A point source of light S is placed at a distance d from a screen; the intensity at the center of the screen is I. When a perfectly reflecting mirror M is placed a distance d behind the source, then intensity at the center of the screen becomes $\frac{10}{P} I$ .What is value of P.

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answer is 9.

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Detailed Solution

Intensity, $I = \frac{k}{r^{2}}$
Intensity at screen without mirror
$I_{1} = \frac{k}{d^{2}}$
Intensity at the screen with mirror
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$I_{2} = \frac{k}{d^{2}} + \frac{k}{{(3 d)}^{2}} = \frac{10 k}{9 d^{2}}$

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