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, \[\frac{1}{f}=(\mu -1)\left[ \frac{1}{R}-\frac{1}{(-R)} \right]\] \[=(\mu -1)\left[ \frac{1}{R}+\frac{1}{R} \right]\] \[=(\mu -1)\frac{2}{R}=\frac{1}{f'}\] \[\Rightarrow \] \[f'=f\] Case II: When lens is cut along\[YOY,\]then each half becomes plano-convex with \[{{R}_{1}}=+R,{{R}_{2}}=\infty \] Thus, \[\frac{1}{f''}=(\mu -1)\left[ \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right]\] \[=(\mu -1)\left( \frac{1}{R}-\frac{1}{\infty } \right)\] \[=\frac{(\mu -1)}{R}=\frac{1}{2f}\] Hence, \[f'=f,f''=2f\]
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