Question

Two identical planoconvex - lenses touching each other at their spherical surface of radius $R$. If refractive index of lense is $1.50$,then the lens behave

• A. Convergent with power $P=\frac{1}{R}$
• B. Convergent with power $P=\frac{1}{6R}$
• C. Divergent with power $P=\frac{1}{3R}$
• D. Divergent with power $P=\frac{1}{6R}$

Answer 1

The correct option is A Convergent with power $P=\frac{1}{R}$

$\begin{array}{c}{R}_1=R\\ R_2=-R\\ \mu =1.5\\ \text{Using lens maker formula}\\ \frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})\\ \frac{1}{f}=(1.5-1)(\frac{1}{R}-\frac{1}{-R})\\ \frac{1}{f}=\left(0.5\right)(\frac{1}{R}+\frac{1}{R})\\ \frac{1}{f}=\frac{0.5\times 2}{R}\\ \frac{1}{f}=\frac{1}{R}\\ f=R\end{array}$

$P=\frac{1}{R}$

Since, focal length is $+Ve$, hence lens are converging.