Blog Log in Get Started Free

Optics MCQs for NEET — Physics Questions with Answers

Practice free Optics (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Language English हिंदी

Chapter Difficulty Show only questions I answered incorrectly Clear Register free for difficulty & keyword filters

● ● ● ● ●

The refractive index of the material of the prism for violet colour is 1.69 and that for red is 1.65. If the refractive index for mean colour is 1.66, the dispersive power of the material of the prism

0.66

0.06

0.65

0.69

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$(2) ω = (\frac{μ_{v} - μ_{R}}{μ_{y} - 1}) = (\frac{1.69 - 1.65}{1.66 - 1}) = \frac{0.04}{0.66} = \frac{2}{33} = 0.06$

● ● ● ● ●

Absolute refractive indices of glass and water are 3/2 and 4/3. The ratio of velocity of light in glass and water will be

4:3

8:7

8:9

3:4

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$(3) μ \propto \frac{1}{V} \propto \frac{1}{λ} \propto \frac{1}{C} \frac{μ_{1}}{μ_{2}} = \frac{V_{2}}{V_{1}} \frac{3 / 2}{4 / 3} = \frac{V_{2}}{V_{1}} = \frac{9}{8} \frac{V_{1}}{V_{2}} = \frac{8}{9}$

● ● ● ● ●

The length of an astronomical telescope adjusted for parallel light is 90 cm. If the magnifying power of the telescope is 17, then the focal length of eyepiece and objective are respectively

10 cm and 80 cm

85 cm and 5 cm

70 cm and 20 cm

5 cm and 85 cm

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$(4) M = \frac{f_{0}}{f_{e}} = 17 L = f_{0} + f_{e} = 90 cm \Rightarrow 17 f_{e} + f_{e} = 90 cm \Rightarrow f_{e} = 5 cm ∴ f_{0} = 85 cm$

● ● ● ● ●

If the wavelength of light used is halved and the numerical aperture of the compound microscope is doubled, then its resolving power will

Remain unchanged

Doubled

Halved

Quadrupled

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$I n i t i a l r e s o l v i n g p o w e r = \frac{2 μ S i n θ}{1.22 λ} (μ S i n θ = n u m e r i c a l a p e r t u r e) F i n a l r e s o l v i n g p o w e r = \frac{2 \times 2 μ S i n θ}{1.22 \frac{λ}{2}} = 4 \times \frac{2 μ S i n θ}{1.22 λ} = 4 \times i n i t i a l r e s o l v i n g p o w e r$

● ● ● ● ●

The sum (diameter D) subtends an angle $θ$ radian at the pole of a concave mirror of focal length f. The diameter of the image of the sun formed by the mirror is:

$fθ$

$2 fθ$

$\frac{2 fθ}{D}$

$Dθ$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \Rightarrow \frac{1}{f} = \frac{1}{v} + \frac{1}{\infty} S o, f = v N o w, θ = \frac{d}{f} \Rightarrow d = f θ$

● ● ● ● ●

When a ray is refracted from one medium to another, the wavelength changes from 6000 $\overset{0}{A}$ to 4000 $\overset{0}{A}$ . The critical angle for the interface will be:

$\cos^{- 1} (\frac{2}{3})$

$\sin^{- 1} (\frac{2}{\sqrt{3}})$

$\sin^{- 1} (\frac{2}{3})$

$\cos^{- 1} (\frac{2}{\sqrt{3}})$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$(3) μ_{2} \sin θ_{c} = μ_{1} \sin θ_{c} = \frac{μ_{1}}{μ_{2}} = \frac{v_{2}}{v_{1}} = \frac{λ_{2}}{λ_{1}} = \frac{4000}{6000} = \frac{2}{3} θ_{c} = \sin^{- 1} (\frac{2}{3})$

● ● ● ● ●

The critical angle for prism is 36 $°$ . The maximum angle of prism for which the emergent ray is possible is:

$18 °$

$36 °$

$72 °$

$144 °$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

● ● ● ● ●

A small telescope has an objective lens of focal length 144 cm and an eye-piece of focal length 6.0 cm. The magnifying power of the telescope is ( when the final image is at infinity)

150

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$(1) M = \frac{f_{0}}{f_{e}} = \frac{144}{6} = 24$

● ● ● ● ●

A square of side 3 cm is placed at a distance of 25 cm from a concave mirror of focal length 10 cm. The centre of the square is at the axis of the mirror and the plane is normal to the axis. The area enclosed by the image of the square is

$4 c m^{2}$

$6 c m^{2}$

$16 c m^{2}$

$36 c m^{2}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(a) $|m| = |\frac{h_{I}}{h_{O}}| = \frac{f}{u - f} = \frac{10}{25 - 10} = \frac{10}{15} = \frac{2}{3}$

$m^{2} = \frac{A_{i}}{A_{o}} \Rightarrow A_{i} = m^{2} \times A_{o} = {(\frac{2}{3})}^{2} \times {(3)}^{2} = 4 c m^{2}$

● ● ● ● ●

A ray of light falls on the surface of a spherical glass paperweight making an angle $α$ with the normal and is refracted in the medium at an angle $β$ . The angle of deviation of the emergent ray from the direction of the incident ray

$(α - β)$

$2 (α - β)$

$(α - β) / 2$

$(β - α)$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

The angle of deviation is the angle between the directions of the incident and emergent rays. It can be shown using the laws of refraction that the angle of deviation is given by 2(Î± - Î²), where Î± is the angle of incidence and Î² is the angle of refraction.

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every Optics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.