Blog Log in Get Started Free

Physics MCQs for NEET — Practice Questions with Answers

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

All Physics Chemistry Botany Zoology

Chapter Language Difficulty Show only questions I answered incorrectly Register free to filter questions

● ● ● ● ●

Two thin lenses of focal length $f_{1} a n d f_{2}$ are in contact and coaxial. The power of the combination is

$\sqrt{\frac{f_{1}}{f_{2}}}$

$\sqrt{\frac{f_{2}}{f_{1}}}$

$\frac{f_{1} - f_{2}}{f_{1} f_{2}}$

$\frac{f_{1} + f_{2}}{f_{1} f_{2}}$

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

Explanation

If two thin lenses of focal lengths $f_{1}, f_{2}$ are placed in contact coaxially, then equivalent focal length of combination is

$\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} - \frac{0}{f_{1} f_{2}} = \frac{1}{f_{1}} + \frac{1}{f_{2}}$

Power for the combination is

$P = \frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{f_{1} + f_{2}}{f_{1} f_{2}}$

● ● ● ● ●

A boy is trying to start a fire by focusing sunlight on a piece of paper using an equiconvex lens of focal length 10 cm. The diameter of the sun is 1.39 $\times 10^{9} m$ and its mean distance from the earth is $1.5 \times 10^{11} m$ . What is the diameter of the sun's image on the paper?

$9.2 \times 10^{- 4} m$

$6.5 \times 10^{- 4} m$

$6.5 \times 10^{- 5} m$

$12.4 \times 10^{- 4} m$

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

Explanation

From the relation

$\frac{I}{O} = \frac{v}{u}$

$H e r e, O = 1.39 \times 10^{9} m, v = 0.1 m, u = 1.5 \times 10^{11} m ∴ I = \frac{0.1}{1.5 \times 10^{11}} \times 1.39 \times 10^{9} = 9.2 \times 10^{- 4} m$

● ● ● ● ●

A parallel beam of moving electrons is incident normal on a narrow slit. A fluorescent screen is placed at a large distance from the slit. If the slit is further narrowed, then which of the following statement is correct?

The diffraction pattern is not observed on the screen in the case of electrons.

The angular width of the central maximum of the diffraction pattern will increase.

The angular width of the central maximum will decrease.

The angular width of the central maximum will remain the same.

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

Explanation

$(2) β = \frac{2 λD}{a} If a ↓ then β ↑ .$

● ● ● ● ●

In Young's double-slit experiment, the slits are 0.4 mm apart and illuminated by photons of two wavelengths $λ_{1}$ = 600 nm and $λ_{2}$ = 700nm. At what minimum distance from the common center, bright fringe from one interference pattern coincides with bright fringe from other? (Given that the distance of the screen from the plane of slits is 80cm.)

7.2 mm

6.0 mm

8.4 mm

9.8 mm

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

Explanation

$L e t n_{1} b r i g h t f r i n g e f r o m t h e f i r s t i n t e r f e r e n c e c o i n c i d e s w i t h t h e n_{2} b r i g h t f r i n g e o f t h e s e c o n d i n t e r f e r e n c e . T h e n, n_{1} λ_{1} = n_{2} λ_{2} \frac{n_{1}}{n_{2}} = \frac{λ_{2}}{λ_{1}} \Rightarrow \frac{n_{1}}{n_{2}} = \frac{700}{600} = \frac{7}{6} S o, 7 t h b r i g h t f r i n g e o f 1 s t w i l l c o i n c i d e w i t h 6 t h b r i g h t f r i n g e o f 2 n d . D i s \tan c e f r o m t h e c e n t r a l m a x i m a o f 1 s t = \frac{7 λ D}{d} = \frac{7 \times 600 \times 10^{- 9} \times 0.8}{4 \times 10^{- 4}} = 8.4 m m$

● ● ● ● ●

A diffraction pattern is observed using a beam of red light. What will happen if the red light is replaced by the blue light?

No change takes place

Diffraction bands become narrower

Diffraction bands become broader

Diffraction pattern disappears

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

Explanation

The angular width of the diffraction pattern is directly proportional to wavelength.

● ● ● ● ●

The phenomenon of polarisation justify which nature of light?

longitudinal

transverse

both

geometrical

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

Explanation

Transverse nature of light.

● ● ● ● ●

Two waves of intensity ratio 9:1 interfere to produce fringes in a young's double-slit experiment, the ratio of intensity at maxima to the intensity at minima is

4:1

9:1

81:1

9:4

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

Explanation

$(1) \frac{I_{1}}{I_{2}} = \frac{a_{1}^{2}}{a_{2}^{2}} = \frac{9}{1} \frac{I_{\max}}{I_{\min}} = \frac{{(a_{1} + a_{2})}^{2}}{{(a_{1} - a_{2})}^{2}} = \frac{{(3 + 1)}^{2}}{{(3 - 1)}^{2}} = 4 : 1$

● ● ● ● ●

Interference fringes are produced using white light in a double-slit arrangement. When a mica sheet of uniform thickness of the refractive index 1.6 (relative to air) is placed in the path of light from one of the slits, the central fringe moves by a distance.

This distance is equal to the width of 30 interference bands. If the light of wavelength 4800 $\overset{o}{A}$ is used, the thickness (in $μ$ m) of mica is —

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

Explanation

Shift of fringe pattern = $(μ - 1) \frac{t D}{d}$

$∴ \frac{30 D (4800 \times 10^{- 10})}{d} = (0.6) t \frac{D}{d} 30 \times 4800 \times 10^{- 10} = 0.6 t t = \frac{30 \times 4800 \times 10^{- 10}}{0.6} = \frac{1.44 \times 10^{- 5}}{0.6} = 24 \times 10^{- 6}$

● ● ● ● ●

Young's double-slit experiment is first performed in air and then in a medium other than air. It is found that 8th bright fringe in the medium lies where 5th dark fringe lies in the air. The refractive index of the medium is nearly

1.25

1.59

1.69

1.78

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

Explanation

(d) According to question, 5th fringe in air = 8th bright fringe in the medium

$(2 \times 5 - 1) \frac{λ D}{2 d} = 8 \frac{λ D}{μ d}$

$9 \frac{λ D}{2 d} = 8 \frac{λ D}{μ d}$

$\Rightarrow \frac{9}{2} = \frac{8}{μ}$

$\Rightarrow μ = \frac{8 \times 2}{9}$

$∴$ Refractive index of the medium.

$μ = \frac{16}{9} = 1.7777 = 1.78$

● ● ● ● ●

Two polaroids $P_{1} a n d P_{2}$ are placed with their axis perpendicular to each other. Unpolarised light $I_{o}$ is incident on $P_{1}$ . A third polaroid $P_{3}$ is kept in between $P_{1} a n d P_{2}$ such that its axis makes an angle $45^{°}$ with that of $P_{1}$ . The intensity of transmitted light through $P_{2}$

$\frac{I_{o}}{2}$

$\frac{I_{o}}{4}$

$\frac{I_{o}}{8}$

$\frac{I_{o}}{16}$

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

Explanation

When unpolarized light is incident on the first polaroid P1, the transmitted light becomes linearly polarized. The intensity of this polarized light is reduced by half. When it passes through P3 which makes an angle of 45Â° with P1, the intensity is further reduced by cos^2(45Â°) = 1/2. Finally, when this light reaches P2, which is perpendicular to P1, the component parallel to the axis of P2 is blocked, reducing the intensity by another 1/2. Thus, the final intensity transmitted through P2 is (Io/2) * (1/2) * (1/2) = Io/8.

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every Physics 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.