Question

A thin lens of focal length $ f $ and aperture diameter $ d $ forms an image of intensity $ I. $ If the central part of the aperture upto diameter $ d/2 $ is blocked by an opaque paper, then the new focal length and intensity of image will be

• A. $ \frac{f}{2},\frac{l}{2} $
• B. $ \frac{f}{2},\frac{3}{4}l $
• C. $ f,\frac{l}{2} $
• D. $ f,\frac{3}{4}l $

Answer 1

Answer: (D)

On blocking the central part of the lens, focal length does not change so f remains same. Intensity of image is directly proportional to the area of lens. Initial area,
$ {{A}_{1}}=\pi {{\left( \frac{d}{2} \right)}^{2}}=\frac{\pi {{d}^{2}}}{4} $
On blocking, the central part of the aperture upto diameter
$ \frac{d}{2}, $ the new area $ {{A}_{2}}=\pi {{\left( \frac{d}{2} \right)}^{2}}-\pi {{\left( \frac{d}{4} \right)}^{2}} $
$ =\frac{\pi {{d}^{2}}}{4}-\frac{\pi {{d}^{2}}}{16} $
$ =\frac{3\pi {{d}^{2}}}{16} $
As $ \frac{{{l}_{2}}}{{{l}_{1}}}=\frac{{{A}_{2}}}{{{A}_{1}}} $
$ =\frac{3\pi {{d}^{2}}.4}{16\pi {{d}^{2}}}=\frac{12}{16}=\frac{3}{4} $
$ \therefore $ $ {{l}_{2}}=\frac{3}{4}{{l}_{1}} $