Choose the correct statement from the following
A) \[f=f.f=f\]
B) \[f=2f,f=2f\]
C) \[f=f,f=2f\]
D) \[f=2f,f=f\]
Correct Answer: C
Solution :
Initially the focal length of equiconvex lens is \[\frac{1}{f}=(\mu -1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] ?(i) \[\frac{1}{f}=(\mu -1)\left( \frac{1}{R}-\frac{1}{-R} \right)=\frac{2(\mu -1)}{R}\] Case I When lens is cut along XOX, then each half is again equiconvex with \[{{R}_{1}}=+R,{{R}_{2}}=-R\] Thus \[\frac{1}{f}(\mu -1)\left[ \frac{1}{R}-\frac{1}{-(R)} \right]=(\mu -1)\frac{2}{R}=\frac{1}{f}\] \[\Rightarrow \] \[f=f\] Case II When lens is cut along XOY, then each half becomes piano - convex with. \[{{R}_{1}}=+R,{{R}_{2}}=\infty \] Thus \[\frac{1}{f}=(\mu -1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] \[=\left( \frac{\mu -1}{R} \right)=\frac{1}{2f}\] Hence \[f=f,f=2f\]
You need to login to perform this action.
You will be redirected in
3 sec
