A) \[\frac{f}{2},\,\,\frac{l}{2}\]
B) \[\frac{f}{2},\,\,\frac{3}{4}l\]
C) \[f,\,\,\frac{l}{2}\]
D) \[f,\,\,\frac{3}{4}l\]
Correct Answer: D
Solution :
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 \[\frac{{{I}_{2}}}{{{I}_{1}}}=\frac{{{A}_{2}}}{{{A}_{1}}}=\frac{3\pi {{d}^{2}}\cdot 4}{16\pi {{d}^{2}}}=\frac{12}{16}=\frac{3}{4}\] \[\therefore \] \[{{I}_{2}}=\frac{3}{4}{{I}_{1}}\]
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