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Physics NEET Questions — Free MCQ Practice

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A star emitting light of wavelength 5896 Å is moving away from the earth with a speed of 3600 km/sec. The wavelength of light observed on earth will

A.

Decrease by 5825.25 Å

B.

Increase by 5966.75 Å

C.

Decrease by 70.75 Å

D.

Increase by 70.75 Å

(c = 3 × 10⁸ m/sec is the speed of light)

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Explanation

$ν' = ν \sqrt{(\frac{1 - \frac{v}{c}}{1 + \frac{v}{c}})} λ = λ \sqrt{(\frac{1 + \frac{v}{c}}{1 - \frac{v}{c}})} = λ \sqrt{(\frac{1 + \frac{3600}{c}}{1 - \frac{3600}{c}})} = λ \sqrt{\frac{1.012}{0.988}} = 1.012 λ = 70.75 \overset{o}{A}$

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A heavenly body is receding away from the earth such that the fractional change in λ is 1, then its velocity is :

A.

C

B.

$\frac{3 C}{5}$

C.

$\frac{C}{5}$

D.

$\frac{2 C}{5}$

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Explanation

$W h e n a p l a n e t r e c e d e s a w a y f r o m t h e e a r t h a n d f i s t h e f r e q u e n c y o f v i b r a t i o n : f = \frac{c}{λ} I f v = v e l o c i t y o f m o v i n g a w a y f r o m t h e e a r t h a n d λ' = a p p a r e n t w a v e l e n g t h t o a n o b e s e r v e r o n t h e e a r t h; λ' = \frac{c + v}{f} \Rightarrow λ' = (\frac{c + v}{c}) λ F r a c t i o n a l c h a n g e i n w a v e l e n t h = \frac{λ' - λ}{λ} = \frac{(\frac{c + v}{c}) λ - λ}{λ} = \frac{v}{c} = 1 \Rightarrow v = c$

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A slit of width a is illuminated by white light. For red light (λ = 6500 Å), the first minima is obtained at θ = 30°. Then the value of a will be

A.

3250 Å

B.

6.5 × 10^–4 mm

C.

1.24 microns

D.

2.6 × 10^–4 cm

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Explanation

For first minima $θ = \frac{λ}{a}$ or $a = \frac{λ}{θ}$

$∴ a = \frac{6500 \times 10^{- 8} \times 6}{π}$ (As 30^o = $\frac{π}{6}$ radian)

$= 1 .24 \times 10^{- 4} c m = 1 .24$ microns

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The radius of central zone of the circular zone plate is 2.3 mm. The wavelength of incident light is $5893 Å .$ Source is at a distance of 6m. Then the distance of the first image will be

A.

9 m

B.

12 m

C.

24 m

D.

36 m

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Explanation

$f_{1} = \frac{r^{2}}{λ} = \frac{{(2 .3 \times 10^{- 3})}^{2}}{5893 \times 10^{- 10}} = 9 m .$

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What will be the angular width of central maxima in Fraunhoffer diffraction when light of wavelength $6000 Å$ is used and slit width is 12×10^–5 cm

A.

2 rad

B.

3 rad

C.

1 rad

D.

8 rad

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Explanation

Angular width $= \frac{2 λ}{d} = \frac{2 \times 6000 \times 10^{- 10}}{12 \times 10^{- 5} \times 10^{- 2}} = 1 r a d .$

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Direction of the first secondary maximum in the Fraunhofer diffraction pattern at a single slit is given by (a is the width of the slit)

A.

$a \sin θ = \frac{λ}{2}$

B.

$a \cos θ = \frac{3 λ}{2}$

C.

$a \sin θ = λ$

D.

$a \sin θ = \frac{3 λ}{2}$

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Explanation

For n^th secondary maxima path difference

$d \sin θ = (2 n + 1) \frac{λ}{2} \Rightarrow a \sin θ = \frac{3 λ}{2}$

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A parallel beam of monochromatic light of wavelength 5000 Å is incident normally on a single narrow slit of width 0.001 mm. The light is focused by a convex lens on a screen placed on the focal plane. The first minimum will be formed for the angle of diffraction equal to

A.

0^o

B.

15^o

C.

30^o

D.

60^o

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Explanation

For the first minima $d \sin θ = λ$

$\Rightarrow \sin θ = \frac{λ}{d} \Rightarrow θ = \sin^{- 1} (\frac{5000 \times 10^{- 10}}{0 .001 \times 10^{- 3}}) = 30^{o}$

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In the far field diffraction pattern of a single slit under polychromatic illumination, the first minimum with the wavelength $λ_{1}$ is found to be coincident with the third maximum at $λ_{2}$ . So

A.

$3 λ_{1} = 0 .3 λ_{2}$

B.

$3 λ_{1} = λ_{2}$

C.

$λ_{1} = 3 .5 λ_{2}$

D.

$0 .3 λ_{1} = 3 λ_{2}$

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Explanation

Position of first minima = position of third maxima i.e., $\frac{1 \times λ_{1} D}{d} = \frac{(2 \times 3 + 1)}{2} \frac{λ_{2} D}{d} \Rightarrow λ_{1} = 3 .5 λ_{2}$

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The angle of polarisation for any medium is 60^o, what will be critical angle for this

A.

$\sin^{- 1} \sqrt{3}$

B.

$\tan^{- 1} \sqrt{3}$

C.

$\cos^{- 1} \sqrt{3}$

D.

$\sin^{- 1} \frac{1}{\sqrt{3}}$

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Explanation

By using $μ = \tan θ_{p} \Rightarrow μ = \tan 60 = \sqrt{3}$ ,

also $C = \sin^{- 1} (\frac{1}{μ}) \Rightarrow C = \sin^{- 1} (\frac{1}{\sqrt{3}})$

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In the propagation of electromagnetic waves, the angle between the direction of propagation and plane of polarisation is

A.

0^o

B.

45^o

C.

90^o

D.

180^o

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Explanation

Plane of polarization is a confinement of the electric/magnetic field vector to a given plane along the direction of propagation. Therefore the angle between them is $0^{0}$

Physics MCQs for NEET — Practice Questions with Answers

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