Question

A point source of $'S'$ at a distance d from the screen $A$ produces light intensity $I_{0}$ at the centre of the screen. If a completely reflecting mirror $M$ is placed at a distance $d$ behind the source as shown in the figure, the intensity at the centre of the screen changes to :

• A. $\frac{9}{10}I_{0}$
• B. $\frac{10}{9}I_{0}$
• C. $\frac{8}{9}I_{0}$
• D. $\frac{9}{8}I_{0}$

Answer 1

Answer: (B)

We know intensity is defined as
$I=\frac{P}{A}=\frac{P}{4 \pi r^{2}}$ .
where
$P\&A=$ Power and area
$r=$ Distance of point from source
Given intensity due to source on screen is
$I_{0}=\frac{P}{4 \pi d^{2}}$
Intensity on screen due to image of source formed in plane mirror will be
$I=\frac{P}{4 \pi \left(3 d\right)^{2}}=\frac{P}{36 \pi \left(d\right)^{2}}=\frac{I_{0}}{9}$
Total intensity will be $I_{0}+\frac{I_{0}}{9}=\frac{10 I_{0}}{9}$