A) 1.33
B) 1.42
C) 1.53
D) 1.60
Correct Answer: D
Solution :
The system is equivalent to combination of three lenses in contact, i.e., \[\frac{1}{F}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}+\frac{1}{{{f}_{3}}}\] But by lens Makers formula \[\frac{1}{{{f}_{1}}}=(1.5-1)\left( \frac{1}{-\infty }-\frac{1}{-12} \right)=0.5\times \frac{1}{12}=\frac{1}{24}\] \[\frac{1}{{{f}_{2}}}=(\mu -1)\left( \frac{1}{-12}-\frac{1}{12} \right)=-\left( \frac{\mu -1}{6} \right)\] \[\frac{1}{{{f}_{3}}}=(1.5-1)\left( \frac{1}{12}-\frac{1}{\infty } \right)=\frac{1}{24}\] Also \[F=-60\,cm\] \[\therefore \] \[\frac{1}{-60}=\frac{1}{24}-\frac{(\mu -1)}{6}+\frac{1}{24}=-\frac{(\mu -1)}{6}+\frac{1}{12}\] or \[\frac{(\mu -1)}{6}=\frac{1}{12}+\frac{1}{60}\] or \[\frac{(\mu -1)}{6}=\frac{5+1}{60}+\frac{6}{60}\] or \[(\mu -1)=\frac{36}{60}=0.6\] or \[\mu =0.6+1=1.6\]You need to login to perform this action.
You will be redirected in
3 sec