A) \[\mu R\]
B) \[\frac{R}{(\mu -1)}\]
C) \[\frac{{{R}^{2}}}{\mu }\]
D) \[\left[ \frac{(\mu +1)}{(\mu -1)} \right]R\]
Correct Answer: B
Solution :
When an object is placed in front of such a lens, the rays are first of all refracted from the convex surface, then reflected from the polished plane surface and again refracted from convex surface. Let\[{{f}_{l}},\,\,{{f}_{m}}\]be focal lengths of convex surface and mirror (plane polished surface) respectively, then effective focal length is \[\frac{1}{F}=\frac{1}{{{f}_{l}}}+\frac{1}{{{f}_{m}}}+\frac{1}{{{f}_{l}}}=\frac{2}{{{f}_{l}}}+\frac{1}{{{f}_{m}}}\] Since, \[{{f}_{m}}=\frac{R}{2}=\infty \] \[\therefore \] \[\frac{1}{F}=\frac{2}{{{f}_{l}}}\] From lens formula \[\frac{1}{{{f}_{l}}}=(\mu -1)\left( \frac{1}{R} \right)\] \[\therefore \] \[\frac{1}{F}=\frac{2(\mu -1)}{R}\] \[\Rightarrow \] \[F=\frac{R}{2(\mu -1)}\] or \[{{R}_{eq}}=2\,\,F=\frac{R}{(\mu -1)}\]You need to login to perform this action.
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