A point source of light S is placed at distance 10 cm from the pole of a convex mirror of local length 10 cm. The intensity of light at the center of screen in the absence of mirror is $49W{m}^{-2}$ . Find the intensity of light on screen in the presence of mirror.

Here, Intensity without mirror is   $I_0=\frac{P_o}{4\pi {\left(0.2\right)}^2}$where $P_0$ is power of source.

 
The intensity at the side of mirror is  $I=\frac{P_0}{4\pi {\left(0.1\right)}^2}=\frac{I_{0}4\pi {\left(0.2\right)}^2}{4\pi {\left(0.1\right)}^2}=4I_0$
The power of incident on mirror is $P_m=\frac{\pi d^2}{4}\times 4I_0=\pi d^2{I}_0$
Here, d is aperture of mirror,
 $\because u=-10cm,f=+10cm$
 $\because \frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
 $\therefore v=5cm$
From similar triangles, $\frac{D}{d}=\frac{35}{5}$
 $\therefore D=7d$
Intensity due to reflected light on the screen is $I_1=\frac{4P_m}{\pi {\left(7d\right)}^2}=\frac{4\pi d^2{I}_0}{\pi 49d^2}=\frac{4I_0}{49}$
Total intensity =  $I_0+\frac{4I_0}{49}=\frac{53I_0}{49}$

                               $=\frac{53\times 49}{49}=53W{m}^{-2}$