5 $. The convex surface has a radius of curvature $ R = 20 \, \text{cm} $, and the source is placed $ 100 \, \text{cm} $ from the convex surface.
The formula used to find the image distance $ v $ due to refraction at a spherical surface is:
$ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} $
Substituting the given values:
$ \frac{1.5}{v} - \frac{1}{-100} = \frac{1.5 - 1}{20} $
This equation allows us to solve for $ v $, the distance from the convex surface to the image. Upon solving, we found that $ v = 100 \, \text{cm} $, which means the image forms $ 100 \, \text{cm} $ on the other side of the convex surface, away from the point of refraction.
The distance from the object to the image isn't just $ v $, the distance from the surface to where the image forms. Since the object is $ 100 \, \text{cm} $ from the convex surface and the image also forms $ 100 \, \text{cm} $ from the convex surface but on the opposite side, the total distance between the object and the image is the sum of these distances:
- Distance from the object to the convex surface: $ 100 \, \text{cm} $
- Distance from the convex surface to the image: $ 100 \, \text{cm} $
Therefore, the total distance between the object and the image is $ 100 \, \text{cm} + 100 \, \text{cm} = 200 \, \text{cm} $.
This calculation accounts for the physical layout where the object and the image are on opposite sides of the convex surface, and to determine the distance between them, you sum the distances from each to the surface. This results in the image being formed $ 200 \, \text{cm} $ from the object, which means if you were to measure directly from the object to its image, the total distance covered would be $ 200 \, \text{cm} $.
In an experiment to measure the focal length $(f)$ of a convex lens, the magnitude of object distance $(x)$ and the image distance $(y)$ are measured with reference to the focal point of the lens. The $y$-$x$ plot is shown in figure.
The focal length of the lens is ________ $\mathrm{cm}$.
Explanation:
$\begin{aligned} & \frac{1}{f+20}-\frac{1}{-(f+20)}=\frac{1}{f} \\ & \frac{2}{f+20}=\frac{1}{f} \quad f=20 \mathrm{~cm} \end{aligned}$
Or $\mathrm{x}_1 \mathrm{x}_2=\mathrm{f}^2$ gives $\mathrm{f}=20 \mathrm{~cm}$
The distance between object and its two times magnified real image as produced by a convex lens is $45 \mathrm{~cm}$. The focal length of the lens used is _______ cm.
Explanation:
$\begin{aligned} & \frac{v}{u}=-2 \\ & v=-2 u \quad\text{... (i)} \\ & v-u=45 \quad\text{... (ii)}\\ & \Rightarrow u=-15 \mathrm{~cm} \\ & v=30 \mathrm{~cm} \\ & \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ & f=+10 \mathrm{~cm} \end{aligned}$
Two immiscible liquids of refractive indices $\frac{8}{5}$ and $\frac{3}{2}$ respectively are put in a beaker as shown in the figure. The height of each column is $6 \mathrm{~cm}$. A coin is placed at the bottom of the beaker. For near normal vision, the apparent depth of the coin is $\frac{\alpha}{4} \mathrm{~cm}$. The value of $\alpha$ is _________.
Explanation:
$h_{\text {app }}=\frac{h_1}{\mu_1}+\frac{h_2}{\mu_2}=\frac{6}{3 / 2}+\frac{6}{8 / 5}=4+\frac{15}{4}=\frac{31}{4} \mathrm{~cm}$
Two equilateral-triangular prisms $\mathrm{P}_1$ and $\mathrm{P}_2$ are kept with their sides parallel to each other, in vacuum, as shown in the figure. A light ray enters prism $\mathrm{P}_1$ at an angle of incidence $\theta$ such that the outgoing ray undergoes minimum deviation in prism $\mathrm{P}_2$. If the respective refractive indices of $\mathrm{P}_1$ and $\mathrm{P}_2$ are $\sqrt{\frac{3}{2}}$ and $\sqrt{3}$, then $\theta=\sin ^{-1}\left[\sqrt{\frac{3}{2}} \sin \left(\frac{\pi}{\beta}\right)\right]$, where the value of $\beta$ is ____.
