Question

A dobleconves lens of focal length $6cm$ is made of glass of refractive index $1.5$ the radius of curvature of of one surface is double that of other surface. The value of small radius of curvature is

• A. $6cm$
• B. $4.5cm$
• C. $9cm$
• D. $4cm$

Answer 1

Answer: (B)

$4.5cm$

$\begin{array}{c}\text{Given}\left|R_1\right|=2\left|R_2\right|=R\\ \left|R_1\right|=R,R_1=R\\ \left|R_2\right|=\frac{R}{2},=R_2=-\frac{R}{2}\\ n=1.5\\ f=6cm\end{array}$

$\text{we know,}$

$\begin{array}{c}\frac{1}{f}=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})\\ \frac{1}{6}=(1.5-1)(\frac{1}{R}+\frac{2}{R})\\ \frac{1}{6}=0.5\times \frac{3}{R}\\ R=9cm\\ \frac{R}{2}=4.5cm\text{(smaller curvature)}\end{array}$