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 :
\[I\propto {{A}^{2}}\Rightarrow \frac{{{I}_{2}}}{{{I}_{1}}}={{\left( \frac{{{A}_{2}}}{{{A}_{1}}} \right)}^{2}}=\frac{\pi {{r}^{2}}-\frac{\pi {{r}^{2}}}{4}}{\pi {{r}^{2}}}=\frac{3}{4}\] \[\Rightarrow {{I}_{2}}=\frac{3}{4}{{I}_{1}}\] and focal length remains unchanged.
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