Explanation:
At surface BC
$\begin{aligned} & \sqrt{\frac{3}{2}} \sin r_2=\sqrt{3} \sin 30 \\ & \sqrt{\frac{3}{2}} \sin r_2=\frac{\sqrt{3}}{2} \\ & \operatorname{sinr}_2=\frac{1}{\sqrt{2}} \\ & r_2=45^{\circ} \\ & r_1=60^{\circ}-45^{\circ}=15^{\circ} \end{aligned}$
At surface $A B$
$\begin{gathered} 1 \sin \theta=\sqrt{\frac{3}{2}} \sin 15^{\circ} \\ \theta=\sin ^{-1}\left[\sqrt{\frac{3}{2}} \sin \frac{\pi}{12}\right] \\ \beta=12 \end{gathered}$
Explanation:
$n = \frac{\sin \frac{A + \delta_m}{2}}{\sin \frac{A}{2}}$
In this case, the refractive index $n = \sqrt{2}$, and since the prism is equilateral, the prism angle $A = 60^\circ$. Now, we can substitute these values into the formula and solve for the angle of minimum deviation $\delta_m$:
$\sqrt{2} = \frac{\sin \frac{60^\circ + \delta_m}{2}}{\sin \frac{60^\circ}{2}}$
First, let's find the sine of half the prism angle:
$\sin \frac{60^\circ}{2} = \sin 30^\circ = \frac{1}{2}$
Now, we can substitute this value into the formula:
$\sqrt{2} = \frac{\sin \frac{60^\circ + \delta_m}{2}}{\frac{1}{2}}$
To isolate the sine term, we multiply both sides by $\frac{1}{2}$:
$\frac{\sqrt{2}}{2} = \sin \frac{60^\circ + \delta_m}{2}$
Now, we can find the angle inside the sine function:
$\frac{60^\circ + \delta_m}{2} = \sin^{-1} \frac{\sqrt{2}}{2} = 45^\circ$
Finally, we can solve for the angle of minimum deviation $\delta_m$:
$60^\circ + \delta_m = 2 \cdot 45^\circ$
$\delta_m = 90^\circ - 60^\circ = 30^\circ$
The angle of minimum deviation for the liquid in the equilateral hollow prism is $30^\circ$.
A bi convex lens of focal length $10 \mathrm{~cm}$ is cut in two identical parts along a plane perpendicular to the principal axis. The power of each lens after cut is ____________ D.
Explanation:
$\frac{1}{f} = (n - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
For the original biconvex lens, both radii of curvature have the same magnitude but opposite signs, so let's denote them as ±R. The focal length of the original lens is given as 10 cm, and the refractive index (n) is constant for both the original lens and the new plano-convex lenses.
For the original lens, the lensmaker's formula becomes:
$\frac{1}{f} = (n - 1) \left(\frac{1}{R} - \frac{1}{-R}\right)$
Since the focal length is given as 10 cm, we can plug in the value:
$\frac{1}{10} = (n - 1) \left(\frac{1}{R} + \frac{1}{R}\right)$
$\frac{1}{10} = (n - 1) \left(\frac{2}{R}\right)$
For each plano-convex lens, one radius of curvature (R1) is infinite (the flat side), and the other radius (R2) is the same as the original lens (R). The lensmaker's formula for the plano-convex lens becomes:
$\frac{1}{f'} = (n - 1) \left(\frac{1}{\infty} - \frac{1}{R}\right)$
$\frac{1}{f'} = (n - 1) \left(-\frac{1}{R}\right)$
Now, we can substitute the expression for (n-1)(2/R) from the original lens equation:
$\frac{1}{f'} = \frac{1}{10} \cdot \frac{1}{2}$ $\frac{1}{f'} = \frac{1}{20}$
So, the focal length of each plano-convex lens (f') is 20 cm. To find the power of each lens, we can use the formula:
$P (\text{in diopters}) = \frac{1}{f (\text{in meters})}$
Converting the focal length to meters and calculating the power:
$P = \frac{1}{0.2}$
$P = 5 \text{ D}$
So, the power of each plano-convex lens after the cut is 5 diopters (D).
A fish rising vertically upward with a uniform velocity of $8 \mathrm{~ms}^{-1}$, observes that a bird is diving vertically downward towards the fish with the velocity of $12 \mathrm{~ms}^{-1}$. If the refractive index of water is $\frac{4}{3}$, then the actual velocity of the diving bird to pick the fish, will be __________ $\mathrm{ms}^{-1}$.
Explanation:
The fish sees the bird diving with a velocity of $12 \ \text{m/s}$. We can write the equation considering the velocities relative to the fish:
$\frac{V_{\text{b/f}}}{\frac{4}{3}} = \frac{-8}{\frac{4}{3}} + \frac{-v}{1}$
Here, $V_{\text{b/f}}$ is the bird's diving velocity relative to the fish, and $v$ is the actual velocity of the bird.
Now, let's solve for $v$:
$\frac{-12}{\frac{4}{3}} = \frac{-8}{\frac{4}{3}} - v$
$v = 3 \ \text{m/s}$
So, the actual velocity of the diving bird to pick the fish relative to the fish is $3 \ \text{m/s}$.
Two convex lenses of focal length $20 \mathrm{~cm}$ each are placed coaxially with a separation of $60 \mathrm{~cm}$ between them. The image of the distant object formed by the combination is at _____________ $\mathrm{cm}$ from the first lens.
