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Wave Optics MCQs for NEET — Physics Questions with Answers

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

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If L is the coherence length and c the velocity of light, the coherent time is

$\frac{L}{c}$

$\frac{c}{L}$

$\frac{1}{L c}$

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Explanation

Coherent time $= \frac{C o h e r e n c e l e n g t h}{V e l o c i t y o f l i g h t} = \frac{L}{c}$

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If the amplitude ratio of two sources producing interference is 3 : 5, the ratio of intensities at maxima and minima is

25 : 16

5 : 3

16 : 1

25 : 9

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Explanation

$\frac{a_{1}}{a_{2}} = \frac{3}{5}$

$∴ \frac{I_{\max}}{I_{\min}} = \frac{{(a_{1} + a_{2})}^{2}}{{(a_{1} - a_{2})}^{2}} = \frac{{(3 + 5)}^{2}}{{(3 - 5)}^{2}} = \frac{16}{1}$

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For constructive interference to take place between two monochromatic light waves of wavelength λ, the path difference should be

$(2 n - 1) \frac{λ}{4}$

$(2 n - 1) \frac{λ}{2}$

$n λ$

$(2 n + 1) \frac{λ}{2}$

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Explanation

For constructive interference path difference is even multiple of $\frac{λ}{2}$ .

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Soap bubble appears coloured due to the phenomenon of

Interference

Diffraction

Dispersion

Reflection

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Explanation

Phenomenon of interference of light takes place.

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Which of the following statements indicates that light waves are transverse

Light waves can travel in vacuum

Light waves show interference

Light waves can be polarized

Light waves can be diffracted

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Explanation

Transverse waves can be polarised.

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Interference was observed in interference chamber when the air was present, now the chamber is evacuated and if the same light is used, a careful observer will see

No interference

Interference with bright bands

Interference with dark bands

Interference in which width of the fringe will be slightly increased

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Explanation

The refractive index of air is slightly more than 1. When chamber is evacuated, refractive index decreases and hence the wavelength increases and fringe width also increases.

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Ray diverging from a point source forms a wave front that is -

Cylindrical

Spherical

Plane

Cubical

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Explanation

A point source emits light in all directions. The wavefronts formed by the rays diverging from a point source are spherical. This is because the distance traveled by each ray from the source is the same, forming a spherical wavefront.

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Two coherent sources have intensity in the ratio of $\frac{100}{1}$ . Ratio of (intensity) max/(intensity) min is

$\frac{1}{100}$

$\frac{1}{10}$

$\frac{10}{1}$

$\frac{3}{2}$

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Explanation

$\frac{I_{1}}{I_{2}} = \frac{100}{1}$

Now $\frac{I_{\max}}{I_{\min}} = {(\frac{\sqrt{\frac{I_{1}}{I_{2}}} + 1}{\sqrt{\frac{I_{1}}{I_{2}}} - 1})}^{2} = {(\frac{\sqrt{100} + 1}{\sqrt{100} - 1})}^{2} = \frac{121}{81} \approx \frac{3}{2}$

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If two waves represented by $y_{1} = 4 \sin ω t$ and $y_{2} = 3 \sin (ω t + \frac{π}{3})$ interfere at a point, the amplitude of the resulting wave will be about

3.5

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Explanation

$ϕ = π / 3, a_{1} = 4, a_{2} = 3$

So, $A = \sqrt{a_{1}^{2} + a_{2}^{2} + 2 a_{1} . a_{2} \cos φ} \Rightarrow A \approx 6$

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Two coherent sources of intensities, I₁ and I₂ produce an interference pattern. The maximum intensity in the interference pattern will be

I₁ + I₂

$I_{1}^{2} + I_{2}^{2}$

(I₁ + I₂)²

${(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$

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Explanation

Resultant intensity $I_{R} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cos ϕ$

For maximum $I_{R}, ϕ = 0^{o}$

⇒ $I_{R} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$

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