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The two coherent sources of equal intensity produce maximum intensity of 100 units at a point. If the intensity of one of the sources is reduced by 36% by reducing its width then the intensity of light at the same point will be

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answer is 81.

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Detailed Solution

Intensity of each source is $I_{0} = \frac{I_{m a x}}{4} = \frac{100}{4} = 25 u n i t$

If the intensity of one of the sources is reduced by 36%, then $I_{1} = 25 u n i t$ and $I_{2} = 25 - 25 \times \frac{36}{100} = 16$ (unit)

Hence resultant intensity at the same point will now be

$I = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} = 25 + 16 + 2 \sqrt{25 \times 16} = 81 u n i t$

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