Explanation:
1. First refraction in L1 (lens 1):
When considering the first lens (L1), the object is at infinity, so the image I1 formed by this lens is at its focal point. Using the lensmaker's equation:
$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$
Given that the object is at infinity, $d_o = \infty$, and the focal length of the first lens is $f = 20 \mathrm{~cm}$. Plugging in these values, we get:
$\frac{1}{20} = \frac{1}{\infty} + \frac{1}{d_i}$
As $\frac{1}{\infty}$ is essentially 0, we have:
$\frac{1}{20} = \frac{1}{d_i}$
This implies that $d_i = 20 \mathrm{~cm}$, meaning that the image I1 is formed 20 cm from the first lens L1.
2. Second refraction in L2 (lens 2):
Now, the image I1 formed by L1 becomes the object for L2. The distance between the lenses is 60 cm, so the object distance for L2 (u) is -40 cm, because the object is to the left of the lens (u is negative). The focal length of L2 (f) is 20 cm. We can use the lensmaker's equation to find the image distance (v) for L2:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Plugging in the values:
$\frac{1}{v} - \frac{1}{(-40)} = \frac{1}{20}$
$\frac{1}{v} = \frac{1}{20} - \frac{1}{40} = \frac{2 - 1}{40}$
$\frac{1}{v} = \frac{1}{40}$
Therefore, $v = 40 \mathrm{~cm}$
Since the image distance (v) is positive, the image I2 is formed on the right side of L2 at a distance of 40 cm. To find the distance of the final image from L1, we add the distance between the lenses (60 cm) and the image distance (v) from L2.
Final image distance from L1 = 60 cm + 40 cm = 100 cm
Thus, the correct answer is 100 cm.
As shown in the figure, a plane mirror is fixed at a height of $50 \mathrm{~cm}$ from the bottom of tank containing water $\left(\mu=\frac{4}{3}\right)$. The height of water in the tank is $8 \mathrm{~cm}$. A small bulb is placed at the bottom of the water tank. The distance of image of the bulb formed by mirror from the bottom of the tank is ___________ $\mathrm{cm}$.
Explanation:
$ \text { Apparent depth of } \mathrm{O}=\frac{\mathrm{d}}{\mu} $$ =\frac{8}{\frac{4}{3}}=6 \mathrm{~cm} $
Distance of object from mirror = 42 + 6 = 48 m
$ \text { Distance between } \mathrm{O} \text { and } \mathrm{I}_2=48+50=98 \mathrm{~cm} $
The radius of curvature of each surface of a convex lens having refractive index 1.8 is $20 \mathrm{~cm}$. The lens is now immersed in a liquid of refractive index 1.5 . The ratio of power of lens in air to its power in the liquid will be $x: 1$. The value of $x$ is _________.
Explanation:
Let's find the focal length of the lens in air and in the liquid. We will use the lens maker's formula:
$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
where $f$ is the focal length, $\mu$ is the refractive index of the lens material, and $R_1$ and $R_2$ are the radii of curvature of the lens surfaces.
Since the lens is convex, both surfaces have the same radius of curvature (positive), so $R_1 = R_2 = 20 \mathrm{~cm}$.
First, let's find the focal length of the lens in air:
$\frac{1}{f_\text{air}} = (1.8 - 1)\left(\frac{1}{20} - \frac{1}{20}\right) = 0.8\left(\frac{1}{20}\right)$
$f_\text{air} = \frac{1}{0.8\left(\frac{1}{20}\right)} = 25 \mathrm{~cm}$
Now, let's find the focal length of the lens in the liquid. The relative refractive index of the lens with respect to the liquid is:
$\mu_\text{rel} = \frac{1.8}{1.5} = 1.2$
$\frac{1}{f_\text{liquid}} = (1.2 - 1)\left(\frac{1}{20}\right)$
$f_\text{liquid} = \frac{1}{0.2\left(\frac{1}{20}\right)} = 100 \mathrm{~cm}$
The power of a lens is given by:
$P = \frac{1}{f}$
Now, we can find the ratio of the power of the lens in air to its power in the liquid:
$\frac{P\text{air}}{P\text{liquid}} = \frac{f\text{liquid}}{f\text{air}} = \frac{100}{25} = 4$
So, the ratio of the power of the lens in air to its power in the liquid is $x:1$, where $x = 4$.
A point object, 'O' is placed in front of two thin symmetrical coaxial convex lenses $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ with focal length $24 \mathrm{~cm}$ and $9 \mathrm{~cm}$ respectively. The distance between two lenses is $10 \mathrm{~cm}$ and the object is placed $6 \mathrm{~cm}$ away from lens $\mathrm{L}_{1}$ as shown in the figure. The distance between the object and the image formed by the system of two lenses is __________ $\mathrm{cm}$.
