Geometrical Optics — Page 6 | JEE Physics

Q51

Your friend is having eye sight problem. She is not able to see clearly a distant uniform window mesh and it appears to her as non-uniform and distorted. The doctor diagnosed the problem as :

A Astigmatism

B Myopia with Astigmatism

C Presbyopia with Astigmatism

D Myopia and Hypermetropia

Correct Answer

Option B

Solution

Myopia with Astigmatism causes distant objects to be blurry and distorted.

Q52

A ray of laser of a wavelength 630 nm is incident at an angle of 30

^\circ

at the diamond-air interface. It is going from diamond to air. The refractive index of diamond is 2.42 and that of air is 1. Choose the correct option.

A angle of refraction is 24.41

^\circ

B angle of refraction is 30

^\circ

C refraction is not possible

D angle of refraction is 53.4

^\circ

Correct Answer

Option C

Solution

\sin {\theta _C} = {1 \over \mu } = {1 \over {2{\mu _2}}} < \sin {\theta _C}

sin $\theta$ > sin $\theta$ C $\theta$ > $\theta$ C Total internal reflection will happen.

Q53

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

A

\theta = {\sin ^{ - 1}}\left( {{3 \over 4}} \right)

B

\theta > {\sin ^{ - 1}}\left( {{2 \over 3}} \right)

C

\theta < {\sin ^{ - 1}}\left( {{3 \over 4}} \right)

D

\theta > {\sin ^{ - 1}}\left( {{3 \over 4}} \right)

Correct Answer

Option D

Solution

{\mu _A} = {{3 \times {{10}^8}} \over {2 \times {{10}^8}}} = 1.5

{\mu _B} = {{3 \times {{10}^8}} \over {1.5 \times {{10}^8}}} = 2

For TIR

\theta > {i_c}

\theta > {\sin ^{ - 1}}\left( {{{1.5} \over 2}} \right)

\theta > {\sin ^{ - 1}}\left( {{3 \over 4}} \right)

Q54

Car B overtakes another car A at a relative speed of 40 ms

-

1. How fast will the image of car B appear to move in the mirror of focal length 10 cm fitted in car A, when the car B is 1.9 m away from the car A?

A 4 ms

-

1

B 0.2 ms

-

1

C 40 ms

-

1

D 0.1 ms

-

1

Correct Answer

Option D

Solution

Here,

{1 \over v} + {1 \over u} = {1 \over f}

.......(1) $\Rightarrow$

{1 \over v} + {1 \over { - 190}} = {1 \over {10}}

$\Rightarrow$ v =

{{19} \over 2}

Differentiating equation (1) w.r.t t we get

- {1 \over {{v^2}}}\left( {{{dv} \over {dt}}} \right) - {1 \over {{u^2}}}\left( {{{du} \over {dt}}} \right) = 0

$\Rightarrow$

\left( {{{dv} \over {dt}}} \right) = - {{{v^2}} \over {{u^2}}}\left( {{{du} \over {dt}}} \right)

$\Rightarrow$

\left( {{{dv} \over {dt}}} \right) = - {\left( {{{{{19} \over 2}} \over {190}}} \right)^2} \times 40

= {1 \over {400}} \times 40 = 0.1

m/s

Q55

Time taken by light to travel in two different materials

A

and

B

of refractive indices

\mu_{A}

and

\mu_{B}

of same thickness is

t_{1}

and

t_{2}

respectively. If

t_{2}-t_{1}=5 \times 10^{-10}

s and the ratio of

\mu_{A}

to

\mu_{B}

is

1: 2

. Then, the thickness of material, in meter is: (Given

v_{\mathrm{A}}

and

v_{\mathrm{B}}

are velocities of light in

A

and

B

materials respectively.)

A

5 \times 10^{-10} \,v_{\mathrm{A}}\, \mathrm{m}

B

5 \times 10^{-10} \mathrm{~m}

C

1.5 \times 10^{-10} \mathrm{~m}

D

5 \times 10^{-10} \,v_{\mathrm{B}} \,\mathrm{m}

Correct Answer

Option A

Solution

{t_2} - {t_1} = 5 \times {10^{ - 10}}

\Rightarrow {d \over {{v_B}}} - {d \over {{v_A}}} = 5 \times {10^{ - 10}}

and,

{{{v_B}} \over {{v_A}}} = {{{\mu _A}} \over {{\mu _B}}} = {1 \over 2}

\Rightarrow d\left( {1 - {{{v_B}} \over {{v_A}}}} \right) = 5 \times {10^{ - 10}} \times {v_B}

\Rightarrow d\left( {1 - {1 \over 2}} \right) = 5 \times {10^{ - 10}} \times {v_B}

\Rightarrow d = 10 \times {10^{ - 10}} \times {v_B}\,m

\Rightarrow d = 5 \times {10^{ - 10}} \times {v_A}\,m

Q56

For an object placed at a distance 2.4 m from a lens, a sharp focused image is observed on a screen placed at a distance 12 cm from the lens. A glass plate of refractive index 1.5 and thickness 1 cm is introduced between lens and screen such that the glass plate plane faces parallel to the screen. By what distance should the object be shifted so that a sharp focused image is observed again on the screen?

