Geometrical Optics — Page 2 | JEE Physics

Q11

Let

x

-

z

plane be the boundary between two transparent media. Medium

1

in

z \ge 0

has a refractive index of

\sqrt 2

and medium

2

with

z < 0

has a refractive index of

\sqrt 3 .

A ray of light in medium

1

given by the vector

\overrightarrow A = 6\sqrt 3 \widehat i + 8\sqrt 3 \widehat j - 10\widehat k

is incident on the plane of separation. The angle of refraction in medium

2

is:

A

{45^ \circ }

B

{60^ \circ }

C

{75^ \circ }

D

{30^ \circ }

Correct Answer

Option A

Solution

Angle of incidence is given by

\cos \left( {\pi - i} \right) = {{\left( {6\sqrt 3 \widehat i + 8\sqrt 3 \widehat j - 10\widehat k} \right).\widehat k} \over {20}}

- \cos \,i = - {1 \over 2}

\angle i = {60^ \circ }

From Snell's law,

\sqrt 2 \sin i = \sqrt 3 \sin r

$\Rightarrow$

\angle r = {45^ \circ }

Q12

A car is fitted with a convex side-view mirror of focal length

20

cm

. A second car

2.8m

behind the first car is overtaking the first car at a relative speed of

15

m/s

. The speed of the image of the second car as seen in the mirror of the first one is :

A

{1 \over {15}}\,m/s

B

10\,m/s

C

15\,m/s

D

{1 \over {10}}\,m/s

Correct Answer

Option A

Solution

From mirror formula

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

so,

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

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

\Rightarrow {{dv} \over {dt}} = {1 \over {15}}m/s

Q13

A monochromatic light is incident at a certain angle on an equilateral triangular prism and suffers minimum deviation. If the refractive index of the material of the prism is

\sqrt 3

, then the angle of incidence is:

A 60o

B 45o

C 90o

D 30o

Correct Answer

Option A

Solution

i = e r1 = r2 =

{A \over 2}

= 30o by Snell's law 1 $\times$ sin i =

\sqrt 3 \times {1 \over 2} = {{\sqrt 3 } \over 2}

i = 60

Q14

An initially parallel cylindrical beam travels in a medium of refractive index

\mu \left( I \right) = {\mu _0} + {\mu _2}\,I,

where

{\mu _0}

and

{\mu _2}

are positive constants and

I

is the intensity of the light beam. The intensity of the beam is decreasing with increasing radius. As the beam enters the medium, it will

A diverge

B converge

C diverge near the axis and converge near the periphery

D travel as a cylindrical beam

Correct Answer

Option B

Solution

In the medium, the refractive index will decreases from the axis forwards the periphery of the beam.

Therefore, the beam will move as one move from the axis to the periphery and hence the beam will converge.

Q15

The focal length f is related to the radius of curvature r of the spherical convex mirror by :

A f = r

B f =

-

r

C f = +

{{1 \over 2}}

r

D f =

-

{{1 \over 2}}

r

Correct Answer

Option C

Solution

For convex mirror, the focal length (f) and radius of curvature (r) are related as

f = + {r \over 2}

.

Q16

An object is placed beyond the centre of curvature C of the given concave mirror. If the distance of the object is d1 from C and the distance of the image formed is d2 from C, the radius of curvature of this mirror is :

A

{{2{d_1}{d_2}} \over {{d_1} - {d_2}}}

B

{{2{d_1}{d_2}} \over {{d_1} + {d_2}}}

C

{{{d_1}{d_2}} \over {{d_1} + {d_2}}}

D

{{{d_1}{d_2}} \over {{d_1} - {d_2}}}

Correct Answer

Option A

Solution

Using Newton's formula

(f + {d_1})(f - {d_2}) = {f^2}

{f^2} + f{d_1} - f{d_2} - {d_1}{d_2} = {f^2}

f = {{{d_1}{d_2}} \over {{d_1} - {d_2}}}

$\therefore$

R = {{2{d_1}{d_2}} \over {{d_1} - {d_2}}}

Q17

The critical angle of a medium for a specific wavelength, if the medium has relative permittivity 3 and relative permeability

{4 \over 3}

for this wavelength, will be :

A 45°

B 15°

C 30°

D 60°

Correct Answer

Option C

Solution

n =

\sqrt {{\varepsilon _r}{\mu _r}} = \sqrt {3 \times {4 \over 3}}

= 2 n sin c = 1 sin 90o sin c =

{1 \over 2}

$\Rightarrow$ c = 30o

Q18

If two mirrors are kept at

{60^ \circ }

to each other, then the number of images formed by them is

A

5

B

6

C

7

D

8

Correct Answer

Option A

Solution

KEY CONCEPT : When two plane mirrors are inclined a each other at an angle $\theta$ then the number of the images of a point object placed between the plane mirrors is

{{{{360}^ \circ }} \over \theta } - 1,\,\,if{{{{360}^ \circ }} \over \theta }

is even $\therefore$ Number of images formed

= {{{{360}^ \circ }} \over {{{60}^ \circ }}} - 1 = 5

Q19

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

Correct Answer

Option B

Solution

KEY CONCEPT : The resolving power of a telescope

R.P = {D \over {122\lambda }}

where

D=

diameter of the objective lens

\lambda =

wavelength of light. Clearly, larger the aperture, larger is the value of

D,

more is the resolving power or resolution.

Q20

Which of the following is used in optical fibres?

A total internal reflection

B scattering

C diffracttion

D refraction

Correct Answer

Option A

Solution

Optical fibers work on the principle of total internal reflection.

When light is transmitted through the fiber, it is reflected off the inner walls of the fiber in such a way that it remains within the fiber, allowing it to carry the light signal over great distances with minimal loss.

So, the correct answer is : Option A : total internal reflection.