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Two coherent sources of intensity ratio 1 : 4 produce an interference pattern. The fringe visibility will be

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0.6

0.8

0.4

answer is B.

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Detailed Solution

$\frac{I_{1}}{I_{2}} = \frac{1}{4} \Rightarrow I_{1} = k and I_{2} = 4 k ∴ Fringe visibility V = \frac{2 \sqrt{I_{1} I_{2}}}{(I_{1} + I_{2})} = \frac{2 \sqrt{k \times 4 k}}{(k + 4 k)} = 0 .8$

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