Explanation:
$\frac{1}{\mathrm{v}}+\frac{1}{6}=\frac{1}{24}$
$ \begin{aligned} \frac{1}{\mathrm{v}} & =\frac{1}{24}-\frac{1}{6}=-\frac{1}{8} \\\\ \mathrm{v} & =-8 \mathrm{~cm} \end{aligned} $
From IInd lens :
$\frac{1}{\mathrm{v}}+\frac{1}{18}=\frac{1}{9}$
$ \begin{aligned} \frac{1}{\mathrm{v}} & =\frac{1}{18} \\\\ \mathrm{v} & =18 \end{aligned} $
So distance between object and its image
= 6 + 10 + 18 = 34 cm
Two transparent media having refractive indices 1.0 and 1.5 are separated by a spherical refracting surface of radius of curvature $30 \mathrm{~cm}$. The centre of curvature of surface is towards denser medium and a point object is placed on the principle axis in rarer medium at a distance of $15 \mathrm{~cm}$ from the pole of the surface. The distance of image from the pole of the surface is ____________ $\mathrm{cm}$.
Explanation:
The refraction at a spherical surface is governed by the formula:
$\frac{1}{v} - \frac{1}{u} = \frac{n_2 - n_1}{R} n_2$
where:
- (v) is the image distance,
- (u) is the object distance,
- ($n_1$) is the refractive index of the medium where the object is,
- ($n_2$) is the refractive index of the medium where the image is formed,
- (R) is the radius of curvature of the refracting surface.
Here, we have:
- (u = -15) cm (the object is on the same side as the light is coming from, so the distance is taken as negative),
- ($n_1$ = 1.0) (refractive index of the rarer medium),
- ($n_2$ = 1.5) (refractive index of the denser medium),
- (R = -30) cm (the center of curvature is in the denser medium, so the radius is taken as negative).
Substituting these values into the formula, we get:
$\frac{1}{v} - \left(-\frac{1}{15}\right) = \frac{1.5 - 1.0}{-30} \cdot 1.5$
Solving for (v), we get:
$v = \frac{1}{\frac{1}{15} + \frac{1.5}{30}} = -30 \, \text{cm}$
Therefore, the image is formed at a distance of 30 cm from the pole of the surface, on the same side as the object. The negative sign indicates that the image is virtual and is formed on the same side of the surface as the light is coming from.
Two vertical parallel mirrors A and B are separated by $10 \mathrm{~cm}$. A point object $\mathrm{O}$ is placed at a distance of $2 \mathrm{~cm}$ from mirror $\mathrm{A}$. The distance of the second nearest image behind mirror A from the mirror $\mathrm{A}$ is _________ $\mathrm{cm}$.
Explanation:
Object distance $=8 \mathrm{~cm}$ [For image by $B]$
$ \begin{aligned} \Rightarrow \text { Required distance } & =8+8+2 \mathrm{~cm} \\\\ & =18 \mathrm{~cm} \end{aligned} $
A pole is vertically submerged in swimming pool, such that it gives a length of shadow $2.15 \mathrm{~m}$ within water when sunlight is incident at angle of $30^{\circ}$ with the surface of water. If swimming pool is filled to a height of $1.5 \mathrm{~m}$, then the height of the pole above the water surface in centimeters is $\left(n_{w}=4 / 3\right)$ ____________.
Explanation:
The pole is vertically submerged in the swimming pool, and the length of the shadow is due to the sunlight that is incident at an angle of $30^{\circ}$ with the surface of water. Hence, the angle of incidence, $i$, is $60^{\circ}$ (since the angle of incidence is measured from the normal to the surface, and the normal is perpendicular to the surface).
Using Snell's law, we find the angle of refraction, $r$, as you correctly did:
$\sin r = \frac{n_{1}}{n_{2}} \sin i = \frac{3}{4} \sin 60^{\circ} = \frac{3 \sqrt{3}}{8}$
This gives $\tan r = \frac{3 \sqrt{3}}{\sqrt{37}}$.
Now, the shadow of the pole in the water forms a right triangle, with the submerged part of the pole as one side, the shadow as the hypotenuse, and the line segment from the water surface to the end of the shadow as the other side. The angle at the water surface is $r$. Thus, we can write the following relationships:
The length of the submerged part of the pole (which I'll call $x$), is given by:
$x = 2.15 \, \text{m} \cdot \cos r$
And the length of the line segment from the water surface to the end of the shadow (which I'll call $y$), is given by:
$y = 2.15 \, \text{m} \cdot \sin r$
The total length of the pole above the water surface is then the sum of $x$ and the depth of the swimming pool (1.5 m), which gives us the equation:
$x \sqrt{3} + 1.5 \, \text{m} \cdot \tan r = 2.15 \, \text{m}$
Solving this for $x$ gives us:
$x = \frac{2.15 \, \text{m}}{\sqrt{3}} - \frac{1.5 \, \text{m} \cdot 3}{\sqrt{37}} = 0.502 \, \text{m} = 50.2 \, \text{cm}$
So, the height of the pole above the water surface is approximately 50 cm.
