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(B) $1$. Without the mirror,the illuminance $I_1$ on the screen due to the point source at a distance $r$ is given by $I_1 = \frac{L}{r^2}$,where $L$

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$2:1$

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(B) $1$. Without the mirror,the illuminance $I_1$ on the screen due to the point source at a distance $r$ is given by $I_1 = \frac{L}{r^2}$,where $L$ is the luminous intensity of the source.$2$. With the mirror,the source is placed at the center of curvature $(u = r)$. For a concave mirror,an object placed at the center of curvature forms a real image at the same position $(v = r)$.$3$. The light reaching the screen now consists of two parts: the direct light from the source and the light reflected from the mirror,which appears to come from the image formed at the same location as the source.$4$. The illuminance due to the reflected light is also $I_{reflected} = \frac{L}{r^2}$.$5$. The total illuminance on the screen with the mirror is $I_2 = I_{direct} + I_{reflected} = \frac{L}{r^2} + \frac{L}{r^2} = \frac{2L}{r^2}$.$6$. Therefore,the ratio of illuminances is $\frac{I_2}{I_1} = \frac{2L/r^2}{L/r^2} = 2:1$.

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