A) \[f/2\]
B) \[f\]
C) \[4f\]
D) \[2f\]
Correct Answer: D
Solution :
When cut horizontally \[\frac{1}{f}\,=(\mu -1)\,\left[ \frac{1}{R}\,-\frac{1}{(-R)} \right]\,=\frac{2(\mu -1)}{R}\] Focal length remains same when cut vertically \[\frac{1}{f'}\,=(\mu -1)\,\left[ \frac{1}{R}\,-\frac{1}{\infty } \right]\,=\frac{(\mu -1)}{R}\,=\frac{1}{2f}\] (from eq.(1) \[f'=2f\]You need to login to perform this action.
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