A 0.8 m

B 3.2 m

C 1.2 m

D 5.6 m

Correct Answer

Option B

Solution

The shift produced by the glass plate is

d = t\left( {1 - {1 \over \mu }} \right) = 1 \times \left( {1 - {1 \over {1.5}}} \right) = {1 \over 3}

cm So final image must be produced at

\left( {12 - {1 \over 3}} \right)

cm

= {{35} \over 3}

cm from lens so that glass plate must shift it to produce image at screen. So

{1 \over {12}} - {1 \over { - 240}} = {1 \over f} = {1 \over {35/3}} - {1 \over u}

{1 \over u} = {3 \over {35}} - {1 \over {12}} - {1 \over {240}}

or

u = - 560

cm so shift

= 5.6 - 2.4 = 3.2

m

Q57

Light travels in two media

M_{1}

and

M_{2}

with speeds

1.5 \times 10^{8} \mathrm{~ms}^{-1}

and

2.0 \times 10^{8} \mathrm{~ms}^{-1}

respectively. The critical angle between them is :

A

\tan ^{-1}\left(\dfrac{3}{\sqrt{7}}\right)

B

\tan ^{-1}\left(\dfrac{2}{3}\right)

C

\cos ^{-1}\left(\dfrac{3}{4}\right)

D

\sin ^{-1}\left(\dfrac{2}{3}\right)

Correct Answer

Option A

Solution

Critical angle between them

\sin {i_c} = {{{\mu _2}} \over {{\mu _1}}} = {{{v_1}} \over {{v_2}}}

\sin {i_c} = {3 \over 4}

\Rightarrow \tan {i_c} = {3 \over {\sqrt 7 }}

{i_c} = {\tan ^{ - 1}}{3 \over {\sqrt 7 }}

Q58

A microscope was initially placed in air (refractive index 1). It is then immersed in oil (refractive index 2). For a light whose wavelength in air is

\lambda

, calculate the change of microscope's resolving power due to oil and choose the correct option.

A Resolving power will be

\dfrac{1}{4}

in the oil than it was in the air.

B Resolving power will be twice in the oil than it was in the air.

C Resolving power will be four times in the oil than it was in the air.

D Resolving power will be

\dfrac{1}{2}

in the oil than it was in the air.

Correct Answer

Option B

Solution

The resolving power of a microscope is determined by the Rayleigh criterion, which states that it is inversely proportional to the wavelength of light used in the medium in which the microscopy is being performed.

Mathematically, the resolving power (RP) can be represented as: $RP = \dfrac{1}{\lambda_n}$ where $\lambda_n$ is the wavelength of light in the medium, which can be found using the formula: $\lambda_n = \dfrac{\lambda}{n}$ Here, $\lambda$ is the wavelength of light in vacuum (or air, for practical purposes, since their refractive indices are close enough), and $n$ is the refractive index of the medium.

In air, the refractive index $n = 1$ , so the wavelength of light in air ( $\lambda_{air}$ ) is equal to $\lambda$ .

In oil, the refractive index $n = 2$ , so the wavelength of light in oil ( $\lambda_{oil}$ ) is $\lambda / 2$ .

Therefore, the change in resolving power when moving from air to oil can be calculated as the ratio of resolving powers in oil to air: $\dfrac{RP_{oil}}{RP_{air}} = \dfrac{\lambda_{air}}{\lambda_{oil}} = \dfrac{\lambda}{\lambda / 2} = 2$ This means that the resolving power in oil is twice that in air.

Thus, the correct option is: Option B: Resolving power will be twice in the oil than it was in the air.

Q59

In normal adujstment, for a refracting telescope, the distance between objective and eye piece is

30 \mathrm{~cm}

. The focal length of the objective, when the angular magnification of the telescope is 2 , will be :

A 20 cm

B 30 cm

C 10 cm

D 15 cm

Correct Answer

Option A

Solution

$\because$

m = {{{f_o}} \over {{f_e}}}

\Rightarrow 2 = {{{f_o}} \over {{f_e}}}

...... (i) and,

l = {f_o} + {f_e}

\Rightarrow 30 = {f_o} + {f_e}

..... (ii)

\Rightarrow 30 = {f_o} + {{{f_o}} \over 2}

\Rightarrow 30 \times {2 \over 3} = {f_o}

\Rightarrow {f_o} = 20

cm

Q60

The power of a lens (biconvex) is

1.25 \mathrm{~m}^{-1}

in particular medium. Refractive index of the lens is 1.5 and radii of curvature are

20 \mathrm{~cm}

and

40 \mathrm{~cm}

respectively. The refractive index of surrounding medium:

A 1.0

B

\dfrac{9}{7}

C

\dfrac{3}{2}

D

\dfrac{4}{3}

Correct Answer

Option B

Solution

$\because$

{1 \over f} = \left( {{{{\mu _2}} \over {{\mu _1}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)

\Rightarrow {{1.25} \over {100}} = \left( {{{1.5} \over {{\mu _1}}} - 1} \right)\left( {{1 \over {20}} + {1 \over {40}}} \right)

\Rightarrow {1 \over {80}} = \left( {{{1.5} \over {{\mu _1}}} - 1} \right) \times {{(4 + 2)} \over {80}}

\Rightarrow {{1.5} \over {{\mu _1}}} - 1 = {1 \over 6}

\Rightarrow {{1.5} \over {{\mu _1}}} = {7 \over 6}

\Rightarrow {\mu _1} = {{1.5 \times 6} \over 7} = {9 \over 7}