question_answer

A thin convex lens of focal length 10 cm and refractive index 1.5 is cut vertically into two equal pieces. They are placed as shown with a liquid of refractive index 3 between them. What is the focal   length of the combination?

A) -10cm.

B) -10/4 cm.

C) -10/3 cm.

D) None of these

Correct Answer: C

Solution :

[c] \[\frac{1}{f}=\left( \mu -1 \right)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right);\] \[R=10cm.\] \[\frac{1}{f'}=\left( 3-1 \right)\left( \frac{1}{-10}-\frac{1}{10} \right),\] \[f'=-\frac{10}{4}\] \[\frac{1}{{{f}_{eq}}}=\frac{1}{20}-\frac{4}{10}+\frac{1}{20}=\frac{2}{20}-\frac{4}{10}\Rightarrow {{f}_{eq}}=-\frac{10}{3}\]