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Wave Optics NEET Questions — Free MCQ Practice

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In a diffraction pattern due to a single slit of width a,the first minimum is observed at an angle $30^{\circ}$ when light of wavelength 5000 $\dot{A}$ is incident on the slit. The first secondary maximum is observed is an angle of

A.

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

B.

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

C.

$\sin^{- 1} (\frac{3}{4})$

D.

$\sin^{- 1} (\frac{1}{4})$

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Explanation

(c) As the first minimum is observed at an angle of $30^{\circ}$ in a diffraction pattern due to a single slit of width a.

i.e., $n = 1, θ = 30^{\circ}$

$∵$ According to bragg's law of diffraction,

$a \sin θ = n λ \Rightarrow a \sin 30^{\circ} = (1) λ (n = 1) \Rightarrow a = 2 λ . . . (i) \{∵ \sin 30^{\circ} = \frac{1}{2}\}$

For Ist secondary maxima

$\Rightarrow a \sin θ_{1} = \frac{3 λ}{2} \Rightarrow \sin θ_{1} = \frac{3 λ}{2 a} . . . (i i)$

Substitute value of a from eq. (i) to eq (ii), we get

$\sin θ_{1} = \frac{3 λ}{4 λ} \Rightarrow \sin θ_{1} = \frac{3}{4} \Rightarrow θ_{1} = \sin^{- 1} (\frac{3}{4})$

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For a parallel beam of monochromatic light of wavelength diffraction is produced by a single slit whose width 'a' is of the order of the wavelength of the light. If 'D' is the distance of the screen from the slit, the width of the central maxima will be

A.

2Dλ/a

B.

Dλ/a

C.

Da/λ

D.

2Da/λ

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In a double-slit experiment, the two slits are 1 mm apart and the screen is placed 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double-slit within the central maxima of a single-slit pattern?

A.

0.2 mm

B.

0.1 mm

C.

0.5 mm

D.

0.02 mm

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Explanation

Given d=1mm=1x10^-3m

D=1m λ=500mm=5x10^-7 m

As width of central maxima=width of 10 maxima

2Dλ/a=10(λD/d)

=> a=d/5=10^-3/5

=0.2x10^-3 m

=0.2mm

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At the first minimum adjacent to the central maximum of a single slit diffraction pattern, the phase difference between the Huygen's wavelet from the edge of the slit and the wavelet from the midpoint of the slit is

A.

π/4 radian

B.

π/2 radian

C.

π radian

D.

π/8 radian

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Explanation

At the first minimum adjacent to the central maximum in a single slit diffraction pattern, the phase difference between the Huygen's wavelet from the edge of the slit and the wavelet from the midpoint of the slit is Ï€ radians. This phase difference leads to destructive interference, resulting in the formation of the first minimum.

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In the Young's double-slit experiment, the intensity of light at a point on the screw (where the path difference is λ ) is K. (λ being the wavelength of light used). The intensity at a point where the path difference is λ /4 will be

A.

K

B.

K/4

C.

K/2

D.

zero

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Explanation

For net intensity

I'=4I_o cos² φ/2 (φ=2π/λxλ)

For the first case,

K=4I_o cos2 [π] K=4I_o ...(i)

For the second case

K'=4I_o cos² (π/2/2) (φ=2π/λxλ/4)

=4I_o cos² (π/2)

K'=2I_o ...(ii)

Comparing Eqs. (i)and(ii)

K'=K/2

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In Young’s double slit experiment. the slits are 2 mm apart and are illuminated by photons of two wavelengths , λ₁= 12000Å and , λ₂= 10000Å. At what minimum distance from the common central bright fringe on the screen 2m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?

A.

8mm

B.

6mm

C.

4 mm

D.

3mm

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A parallel beam of fast-moving electrons is incident normally on a narrow slit. A fluorescent screen is placed at a large distance from the slit. If the speed of the electrons is increased, then which of the following statements is correct?

A.

Direction pattern is not observed on the screen in the case of electrons

B.

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

C.

The angular width of the central maximum will decrease

D.

The angular width of the central maximum will be unaffected

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Explanation

(c)

$As, λ = \frac{h}{mv} \Rightarrow λ \propto \frac{1}{v} Therefore, as speed of electron increases, de - broglie wavelngth of electron decreases . Angular width of central maxima, w \propto λ \propto \frac{1}{v} Then, width of central maxima decreases as speed of electron increases .$

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Two coherent sources of light can be obtained by

A.

Two different lamps

B.

Two different lamps but of the same power

C.

Two different lamps of the same power and having the same colour

D.

None of the above

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Explanation

The coherent source cannot be obtained from two different light sources.

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By Huygen's wave theory of light, we cannot explain the phenomenon of

A.

Interference

B.

Diffraction

C.

Photoelectric effect

D.

Polarisation

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Explanation

Huygen's wave theory fails to explain the particle nature of light (i.e. photoelectric effect)

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Two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible intensities in the resulting beam are

A.

5I and I

B.

5I and 3I

C.

9I and I

D.

9I and 3I

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Explanation

$I_{\max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} = {(\sqrt{I} + \sqrt{4 I})}^{2} = 9 I$

$I_{\min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2} = {(\sqrt{I} - \sqrt{4 I})}^{2} = I$

Wave Optics MCQs for NEET — Physics Questions with Answers

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