A lens having focal length f and aperture of diameter d forms an image of intensity $I$ . Aperture of diameter $\frac{d}{2}$ in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively:

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$f a n d \frac{3 I}{4}$

$\frac{3 f}{4} a n d \frac{I}{2}$

$f a n d \frac{I}{4}$

$\frac{f}{2} a n d \frac{I}{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} F o c a l l e n g t h o f t h e l e n s r e m a i n s s a m e . I n t e n s i t y o f i m a g e f o r m e d b y l e n s i s p r o p o r t i o n a l \\ t o a r e a e x p o s e d t o i n c i d e n t l i g h t f r o m o b j e c t . \\ ∴ \frac{I_{2}}{I_{1}} = \frac{A_{2}}{A_{1}} \\ I n i t i a l l y, A_{1} = π {(\frac{d}{2})}^{2} = \frac{π d^{2}}{4} \\ F i n a l l y, a f t e r b l o c k i n g e x p o s e d a r e a i s \\ A_{2} = \frac{π d^{2}}{4} - π {(\frac{d}{4})}^{2} = \frac{3 π d^{2}}{16} \\ ∴ \frac{I_{2}}{I_{1}} = \frac{A_{2}}{A_{1}} = \frac{\frac{3 π d^{2}}{16}}{\frac{π d^{2}}{4}} = \frac{3}{4} \\ \Rightarrow I_{2} = \frac{3}{4} I_{1} = \frac{3}{4} I \\ H e n c e, f o c a l l e n g t h o f a l e n s = f \\ a n d I n t e n s i t y o f t h e i m a g e = 3 I / 4 \end{array}$