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Wave Optics

Question

Wired Faculty App

Two coherent light beams of intensities I and 41 are superposed. The maximum and minimum possible intensities in the resulting beam are

5I and I

5I and 3I

9I and I

9I and 3I

Solution

C.

9I and I

Given:- I₁ = I, I₂ = 4I

Resultant intensity = $I_{1} + I_{2} 2 \sqrt{I_{1} I_{2}} cosδ$

where δ is the phase difference between two waves.

intensity will be maximum when δ = 0 ....(cos 0^o = 1)

I_max = $4 I + I + 2 \sqrt{4 I \times I} \cos (0)$

= $4 I + I + 2 \sqrt{4 I^{2}}$

= $4 I + I + 2 (2 I)$

I_max = 9I

Intensity will be minimum when δ = 0

and I_min = $4 I + I + 2 \sqrt{4 I \times I} \cos (180)$

= $4 I + I + 2 (2 I) (- 1) A$ .....(cos 180^o = -1)

I_min = I

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