First slide

Spherical/Curved mirrors

Question

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 $49 W m^{- 2}$ . Find the intensity of light on screen in the presence of mirror.

Very difficult

A

$49 W m^{- 2}$
B

$53 W m^{- 2}$
C

$80 W m^{- 2}$
D

$98 W m^{- 2}$

Solution

Here, Intensity without mirror is $I_{0} = \frac{P_{o}}{4 π {(0.2)}^{2}}$ where $P_{0}$ is power of source.

The intensity at the side of mirror is $I = \frac{P_{0}}{4 π {(0.1)}^{2}} = \frac{I_{0} 4 π {(0.2)}^{2}}{4 π {(0.1)}^{2}} = 4 I_{0}$
The power of incident on mirror is $P_{m} = \frac{π d^{2}}{4} \times 4 I_{0} = π d^{2} I_{0}$
Here, d is aperture of mirror,
$∵ u = - 10 c m, f = + 10 c m$
$∵ \frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
$∴ v = 5 c m$
From similar triangles, $\frac{D}{d} = \frac{35}{5}$
$∴ D = 7 d$
Intensity due to reflected light on the screen is $I_{1} = \frac{4 P_{m}}{π {(7 d)}^{2}} = \frac{4 π d^{2} I_{0}}{π 49 d^{2}} = \frac{4 I_{0}}{49}$
Total intensity = $I_{0} + \frac{4 I_{0}}{49} = \frac{53 I_{0}}{49}$
$= \frac{53 \times 49}{49} = 53 W m^{- 2}$

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