Question

A point source of ${}^{\prime }{S}^{\prime }$ 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
$I=\frac{P}{4\pi {\left(3d\right)}^2}=\frac{P}{36\pi {\left(d\right)}^2}=\frac{I_0}{9}$
Total intensity will be $I_0+\frac{I_0}{9}=\frac{10I_0}{9}$