A) \[\frac{f}{2}\]
B) \[f\]
C) \[\frac{3f}{2}\]
D) \[2f\]
Correct Answer: D
Solution :
For equiconvex lens \[{{R}_{1}}=-{{R}_{2}}=R,\] so that \[\frac{1}{f}=(\mu -1)\left\{ \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right\}=(\mu -1)\frac{2}{R}\] If the lens in cut along AB, we have two piano convex lenses. For a planoconvex lens, \[{{R}_{2}}=\infty .\] Therefore, its focal length \[f\]is \[\frac{1}{f'}=(\mu -1)\frac{1}{R}=\frac{1}{2f},\] \[\therefore \] \[f'=2f.\] Hence, the correction option is [d].
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