Question

A point source of light $S$ is placed in front of a perfectly reflecting mirror as shown in the figure. $\sum$ is a screen. The intensity at the center of screen is found to be $I$.
If the mirror is removed, then the intensity at the center of screen would be

• A. $I$
• B. $\frac{I}{9}$
• C. $\frac{9I}{10}$
• D. $2I$

Answer 1

Answer: (C)

$\frac{9I}{10}$

Let power of light source be $P$, then intensity at any point on the screen is due to light rays directly received from source and that due to light rays after reflection from the mirror.
$I=\frac{P}{4\pi a^2}+\frac{P}{4\pi \times (3a{)}^2}$
When mirror is taken away,
$I_1=\frac{P}{4\pi a^2}=\frac{9I}{10}$.