Question

What is the radius of curvature of an equiconvex lens made of glass of refractive index $\frac{5}{3},$ if its focal length in air is $40\text{cm}$ ?

• A. $10\text{cm}$
• B. $\frac{160}{3}\text{cm}$
• C. $20\text{cm}$
• D. $\frac{140}{3}\text{cm}$

Answer 1

The correct option is B $\frac{160}{3}\text{cm}$

By Lens maker’s formula
$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$
$\frac{1}{f}=(\frac{5}{3}-1)\left(\frac{2}{R}\right)$ (As lens is equiconvex)
$\Rightarrow \frac{1}{40}=\frac{4}{3R}$
$R=\frac{160}{3}\text{cm}$