A thin cylindrical rod of length $10 \mathrm{~cm}$ is placed horizontally on the principle axis of a concave mirror of focal length $20 \mathrm{~cm}$. The rod is placed in a such a way that mid point of the rod is at $40 \mathrm{~cm}$ from the pole of mirror. The length of the image formed by the mirror will be $\frac{x}{3} \mathrm{~cm}$. The value of $x$ is _____________.
Explanation:
$\text { A: }$
$ \begin{aligned} & \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\\\ & \Rightarrow \frac{1}{v}+\frac{1}{-45}=\frac{1}{-20} \\\\ & \Rightarrow \frac{1}{v}=\frac{1}{45}-\frac{1}{20}=\frac{4-9}{180}=-\frac{1}{36} \\\\ & \Rightarrow v=-36 \mathrm{~cm} \end{aligned} $
B: $\frac{1}{v}+\frac{1}{-35}=\frac{1}{-20}$
$\Rightarrow \frac{1}{v}=\frac{1}{35}-\frac{1}{20}=\frac{4-7}{140}$
$\Rightarrow \quad v=-\frac{140}{3}$
$\Rightarrow$ length of image $=\frac{140}{3}-36=\frac{32}{3} \mathrm{~cm}$
$\Rightarrow x=32$
In an experiment for estimating the value of focal length of converging mirror, image of an object placed at $40 \mathrm{~cm}$ from the pole of the mirror is formed at distance $120 \mathrm{~cm}$ from the pole of the mirror. These distances are measured with a modified scale in which there are 20 small divisions in $1 \mathrm{~cm}$. The value of error in measurement of focal length of the mirror is $\frac{1}{\mathrm{~K}} \mathrm{~cm}$. The value of $\mathrm{K}$ is __________.
Explanation:
${1 \over v} + {1 \over u} = {1 \over f}$ ...... (1)
$ \Rightarrow - {1 \over {{f^2}}}df = - {1 \over {{v^2}}}dv - {1 \over {{u^2}}}du$
$ \Rightarrow {{df} \over {{f^2}}} = {{dv} \over {{v^2}}} + {{du} \over {{u^2}}}$ ..... (2)
From (1) : $ - {1 \over {120}} - {1 \over {40}} = {1 \over f} \Rightarrow f = - 30$ cm
Also, least count $ = {{1\,\mathrm{cm}} \over {20}} = 0.05$ cm
$ \Rightarrow df = \left[ {{{0.05} \over {{{120}^2}}} + {{0.05} \over {{{40}^2}}}} \right] \times {30^2}$
$ = 0.05\left[ {{1 \over {16}} + {9 \over {16}}} \right] = {5 \over 8} \times {5 \over {100}} = {1 \over {32}}$ cm
$ \Rightarrow k = 32$
In an experiment of measuring the refractive index of a glass slab using travelling microscope in physics lab, a student measures real thickness of the glass slab as 5.25 mm and apparent thickness of the glass slab as 5.00 mm. Travelling microscope has 20 divisions in one cm on main scale and 20 divisions on vernier scale is equal to 49 divisions on main scale. The estimated uncertainty in the measurement of refractive index of the slab is $\frac{x}{10}\times10^{-3}$, where $x$ is ___________
Explanation:
$\mu=\frac{\mathrm{real\,depth}\,(l_1)}{\mathrm{apparent\,depth}\,(l_2)}$
$=\frac{5.25}{5}=1.05$
$\frac{d\mu}{\mu}=\frac{dl_1}{l_1}+\frac{dl_2}{l_2}$
$d\mu = \left( {{{d{l_1}} \over {{l_1}}} + {{d{l_2}} \over {{l_2}}}} \right)\mu $
$ = \left( {{{0.01} \over {5.25}} + {{0.01} \over {5.00}}} \right) \times 1.05$
$ = {{41} \over {10}} \times {10^{ - 3}}$
so $x = 41$
An object is placed on the principal axis of convex lens of focal length 10cm as shown. A plane mirror is placed on the other side of lens at a distance of 20 cm. The image produced by the plane mirror is 5cm inside the mirror. The distance of the object from the lens is ___________ cm.
Explanation:
$I_{1}$ is image formed by lens and $I_{2}$ is image formed by mirror.
Location of $I_{1}$ and $I_{2}$ from mirror will be equal $=5 \mathrm{~cm}$ Hence $\mathrm{I}_{1}=15 \mathrm{~cm}$ from lens
From $\frac{1}{v}-\frac{1}{u}=\frac{1}{10}$
$ u=-x, v=15 $
$\frac{1}{x}=\frac{1}{10}-\frac{1}{15} \Rightarrow x=30 \mathrm{~cm}$
A ray of light is incident from air on a glass plate having thickness $\sqrt3$ cm and refractive index $\sqrt2$. The angle of incidence of a ray is equal to the critical angle for glass-air interface. The lateral displacement of the ray when it passes through the plate is ____________ $\times$ 10$^{-2}$ cm. (given $\sin 15^\circ = 0.26$)
Explanation:
$\Rightarrow$ at point (1)
$\mu \sin r=\sin i=\frac{1}{\sqrt{2}}$
$\sin r=\frac{1}{2} \quad \Rightarrow \quad r=30^{\circ}$
Lateral displacement
$=\frac{t}{\cos r} \sin \left(15^{\circ}\right)=\frac{\sqrt{3}}{\left(\frac{\sqrt{3}}{2}\right)} \times 0.26$
$=2 \times 0.26$
$=0.52 \mathrm{~cm}$ $=52 \times 10^{-2} \mathrm{~cm}$
A convex lens of refractive index 1.5 and focal length 18cm in air is immersed in water. The change in focal length of the lens will be ___________ cm.
