Geometrical Optics — Page 3 | NEET Physics

Q21

A convex lens 'A' of focal length 20 cm and a concave lens 'B' of focal length 5 cm are kept along the same axis with a distance 'd' between them. If a parallel beam of light falling on 'A' leaves 'B' as a parallel beam, then the distance 'd' in cm will be :

A 30

B 25

C 15

D 50

Correct Answer

Option C

Solution

Parallel beam of light after refraction from convex lens converge at the focus of convex lens.

In question it is given light after refraction pass through concave lens becomes parallel.

Therefore light refracted from convex lens virtually meet at focus of concave lens.

According to above ray diagram, d = f A $-$ f B = 20 $-$ 5 = 15 cm

Q22

A lens of large focal length and large aperture is best suited as an objective of an astronomical telescope since :

A a large aperture contributes to the quality and visibility of the images.

B a large area of the objective ensures better light gathering power.

C a large aperture provides a better resolution.

D All the above.

Correct Answer

Option D

Solution

With larger aperture of objective lens, the light gathering power in telescope is high.

Also, the resolving power or the ability to observe two objects distinctly also depends on the diameter of the objective.

Thus objective of large diameter is preferred.

Also, with large diameters fainter objects can be observed.

Hence it also contributes to the better quality and visibility of images.

Hence, all options are correct.

Q23

A point object is placed at a distance of 60 cm from a convex lens of focal length 30 cm. If a plane mirror were put perpendicular to the principal axis of the lens and at a distance of 40 cm from it, the final image would be formed at a distance of -

A 20 cm from the plane mirror, it would be a virtual image.

B 20 cm from the lens, it would be a real image.

C 30 cm from the lens, it would be a real image.

D 30 cm from the plane mirror, it would be a virtual image.

Correct Answer

Option A

Solution

First for image formation from lens u = $-$ 60 cm f = +30 cm

\Rightarrow v = {{uf} \over {u + f}} = {{ - 60 \times 30} \over { - 60 + 30}} = 60

cm This real image formed by lens acts as virtual object for mirror.

Real image from plane mirror is formed 20 cm in front of mirror, hence at 20cm distance from lens.

Now for second refraction from lens, u = $-$ 20 cm f = +30 cm

v = {{uf} \over {u + f}} = {{ - 20 \times 30} \over { - 20 + 30}} = - 60

cm So, final virtual image is 60 cm from lens, or 20 cm behind mirror.

Q24

A ray is incident at an angle of incidence i on one surface of a small angle prism (with angle of prism A) and emerges normally from the opposite surface. If the refractive index of the material of the prism is

\mu

, then the angle of incidence is nearly equal to :

A

{{2A} \over \mu }

B

\mu

A

C

{{\mu A} \over 2}

D

{A \over {2\mu }}

Correct Answer

Option B

Solution

Since the light emerges normally from the other surface, the angle of, emergence e = 0 For the triangular prism, we know r 1 + r 2 = A But, e = r 2 = 0 So, r 1 = A For surface 1, Snell's Law is

\sin i = \mu .\sin {r_1}

\sin i = \mu .\sin {A}

For small angles $\theta$ = sin $\theta$ So, i = $\mu$ .A

Q25

The Brewster's angle i b for an interface should be :

A 30

^\circ

< i b < 45

^\circ

B 45

^\circ

< i b < 90

^\circ

C i b = 90

^\circ

D 0

^\circ

< i b < 30

^\circ

Correct Answer

Option B

Solution

Refractive index $\mu$ is equal to tangent of Brewster's angle i b $\mu$ = tan i b 1 < $\mu$ < $\infty$ 1 < tan i b < $\infty$ tan -1 (1) < i b < tan -1 ( $\infty$ ) 45

^\circ

< i b < 90

^\circ

Q26

Pick the wrong answer in the context with rainbow.

A An observer can see a rainbow when his front is towards the sun.

B Rainbow is a combined effect of dispersion refraction and reflection of sunlight.

C When the light rays undergo two internal reflections in a water drop, a secondary rainbow is formed.

D The order of colours is reversed in the secondary rainbow.

Correct Answer

Option A

Solution

When observer faces infront of Sun. He will not observe rainbow.

Q27

Two similar thin equi-convex lenses, of focal length f each, are kept coaxially in contact with each other such that the focal length of the combination is F 1 . When the space between the two lenses is filled with glycerin (which has the same refractive index (m = 1.5) as that of glass) then the equivalent focal length is F 2 . The ratio F 1 : F 2 will be :

A 1 : 2

B 2 : 3

C 3 : 4

D 2 : 1

Correct Answer

Option A

Solution

Equivalent focal length (in air)

{1 \over {{F_1}}} = {1 \over f} + {1 \over f} = {2 \over f}

When glycerin is filled inside, glycerin lens behaves like a diverging lens of focal length (–f)

{1 \over {{F_2}}} = {1 \over f} + {1 \over f} - {1 \over f} = {1 \over f}

$\therefore$

{{{F_1}} \over {{F_2}}} = {1 \over 2}

Q28

In total internal reflection when the angle of incidence is equal to the critical angle for the pair of media in contact, what will be angle of refraction?

A equal to angle of incidence

B 90 o

C 180 o

D 0 o

Correct Answer

Option B

Solution

Let i c be the critical angle.

Hence, at i = i c , refracted ray grazes with the surface.

So, angle of refraction is 90°.

Q29

The refractive index of the material of a prism is

\sqrt 2

and the angle of the prism is 30°. One of the two refracting surfaces of the prism is made a mirror inwards, by silver coating. A beam of monochromatic light entering the prism from the other face will retrace its path (after reflection from the silvered surface) if its angle of incidence on the prism is

A 60 0

B 45 0

C 30 0

D zero

Correct Answer

Option B

Solution

For retracing the path, light ray should be normally incident on silvered face. Applying Snell’s law at point M,

{{\sin i} \over {\sin 30^\circ }} = {{\sqrt 2 } \over 1}

$\Rightarrow$ sin i =

\sqrt 2 \times {1 \over 2}

=

{1 \over {\sqrt 2 }}

$\Rightarrow$ i = 45 o

Q30

An astronomical refracting telescope will have large angular magnification and high angular resolution, when it has an objective lens of

A small focal length and large diameter

B large focal length and small diameter

C large focal length and large diameter

D small focal length and small diameter

Correct Answer

Option C

Solution

For telescope, angular magnification =

{{{f_0}} \over {{f_e}}}

, which shows that focal length of objective lens to be large. Angular resolution =

{D \over {1.22\lambda }}

which is large, hence objective should have large focal length and larger diameter.