﻿ Medical Entrance Exam Question and Answers | Ray Optics and Optical Instruments - Zigya

## Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

## Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

## Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

# Ray Optics and Optical Instruments

#### Multiple Choice Questions

1.

An astronaut is looking down on earth's surface from a space shuttle at an altitude of 400 km.
Assuming that the astronaut's pupil diameter is 5 mm and the wavelength of visible light is 500 nm, the astronaut will be able to resolve linear objects of the size about

• 0.5 m

• 5 m

• 50 m

• 500 m

C.

50 m

The resolving power of the eye is given by

R = 1.22 $\frac{\mathrm{\lambda }}{\mathrm{b}}$ × D

where λ is the wavelength of light

b is the diameter of the lens

D is the distance from the lens at which light is focussed

d = 400 km = 400 × 103 m

d = 5 mm = 5 × 10-3 m

∴       R = 1.22 ×

R = 48.8

R = 50 m

2.

A diver, at a depth of 12 metre from the surface of the water, sees the sky within a cone whose half vertex angle is : $\left({\mathrm{a}}^{{\mathrm{\mu }}_{\mathrm{w}}}=\frac{4}{3}\right)$

• cos−1$\left(\frac{4}{3}\right)$

• sin−1$\left(\frac{4}{3}\right)$

• 90$°$

• sin−1$\left(\frac{3}{4}\right)$

D.

sin−1$\left(\frac{3}{4}\right)$

Let diver is placed at S. He will see a cone of maximum radius r. In this condition

sin(ic) = $\frac{1}{{\mathrm{a}}^{{\mathrm{\mu }}_{\mathrm{w}}}}$

$⇒$      sin(ic)= $\frac{1}{4/3}$

$\mathrm{⇒}$       sin(ic) = $\frac{3}{4}$

$\therefore$            ic = sin1$\left(\frac{3}{4}\right)$

3.

In an experiment to find the focal length of a concave mirror a graph is drawn between the magnitudes of u and v. The graph looks like

• • • • C. In case of concave mirror, when we move the object  from infinity to focus of the mirror ( i.e u decrease from  ∞ to f ), the image moves from focus to ∞ ( i.e v increase from f to  ∞). Therefore, the graph looks like the curve shown in fig (c).

4.

Assertion: Owls can move freely during night.

Reason: They have large number of rods on their retina.

• if both assertion and reason are true and the reason is the correct explanation of the assertion

• if both assertion and reason are true and the reason is not the correct explanation of the assertion

• if assertion is true but reason is false

• if both assertion and reason are false statements

C.

if assertion is true but reason is false

There are two types of receptors in the eyes. One is called rods and other is cons. The rods are sensitive for dark light. As the owl has large number of cones in its retina and so the owl can move freely in the dark.

5.

A light wave travels from glass to water. The refractive indices for glass and water are $\frac{3}{2}$ and $\frac{4}{3}$ respectively. The value of the critical angle will be :

• ${\mathrm{sin}}^{-1}\left(\frac{1}{2}\right)$

• ${\mathrm{sin}}^{-1}\left(\frac{9}{8}\right)$

• ${\mathrm{sin}}^{-1}\left(\frac{8}{9}\right)$

• ${\mathrm{sin}}^{-1}\left(\frac{5}{7}\right)$

C.

${\mathrm{sin}}^{-1}\left(\frac{8}{9}\right)$

From the relation, or

where C = critical angle

${\mathrm{\mu }}_{\mathrm{g}}$ = refractive index of glass = 3/2

${\mathrm{\mu }}_{\mathrm{\omega }}$ = refractive index of water= 4/3

$\therefore$                 sin C = $\frac{{\mathrm{\mu }}_{\mathrm{\omega }}}{{\mathrm{\mu }}_{\mathrm{g}}}$

sin C = $\frac{4/3}{3/2}$

sin C = $\frac{8}{9}$

Hence,                C = ${\mathrm{sin}}^{-1}\left(\frac{8}{9}\right)$

6.

The Cauchy's dispersion formula is

• n = A + Bλ-2 + Cλ-4

• n = A + Bλ-2 + Cλ4

• n = A + Bλ2 + Cλ-4

• n = A + Bλ2 + Cλ4

A.

n = A + Bλ-2 + Cλ-4

The Cauchy's dispersion formula represents the dispersion of most of the substances with considerable accuracy.

From Cauchy's dispersion formula if n is refractive index and A, B and C are constants for a given medium

The most general form of Cauchy's equation is

n(λ) = A +  + ........

n = A + Bλ-2 + Cλ-4

7.

The magnification of an astronomical telescope is 10 and the focal length of eye-piece is 20 cm. The focal length of objective lens is

• 2 cm

• 200 cm

• 100 cm

C.

200 cm

Magnification of astronomical telescope

m = $\frac{{\mathrm{f}}_{\mathrm{o}}}{{\mathrm{f}}_{\mathrm{e}}}$

where, fo = focal length of objective lens

fe = focal length of eye piece

$\therefore$       10 = $\frac{{\mathrm{f}}_{\mathrm{o}}}{20}$

$⇒$        fo = 10 × 20

$⇒$        fo = 200 cm

8.

A convex lens of focal length 12 cm is made up of a glass of refractive index $\frac{3}{2}$.When it is immersed in a liquid of refractive index $\frac{5}{4}$, its focal length will be

• 15 cm

• 6 cm

• 30 cm

• 24 cm

C.

30 cm

Here: an$\frac{3}{2}$anw=$\frac{5}{4}$ , f1 = 12 cm

$\therefore$      wng

or

or           f2  =  $\left(\frac{1/2}{1/5}\right)×12$

Hence,    f2   = 30 cm

9.

Golden view of sea shell is due to

• diffraction

• polarisation

• dispersion

• reflection

B.

polarisation

When a ray of light falls on sea shell, then its small amount first gets refracted (slightly polarised) and then almost gets reflected back (fully polarised). That is why, view of sea shell is golden due to polarisation.

10.

Brilliance of diamond is due to

• shape

• reflection

• cutting

• total internal reflection

D.

total internal reflection

Total internal reflection can occur only when a ray is incident on the surface of a medium whose refractive index is smaller than that of the medium in which the ray is travelling. Since the refractive index of air is  1.00029  and that of a  diamond is 2.42, therefore brilliance of diamond is due to total internal reflection.