42. A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperture of diameter d/2 in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively ?
(a) \( f \ \text{and} \ \dfrac{l}{4} \)
(b) \( \dfrac{3f}{4} \ \text{and} \ \dfrac{I}{2} \)
(c) \( f \ \text{and} \ \dfrac{3I}{4} \)
(d) \( \dfrac{f}{2} \ \text{and} \ \dfrac{I}{2} \)
Answer : (c) \( f \ \text{and} \ \dfrac{3I}{4} \)
\( \text{Area} \quad \frac{\pi}{4} d^2 \ \rightarrow \ I \)
\( \text{Area} = \frac{\pi}{4} d^2 - \frac{\pi}{4} \left( \frac{d}{2} \right)^2 \)
\( = \frac{\pi}{4} d^2 \left[ 1 - \frac{1}{4} \right] \)
\( = \frac{3}{4} \cdot \frac{\pi}{4} d^2 \)
\( \text{Intensity} = \frac{3}{4} I \)
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