Geometrical Optics
The speed of light in media 'A' and 'B' are $2.0 \times {10^{10}}$ cm/s and $1.5 \times {10^{10}}$ cm/s respectively. A ray of light enters from the medium B to A at an incident angle '$\theta$'. If the ray suffers total internal reflection, then
The refracting angle of a prism is A and refractive index of the material of the prism is cot (A/2). Then the angle of minimum deviation will be -
The aperture of the objective is 24.4 cm. The resolving power of this telescope, if a light of wavelength 2440 $\mathop A\limits^o $ is used to see th object will be :
A convex lens has power P. It is cut into two halves along its principal axis. Further one piece (out of the two halves) is cut into two halves perpendicular to the principal axis (as shown in figures). Choose the incorrect option for the reported pieces.
Consider a light ray travelling in air is incident into a medium of refractive index $\sqrt{2n}$. The incident angle is twice that of refracting angle. Then, the angle of incidence will be :
A light wave travelling linearly in a medium of dielectric constant 4, incidents on the horizontal interface separating medium with air. The angle of incidence for which the total intensity of incident wave will be reflected back into the same medium will be :
(Given : relative permeability of medium $\mu$r = 1)
The difference of speed of light in the two media A and B (vA $-$ vB) is 2.6 $\times$ 107 m/s. If the refractive index of medium B is 1.47, then the ratio of refractive index of medium B to medium A is : (Given : speed of light in vacuum c = 3 $\times$ 108 ms$-$1)
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 $
A light ray is incident, at an incident angle $\theta$1, on the system of tow plane mirrors M1 and M2 having an inclination angle 75$^\circ$ between them (as shown in figure). After reflecting from mirror M1 it gets reflected back by the mirror M2 with an angle of reflection 30$^\circ$. The total deviation of the ray will be _____________ degree.
Explanation:
On first reflection angel of deviation is 90° and on second reflection angle of deviation is 120°
So total deviation is $\delta $ = 90° +120° = 210°
A ray of light is incident at an angle of incidence 60$^\circ$ on the glass slab of refractive index $\sqrt3$. After refraction, the light ray emerges out from other parallel faces and lateral shift between incident ray and emergent ray is 4$\sqrt3$ cm. The thickness of the glass slab is __________ cm.
Explanation:
$1 \times \sin 60^\circ = \sqrt 3 \times \sin r$
$ \Rightarrow r = 30^\circ $
$\therefore$ ${l_1} = 4\sqrt 3 \times 2$
$ = 8\sqrt 3 $ cm
$\therefore$ Thickness, $t = {l_1}\cos 30^\circ $
$ = 8\sqrt 3 \times {{\sqrt 3 } \over 2}$
$ = 4 \times 3$
$ = 12$ cm
Two identical thin biconvex lens of focal length 15 cm and refractive index 1.5 are in contact with each other. The space between the lenses is filled with a liquid of refractive index 1.25. The focal length of the combination is __________ cm.
Explanation:
${1 \over {{f_l}}} = \left( {{{{\mu _e}} \over {{\mu _m}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$
here $|{R_1}| = |{R_2}| = R$
$ \Rightarrow {1 \over {{f_{{l_1}}}}} = (1.5 - 1)\left( {{2 \over R}} \right) = {1 \over {15}}$
$ \Rightarrow {1 \over R} = {1 \over {15}}$ or $R = 15$ cm
for the concave lens made up of liquid
${1 \over {{f_{{l_2}}}}} = (1.25 - 1)\left( { - {2 \over R}} \right) = - {1 \over {30}}$ cm
now for equivalent lens
${1 \over {{f_e}}} = {2 \over {{f_{{l_1}}}}} + {1 \over {{f_{{l_2}}}}}$
$ = {2 \over {15}} - {1 \over {30}} = {3 \over {30}} = {1 \over {10}}$
or ${f_e} = 10$ cm
Three plane mirrors form an equilateral triangle with each side of length $L$. There is a small hole at a distance $l>0$ from one of the corners as shown in the figure. A ray of light is passed through the hole at an angle $\theta$ and can only come out through the same hole. The cross section of the mirror configuration and the ray of light lie on the same plane.
Which of the following statement(s) is(are) correct?
