Question

A biconvex lens has a focal length $\frac{2}{3}$ times the radius of curvature of either surface. The refractive index of the lens material is :

Answer 1

Step 1: Given data

It is a biconvex lens. There are $2$ radii of curvatures

The radii of curvatures are $R_1\&R_2$

$R_1=+R$ and

$R_2=-R$ here, the same value and negative sign indicate that the radius of curvature of both the surface of the lens is same but the direction of measurement is opposite.

The refractive index be, $\mu$

Step 2: Formula used

Focal length, $f$=$\frac{2}{3}R$

Applying the lens maker formula ,

$$\begin{gathered} \frac{1}{f}=(\mu -1)[\frac{1}{R_1}-\frac{1}{R_2}] \\ \Rightarrow \frac{1}{\left({\displaystyle \frac{2}{3}}R\right)}=(\mu -1)[\frac{1}{R}+\frac{1}{R}] \\ \Rightarrow \frac{3}{{\displaystyle 2R}}=\frac{(\mu -1)2}{R} \\ \Rightarrow \mu -1=\frac{3}{4} \\ \Rightarrow \mu =\frac{3}{4}+1 \\ \Rightarrow \mu =1.75 \end{gathered}$$

Thus, the refractive index of the material is equal to $1.75$.