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(D) The illuminance $E$ at a point due to a source is given by $E = \frac{I \cos 	heta}{r^2}$, where $I$ is the luminous intensity, $r$ is the distan

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$B = C < A$

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(D) The illuminance $E$ at a point due to a source is given by $E = \frac{I \cos 	heta}{r^2}$, where $I$ is the luminous intensity, $r$ is the distance from the source, and $	heta$ is the angle between the normal to the surface and the direction of the light rays.Point $A$ is directly below the center of the tube, making it the closest point to the source. Thus, the distance $r_A$ is the minimum.Points $B$ and $C$ are located symmetrically at the same distance from the center of the tube, so $r_B = r_C > r_A$.Since the illuminance is inversely proportional to the square of the distance, the illuminance at $A$ is greater than the illuminance at $B$ and $C$.Therefore, $E_A > E_B = E_C$, which can be written as $B = C < A$.

The figure shows a glowing mercury tube. The illuminances at points $A$, $B$, and $C$ are related as

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