Question

The focal length of a bioconvaxe lens of refractiive index $1.5$ is $0.06m$. Radii of curvature are in the ratio $1:2$. Then radii of curvature of two lens surfaces are

• A. $0.045m,0.09m$
• B. $0.09m,0.18m$
• C. $0.04cm,0.08cm$
• D. $0.06m,0.12m$

Answer 1

The correct option is C $0.04cm,0.08cm$

$\begin{array}{c}\frac{1}{f}=\left(n-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\frac{R_1}{R_2}=\frac{1}{2}\\ \Rightarrow \frac{1}{f}=\left(1.5-1\right)\left(\frac{1}{R_1}-\frac{1}{2R_2}\right)R_2=2R_1\\ \Rightarrow 0.06=0.5\left(\frac{1}{2R_1}\right)\\ R_1=\frac{0.5}{2\times 0.06}\\ =0.04cm\\ \therefore R_2=0.08cm\end{array}$

$\therefore$ Option $C$ is correct.