Consider a configuration of $n$ identical units, each consisting of three layers. The first layer is a column of air of height $h=\frac{1}{3} \mathrm{~cm}$, and the second and third layers are of equal thickness $d=$ $\frac{\sqrt{3}-1}{2} \mathrm{~cm}$, and refractive indices $\mu_{1}=\sqrt{\frac{3}{2}}$ and $\mu_{2}=\sqrt{3}$, respectively. A light source 0 is placed on the top of the first unit, as shown in the figure. A ray of light from 0 is incident on the second layer of the first unit at an angle of $\theta=60^{\circ}$ to the normal. For a specific value of $n$, the ray of light emerges from the bottom of the configuration at a distance $l=\frac{8}{\sqrt{3}} \mathrm{~cm}$, as shown in the figure. The value of $n$ is ________.
Explanation:
$ \begin{aligned} &x_1 =\frac{1}{3} \times \tan 60 \\\\ & =\frac{1}{\sqrt{3}} \mathrm{~cm} \end{aligned} $
By Snell's law
$ \begin{aligned} &1 \times \sin 60 =\frac{\sqrt{3}}{\sqrt{2}} \sin \theta_2 \\\\ & \Rightarrow \theta_2 =45^ \mathrm{o} \\\\ &\therefore x_2 =d \end{aligned} $
Again, by applying Snell's law
$ \begin{aligned} &\sqrt{\frac{3}{2} \times \frac{1}{\sqrt{2}}} =\sqrt{3} \times \sin \theta_3 \\\\ &\Rightarrow \theta_2 =30^{\circ} \\\\ & \therefore x_3 =\frac{d}{\sqrt{3}} \end{aligned} $
Also
$ \begin{aligned} & x_1+x_2+x_3 =\frac{1}{\sqrt{3}}+d+\frac{d}{\sqrt{3}} \\\\ & =\frac{1}{\sqrt{3}}+\left(\frac{\sqrt{3}-1}{2}\right)+\frac{\sqrt{3}-1}{2 \sqrt{3}} \\\\ & =\frac{2+(3-\sqrt{3})+\sqrt{3}-1}{2 \sqrt{3}} \\\\ & =\frac{4}{2 \sqrt{3}} \\\\ & x_1+x_2+x_3 =\frac{2}{\sqrt{3}} \mathrm{~cm} \\\\ & \text {And,}~~ \eta =\frac{l}{x_1+x_2+x_3} \\\\ &\eta =\frac{8 / \sqrt{3}}{2 / \sqrt{3}}=4 \\\\ &\eta =4 \end{aligned} $
An object and a concave mirror of focal length $f=10 \mathrm{~cm}$ both move along the principal axis of the mirror with constant speeds. The object moves with speed $V_{0}=15 \mathrm{~cm} \mathrm{~s}^{-1}$ towards the mirror with respect to a laboratory frame. The distance between the object and the mirror at a given moment is denoted by $u$. When $u=30 \mathrm{~cm}$, the speed of the mirror $V_{m}$ is such that the image is instantaneously at rest with respect to the laboratory frame, and the object forms a real image. The magnitude of $V_{m}$ is _________ $\mathrm{cm} \,\mathrm{s}^{-1}$.
Explanation:
Magnification,
$ \begin{aligned} & m =\frac{f}{u-f} \\\\ & =\frac{10}{30-10}=\frac{1}{2} \end{aligned} $
Also $\quad m^2=\frac{v_m}{v_0-v_m}$ [where, $v_m=$ speed of mirror]
$ \begin{aligned} & m^2\left(v_{\mathrm{o}}-v_{\mathrm{m}}\right) =v_m \\\\ & \Rightarrow \left(v_0-v_m\right) \times \frac{1}{4} =v_{\mathrm{m}} \\\\ & \Rightarrow v_0 =5 v_m \\\\ & \Rightarrow v_m =\frac{v_0}{5}=\frac{15}{5} \\\\ & \Rightarrow v_m =3 \mathrm{~cm} / \mathrm{s} \end{aligned} $
A rod of length $2 \mathrm{~cm}$ makes an angle $\frac{2 \pi}{3} \mathrm{rad}$ with the principal axis of a thin convex lens. The lens has a focal length of $10 \mathrm{~cm}$ and is placed at a distance of $\frac{40}{3} \mathrm{~cm}$ from the object as shown in the figure. The height of the image is $\frac{30 \sqrt{3}}{13} \mathrm{~cm}$ and the angle made by it with respect to the principal axis is $\alpha$ rad. The value of $\alpha$ is $\frac{\pi}{n} r a d$, where $n$ is __________ .
