A point source of light S at a distance d fro | vedclass

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(B) Let the power of the point source $S$ be $P$. The intensity $I_0$ at the centre of the screen due to the source $S$ at distance $d$ is given by $I

Vedclass · Accepted Answer

$\frac{10}{9} I_0$

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(B) Let the power of the point source $S$ be $P$. The intensity $I_0$ at the centre of the screen due to the source $S$ at distance $d$ is given by $I_0 = \frac{P}{4 \pi d^2}$.The mirror $M$ is placed at a distance $d$ behind the source. The image $S'$ of the source $S$ formed by the mirror will be at a distance $d$ behind the mirror. Thus,the total distance of the image $S'$ from the screen $A$ is $d + d + d = 3d$.The intensity $I'$ at the centre of the screen due to the image $S'$ is $I' = \frac{P}{4 \pi (3d)^2} = \frac{P}{4 \pi (9d^2)} = \frac{1}{9} \left( \frac{P}{4 \pi d^2} \right) = \frac{1}{9} I_0$.Since the object and its image are incoherent,the total intensity $I$ at the centre of the screen is the sum of the intensities due to the source and its image:$I = I_0 + I' = I_0 + \frac{1}{9} I_0 = \frac{10}{9} I_0$.

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