Question

The ratio to radii of curvature of a biconvex lens $R_1:R_2$ = 2 : 3. Its refractive index and focal length is $1.5$ and $12cm$ respectively. What is the value of $R_1$? Consider $R_1$ and $R_2$ are positive and negative respectively.

• A. $5cm$
• B. $10cm$
• C. $15cm$
• D. $20cm$

Answer 1

The correct option is B $10cm$

We know by lens maker's formula,
$\frac{1}{f}=(\mu -1)[\frac{1}{R_1}-\frac{1}{R_2}]$

where $f$ if the focal length, $\mu$ is the refractive index and $R_1,R_2$ are the radii of curvature.

Given,
$\frac{R_1}{R_2}=\frac{2}{3}$

Take $R_1$ = $x$ and $R_2$ = $-3x/2$

$\frac{1}{12}=(1.5-1)[\frac{1}{x}+\frac{2}{3x}]\frac{1}{12}=0.5\left[\frac{5}{3x}\right]x=\frac{12\times 2.5}{3}=10cm$