The separation between the screen and a plane | vedclass

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Question

(C) Let $L$ be the luminous intensity of the point source.Without the mirror,the distance between the source and the screen is $r$. The illuminance $I

Vedclass · Accepted Answer

$10:9$

Vedclass · Answer

(C) Let $L$ be the luminous intensity of the point source.Without the mirror,the distance between the source and the screen is $r$. The illuminance $I_1$ on the screen is given by $I_1 = \frac{L}{r^2}$.With the mirror,the screen receives light from the source directly and from the image formed by the mirror. The source is at a distance $r$ from the mirror,so its image is formed at a distance $r$ behind the mirror. The distance from the image to the screen is $r + r + r = 3r$.The illuminance $I_2$ on the screen with the mirror is the sum of the illuminance from the source and the illuminance from the image:$I_2 = \frac{L}{r^2} + \frac{L}{(3r)^2} = \frac{L}{r^2} + \frac{L}{9r^2} = \frac{L}{r^2} \left(1 + \frac{1}{9}\right) = \frac{10}{9} \frac{L}{r^2}$.Therefore,the ratio of the illuminances is:$\frac{I_2}{I_1} = \frac{\frac{10}{9} \frac{L}{r^2}}{\frac{L}{r^2}} = \frac{10}{9} = 10:9$.

The separation between the screen and a plane mirror is $2r$. An isotropic point source of light is placed exactly midway between the mirror and the screen. Assume that the mirror reflects $100\%$ of the incident light. Then the ratio of illuminances on the screen with and without the mirror is

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