A) \[\frac{f}{2}and\frac{I}{2}\]
B) \[f\,and\frac{I}{4}\]
C) \[\frac{3f}{4}\,and\frac{I}{2}\]
D) \[f\,and\frac{3I}{4}\]
Correct Answer: D
Solution :
On blocking the central part,\[f\]is not affected. Only the intensity decreases original area of lens \[{{A}_{1}}=\pi \frac{{{d}^{2}}}{4}\] Now, area of lens that transmits light \[{{A}_{1}}=\pi \frac{{{d}^{2}}}{4}-\frac{\pi {{(d/2)}^{2}}}{4}=\frac{3\pi {{d}^{2}}}{16}\] \[\therefore \] \[\frac{{{A}_{2}}}{{{A}_{1}}}=\frac{3\pi {{d}^{2}}}{16},\frac{4}{\pi {{d}^{2}}}=\frac{3}{4}\] \[\therefore \] \[\frac{{{I}_{2}}}{{{I}_{1}}}=\frac{3}{4},\]i.e., \[{{I}_{2}}=\frac{3}{4}{{I}_{1}}=\frac{3}{4}I\]
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