A) \[0 < \mu < 1\]
B) \[1 < \mu < 1.5\]
C) \[\mu > 2\]
D) None of these
Correct Answer: B
Solution :
As, \[\frac{1}{f}=(\mu -1)\left( \frac{2}{R} \right)\,for\,\,equiconvex\,\,lens\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f=\frac{R}{2\left( \mu -1 \right)}\] As, \[\operatorname{f} > R\] \[\Rightarrow \,\,\,\,\,\frac{R}{2(\mu -1)}>R\,\,\,\Rightarrow \,\,\,\frac{1}{2(\mu -1)}\,\,>\,\,1\] \[\Rightarrow \,\,\,\,\,2(\mu -1)<1\] \[\Rightarrow \,\,\,\,\,\,\,\,\mu -1<\frac{1}{2}\,\,\,\Rightarrow \,\,\,\mu <1+\frac{1}{2}\,\,\,\Rightarrow \,\,\,\mu \,\,<\,\,1.5\]You need to login to perform this action.
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