(Given refractive index of water $=\frac{4}{3}$)
Explanation:
$ \frac{1}{f}=\left(\frac{\mu_{\text {lens }}}{\mu_{\text {mrdium }}}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right] $
when in air
$ \frac{1}{18}=\left(\frac{1.5}{1}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]\quad...(1) $
$\mu_{\text {lense }}=1.5, \mu_{\text {air }}=1$.
when in water
$ \frac{1}{f}=\left(\frac{1.5}{4 / 3}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]\quad...(2) $
from (1) & (2)
$f=72$
Change in focal length $=72-18 = 54$
As shown in the figure, a combination of a thin plano concave lens and a thin plano convex lens is used to image an object placed at infinity. The radius of curvature of both the lenses is 30 cm and refraction index of the material for both the lenses is 1.75. Both the lenses are placed at distance of 40 cm from each other. Due to the combination, the image of the object is formed at distance $x=$ _________ cm, from concave lens.
Explanation:
$f_{\text {concave }}=-40 \mathrm{~cm}$
$\frac{1}{f_{\text {convex }}}=(1.75-1)\left(\frac{1}{30}-\frac{1}{\infty}\right)=\frac{0.75}{30}$
$f_{\text {convex }}=40 \mathrm{~cm}$
Let the first image is formed at $v_{1}$ so
$\frac{1}{v_{1}}-\frac{1}{\infty}=\frac{1}{f_{\text {concave }}}=-\frac{1}{40}$
$\Rightarrow v_{1}=-40 \mathrm{~cm}$
for second image
$\frac{1}{x-40}-\frac{1}{-80}=\frac{1}{40}$
$\Rightarrow x=120 \mathrm{~cm}$
Explanation:
Given that the object is positioned at the focal point, the resulting light rays will transition to a parallel alignment.
Upon the first reflection at Mirror $ M_1 $:
As the incoming light rays are parallel, the resultant image will be formed at the focal point.
$\mathrm{II}^{\text {nd }}$ refraction from $\mathrm{L}$ :-
$ \begin{aligned} & \mathrm{u}=-(\mathrm{d}-10) \\\\ & \mathrm{f}=10 \mathrm{~cm} \\\\ & \frac{1}{\mathrm{v}}-\frac{1}{\mu}=\frac{1}{\mathrm{f}} \\\\ & \frac{1}{\mathrm{v}}+\frac{1}{\mathrm{~d}-10}=\frac{1}{10} \\\\ & \frac{1}{\mathrm{v}}=\frac{1}{10}-\frac{1}{(\mathrm{~d}-10)} .......(i) \end{aligned} $
This $v$ will be object for $M_2$, and image should be at $10 \mathrm{~cm}$
$ \begin{aligned} & \frac{1}{\mu}+\frac{1}{v_1}=\frac{1}{f} \\\\ & -\frac{1}{(20-v)}-\frac{1}{10}=-\frac{1}{12} \\\\ & \frac{1}{12}-\frac{1}{10}=\frac{1}{20-v} \\\\ & -\frac{2}{120}=\frac{1}{20-v} \\\\ & 20-v=-60 \\\\ & v=80 \mathrm{~cm} \end{aligned} $
From equation (i), we get
$ \begin{aligned} & \frac{1}{80}=\frac{1}{10}-\frac{1}{d-10} \\\\ & \frac{1}{d-10}=\frac{1}{10}-\frac{1}{80} \\\\ & \frac{1}{d-10}=\frac{80-10}{800}=\frac{70}{800} \\\\ & d-10=\frac{80}{7} \Rightarrow d=10+\frac{80}{7}=\frac{150}{7} \\\\ & n=150 \end{aligned} $
Case-2 : If $1^{\text {st }}$ reflection on mirror $m_2$ :
For $\mathrm{m}_2$
$ \begin{aligned} & \frac{1}{\mathrm{~V}_1}+\frac{1}{-10}=\frac{1}{-12} \\\\ & \mathrm{~V}_1=60 \mathrm{~cm} \end{aligned} $
Then refraction on lens $\mathrm{L}$
$ \begin{aligned} & \mathrm{u}_2=80 \mathrm{~cm} \\\\ & \frac{1}{\mathrm{~V}_2}-\frac{1}{-60}=\frac{1}{10} \\\\ & \mathrm{~V}_2=\frac{80}{7} \end{aligned} $
Then reflection on $\mathrm{m}_2$
Either $V_2$ is at centre (normal incidence)
$ \begin{aligned} & \mathrm{d}-\frac{80}{7}=20 \\\\ & \mathrm{~d}=\frac{220}{7} \\\\ & \frac{\mathrm{n}}{7}=\frac{220}{7} \\\\ & \mathrm{n}=220 \end{aligned} $
$V_2$ is at pole of $m_2$
