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A thin lens focal length $f_{1}$ and its aperture has diameter d. It forms an image of intensity I. Now the central part of the aperture up to diameter, d/2 is blocked by an opaque paper. The focal length and image intensity will change to

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$\frac{f}{2} and \frac{I}{2}$

$f and \frac{I}{4}$

$\frac{3 f}{4} and \frac{I}{2}$

$f and \frac{3 I}{4}$

answer is D.

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Detailed Solution

$I \propto A^{2} \Rightarrow \frac{I_{2}}{I_{1}} = {(\frac{A_{2}}{A_{1}})}^{2} = \frac{{πr}^{2} - \frac{{πr}^{2}}{4}}{{πr}^{2}} = \frac{3}{4} \Rightarrow I_{2} = \frac{3}{4} I_{1}$ and focal length remains unchanged.

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