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Physics MCQs for NEET — Practice Questions with Answers

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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

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Explanation

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

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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

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Explanation

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

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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

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Explanation

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

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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:

$fθ$

$2 fθ$

$\frac{2 fθ}{D}$

$Dθ$

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Explanation

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

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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:

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

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

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

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

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Explanation

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

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

$18 °$

$36 °$

$72 °$

$144 °$

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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

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Explanation

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

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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

$4 c m^{2}$

$6 c m^{2}$

$16 c m^{2}$

$36 c m^{2}$

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Explanation

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

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

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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

$(α - β)$

$2 (α - β)$

$(α - β) / 2$

$(β - α)$

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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.

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Light enters at an angle of incidence in a transparent rod of refractive index n. For what value of the refractive index of the material of the rod the light once entered into it will not leave it through its lateral face what so ever be the value of angle of incidence

$n > \sqrt{2}$

$n = 1$

$n = 1.1$

$n = 1.3$

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Explanation

For total internal reflection to occur, the angle of incidence must be greater than the critical angle. The condition for this is that the refractive index of the material (n) must be greater than âˆš2 (approximately 1.414).

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