$ \begin{aligned} & \mathrm{d}-\frac{80}{7}=0 \\\\ & \mathrm{~d}=\frac{80}{7} \\\\ & \frac{\mathrm{n}}{7}=\frac{80}{7} \\\\ & \mathrm{n}=80 \end{aligned} $
The X-Y plane be taken as the boundary between two transparent media $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$. $\mathrm{M}_{1}$ in $Z \geqslant 0$ has a refractive index of $\sqrt{2}$ and $M_{2}$ with $Z<0$ has a refractive index of $\sqrt{3}$. A ray of light travelling in $\mathrm{M}_{1}$ along the direction given by the vector $\overrightarrow{\mathrm{P}}=4 \sqrt{3} \hat{i}-3 \sqrt{3} \hat{j}-5 \hat{k}$, is incident on the plane of separation. The value of difference between the angle of incident in $\mathrm{M}_{1}$ and the angle of refraction in $\mathrm{M}_{2}$ will be __________ degree.
Explanation:
Normal will be $ - \widehat k$ so
$\cos i = {{\overleftarrow P \,.\,\widehat n} \over {\left| {\overleftarrow P } \right|\,.\,\left| {\widehat n} \right|}}$
${5 \over {10}} = {1 \over 2}$
$ \Rightarrow i = 60^\circ $
and using Snells law
$\sqrt 2 \sin 60^\circ = \sqrt 3 \sin r$
${{\sqrt 3 } \over {\sqrt 2 }} = \sqrt 3 \sin r$
$ \Rightarrow r = 45^\circ $
So, $i - r = 15^\circ $
An object 'O' is placed at a distance of $100 \mathrm{~cm}$ in front of a concave mirror of radius of curvature $200 \mathrm{~cm}$ as shown in the figure. The object starts moving towards the mirror at a speed $2 \mathrm{~cm} / \mathrm{s}$. The position of the image from the mirror after $10 \mathrm{~s}$ will be at _________ $\mathrm{cm}$.
Explanation:
The object after 10 second will be at $u = - 80$ cm.
So ${1 \over v} - {1 \over {80}} = - {1 \over {100}} $
$\Rightarrow v = {{8000} \over { + 20}} = 400$ cm
In an experiment with a convex lens, The plot of the image distance $\left(v^{\prime}\right)$ against the object distance ($\left.\mu^{\prime}\right)$ measured from the focus gives a curve $v^{\prime} \mu^{\prime}=225$. If all the distances are measured in $\mathrm{cm}$. The magnitude of the focal length of the lens is ___________ cm.
Explanation:
Using Newton's formula for lenses,
$v'\mu ' = {f^2} = 225 \Rightarrow f = 15$
A thin prism of angle $6^{\circ}$ and refractive index for yellow light $\left(\mathrm{n}_{\mathrm{Y}}\right) 1.5$ is combined with another prism of angle $5^{\circ}$ and $\mathrm{n}_{\mathrm{Y}}=1.55$. The combination produces no dispersion. The net average deviation $(\delta)$ produced by the combination is $\left(\frac{1}{x}\right)^{\circ}$. The value of $x$ is ____________.
Explanation:
${\delta _{net}} = {\delta _1} + {\delta _2}$
$ = |({\mu _1} - 1){A_1} - ({\mu _2} - 1){A_2}|$
$ = |3^\circ - 2.75^\circ |$
${\delta _{net}} = {{1^\circ } \over 4}$
$ \Rightarrow x = 4$
In the given figure, the face $A C$ of the equilateral prism is immersed in a liquid of refractive index '$n$'. For incident angle $60^{\circ}$ at the side $A C$, the refractive light beam just grazes along face $A C$. The refractive index of the liquid $n=\frac{\sqrt{x}}{4}$. The value of $x$ is ____________.
(Given refractive index of glass $=1.5$ )
Explanation:
On first surface light is passing straight without any deviation, hence angle of refraction $\mathrm{r}_1=0^{\circ}$.
$ \begin{aligned} & r_1+i_2=\mathrm{A} \\\\ & 0+i_2=60^{\circ} \\\\ & \Rightarrow i_2=60^{\circ} \end{aligned} $
On $2^{\text {nd }}$ surface ray, refracting ray is becoming parallel to the junction of surfaces, so $i_2$ will act as critical angle.
