A point source of light is 60, cm from a scre | vedclass

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Question

(D) Let the power of the point source be $P$. When the mirror is removed,the light from the source falls directly on the screen at a distance $d = 60\

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(D) Let the power of the point source be $P$. When the mirror is removed,the light from the source falls directly on the screen at a distance $d = 60\, cm$. The intensity $I_1$ is given by $I_1 = \frac{P}{4\pi d^2} = \frac{P}{4\pi (60)^2}$. When the mirror is present,the light reaches the screen in two ways: $1$. Directly from the source: $I_{direct} = \frac{P}{4\pi (60)^2}$. $2$. Reflected from the mirror: Since the source is at the focus $(f = 20\, cm)$,the reflected rays are parallel to the principal axis. These rays form a beam of the same radius as the mirror aperture. Assuming the mirror aperture radius is $h$,the reflected light is concentrated on an area $\pi h^2$. The intensity of reflected light is $I_{reflected} = \frac{P}{4\pi h^2}$. Given the geometry,$	an 	heta = \frac{h}{20} \approx 	heta$ for small angles. Also,from the direct path,$	an 	heta = \frac{h}{60}$. Using the ratio of intensities: $I_{total} = I_{direct} + I_{reflected} = \frac{P}{4\pi (60)^2} + \frac{P}{4\pi (20)^2}$. The ratio is $\frac{I_{total}}{I_1} = \frac{\frac{P}{4\pi (60)^2} + \frac{P}{4\pi (20)^2}}{\frac{P}{4\pi (60)^2}} = 1 + \frac{(60)^2}{(20)^2} = 1 + 3^2 = 1 + 9 = 10$. Thus,the ratio is $10 : 1$.

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