Explanation:
$\frac{\mathrm{h}_{\mathrm{i}}}{\mathrm{h}_0}=\frac{\mathrm{v}}{\mathrm{u}} \Rightarrow \frac{-\frac{30 \sqrt{3}}{13}}{\sqrt{3}}=\frac{\mathrm{v}}{-\frac{43}{3}} \Rightarrow \mathrm{v}_1=\frac{430}{13} \mathrm{~cm}$
Now using,
$ \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{10}-\frac{3}{40} \Rightarrow \mathrm{v}=40 \mathrm{~cm}$
$ \therefore $ $ \mathrm{x}=40-\frac{430}{13}=\frac{90}{13} \mathrm{~cm}$
$\tan \alpha=\frac{\frac{30 \sqrt{3}}{\frac{90}{13}}}{\mathrm{~N}}=\frac{1}{\sqrt{3}} \Rightarrow \alpha=30^{\circ}=\frac{\pi}{6}$
$ \therefore $ N = 6
List I contains four combinations of two lenses (1 and 2) whose focal lengths (in $\mathrm{cm}$ ) are indicated in the figures. In all cases, the object is placed $20 \mathrm{~cm}$ from the first lens on the left, and the distance between the two lenses is $5 \mathrm{~cm}$. List II contains the positions of the final images.
| List-I | List-II |
|---|---|
(I)  |
(P) Final image is formed at $7.5 \mathrm{~cm}$ on the right side of lens 2 . |
(II)  |
(Q) Final image is formed at $60.0 \mathrm{~cm}$ on the right side of lens 2 . |
(III)  |
(R) Final image is formed at $30.0 \mathrm{~cm}$ on the left side of lens $2 .$ |
(IV)  |
(S) Final image is formed at $6.0 \mathrm{~cm}$ on the right side of lens 2 . |
| (T) Final image is formed at $30.0 \mathrm{~cm}$ on the right side of lens 2 . |
Which one of the following options is correct?
The refractive indices of the material of the prism for red, green and blue wavelength are 1.27, 1.42 and 1.49 respectively. The colour of the ray(s) emerging out of the face PR is :
(A) Incident ray and emergent ray are symmetric to the prism.
(B) The refracted ray inside the prism becomes parallel to its base
(C) Angle of incidence is equal to that of the angle of emergence
(D) When angle of emergence is double the angle of incidence
Choose the correct answer from the options given below :
Assertion A : For a simple microscope, the angular size of the object equals the angular size of the image.
Reason R : Magnification is achieved as the small object can be kept much closer to the eye than 25 cm and hence it subtends a large angle.
In the light of the above statements, choose the most appropriate answer from the options given below :
Explanation:
${{ \delta } _{\min }}$ = 2i $-$ A
= 2 $\times$ 60$^\circ$ $-$ 60$^\circ$ = 60$^\circ$
$\mu = {{{{\sin }^{ - 1}}\left( {{{{\delta _{\min }} + A} \over 2}} \right)} \over {{{\sin }^{ - 1}}\left( {{A \over 2}} \right)}}$
$ = \sqrt 3 $
Vprism $ = {{3 \times {{10}^8}} \over {\sqrt 3 }}$
AP = 10 $\times$ 10$-$2 $\times$ ${{\sqrt 3 } \over 2}$
time $ = {{5 \times {{10}^{ - 2}}} \over {3 \times {{10}^8}}} \times \sqrt 3 \times \sqrt 3 $
= 5 $\times$ 10$-$10 sec
Ans. = 5
When the convex mirror is removed, a real and inverted image is formed at a position. The distance of the image from the object will be .............. (cm)
Explanation:
For the object to coincide with image, the light must fall perpendicularly to mirror. Which means that the light will have to converge at C of mirror. Without the mirror also, the light would coverage at C.
So the distance is : 12 + 8 + 30 = 50 cm