From snell's law $\mu_1 \sin \theta_1=\mu_2 \sin \theta_2$
$ \begin{aligned} & \Rightarrow \frac{3}{2} \sin i_2=\mu \sin \theta_2 \\\\ & \Rightarrow \mu=\frac{3}{2} \times \frac{\sin 60^{\circ}}{\sin 90^{\circ}}=\frac{3}{2} \times \frac{\sqrt{3}}{2}=\frac{3 \sqrt{3}}{4} \text { or } \frac{\sqrt{27}}{4} \end{aligned} $
$\Rightarrow$ Comparing above equation with
given $n=\frac{\sqrt{x}}{4}$, we get $x=27$
The graph between $\frac{1}{u}$ and $\frac{1}{v}$ for a thin convex lens in order to determine its focal length is plotted as shown in the figure. The refractive index of lens is $1.5$ and its both the surfaces have same radius of curvature $R$. The value of $R$ will be ____________ $\mathrm{cm} .$ (where $u=$ object distance, $v=$ image distance)
Explanation:
$f = 10$ cm
${1 \over f} = (\mu - 1)\left( {{1 \over R} - {1 \over { - R}}} \right)$
${1 \over {10}} = {{1.5 - 1} \over 1} \times {2 \over R}$
$ {1 \over {10}} = {1 \over R}$
$R = 10$ cm
A convex lens of focal length 20 cm is placed in front of a convex mirror with principal axis coinciding each other. The distance between the lens and mirror is 10 cm. A point object is placed on principal axis at a distance of 60 cm from the convex lens. The image formed by combination coincides the object itself. The focal length of the convex mirror is ____________ cm.
Explanation:
${1 \over v} + {1 \over u} = {1 \over f}$
${1 \over v} - {1 \over { - 60}} = {1 \over {20}}$
${1 \over v} = - {1 \over {60}} + {1 \over {20}} = {{ - 1 + 3} \over {60}} = {2 \over {60}}$
$ \Rightarrow v = \, +\, 30$ cm
$\therefore$ Radius of curvature of mirror = 30 $-$ 10 = 20 cm
$ \Rightarrow {f_{mirror}} = {{20} \over 2} = 10$ cm
The refractive index of an equilateral prism is $\sqrt 2 $. The angle of emergence under minimum deviation position of prism, in degree, is ___________.
Explanation:
$ \begin{aligned} & \mu=\frac{\sin \left(\frac{A+\delta \sin }{2}\right)}{\sin \left(\frac{A}{2}\right)} \Rightarrow \sqrt{2}=\sin \frac{\left(\frac{60+\delta \min }{2}\right)}{\sin 30^{\circ}} \\\\ & \Rightarrow \frac{1}{2}=\sin \left(\frac{60+\delta \min }{2}\right) \Rightarrow 45^{\circ}=\frac{60+\delta \min }{2} \\\\ & \Rightarrow \delta \min =30^{\circ} \\\\ & \delta=\mathrm{i}+\mathrm{e}-\mathrm{A} \\\\ & \text { Here, } e=i \\\\ & \text { So, } \delta \mathrm{min}=2 \mathrm{e}-\mathrm{A} \Rightarrow 2 \mathrm{e}=\delta \min +\mathrm{A} \\\\ & e=\frac{\delta \min +A}{2}=\frac{30^{\circ}+60^{\circ}}{2}=45^{\circ} \end{aligned} $
A parallel beam of light is allowed to fall on a transparent spherical globe of diameter 30 cm and refractive index 1.5. The distance from the centre of the globe at which the beam of light can converge is _____________ mm.
Explanation:
1st refraction : ${{1.5} \over {{v_1}}} - 0 = {{0.5} \over {15}}$
$\Rightarrow$ v1 = 45 cm
2nd refraction : ${1 \over {{v_2}}} - {{1.5} \over {15}} = {{ - 0.5} \over { - 15}}$
$ \Rightarrow {1 \over {{v_2}}} = {1 \over {30}} + {1 \over {10}}$
$ = {4 \over {30}}$
$\Rightarrow$ v2 = + 7.5 cm
$\Rightarrow$ Distance from centre = 22.5 cm = 225 mm
A small bulb is placed at the bottom of a tank containing water to a depth of $\sqrt7$ m. The refractive index of water is ${4 \over 3}$. The area of the surface of water through which light from the bulb can emerge out is x$\pi$ m2. The value of x is __________.
Explanation:
So $r = h{{\sin {i_c}} \over {\sqrt {1 - {{\sin }^2}{i_c}} }}$
So $A = \pi {r^2}$
$ = {{\pi {h^2}{{\sin }^2}{i_c}} \over {1 - {{\sin }^2}{i_c}}}$
$ = {{\pi 7 \times {9 \over {16}}} \over {1 - {9 \over {16}}}} = {{\pi \times 7 \times 9} \over 7} = 9\pi $