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(D) Let the radius of the circular tunnel be $r$. The lamp is at the top of the tunnel. The distance of point $A$ from the lamp is $2r$.Intensity at $

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$1:\sqrt{2}$

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(D) Let the radius of the circular tunnel be $r$. The lamp is at the top of the tunnel. The distance of point $A$ from the lamp is $2r$.Intensity at $A$ is $I_A = \frac{L}{(2r)^2} = \frac{L}{4r^2}$,where $L$ is the luminous intensity of the lamp.For point $B$,the distance from the lamp is $d = \sqrt{r^2 + r^2} = r\sqrt{2}$. The angle $	heta$ between the normal at $B$ and the light ray is $45^\circ$.Intensity at $B$ is $I_B = \frac{L}{d^2} \cos 	heta = \frac{L}{(r\sqrt{2})^2} \cos 45^\circ = \frac{L}{2r^2} \cdot \frac{1}{\sqrt{2}} = \frac{L}{2\sqrt{2}r^2}$.Therefore,the ratio is $\frac{I_A}{I_B} = \frac{L/4r^2}{L/2\sqrt{2}r^2} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.

An electric lamp is fixed at the ceiling of a circular tunnel as shown in the figure. What is the ratio of the intensities of light at base $A$ and a point $B$ on the wall?

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