Question

The radii of curvature of the surfaces of a double convex lens are $20\mathrm{cm}$ and $40\mathrm{cm}$ respectively, and its focal length is $20\text{cm}$. What is the refractive index of the material of the lens?

Answer 1

Step 1: Given parametersRadii of curvature of the surfaces of a double convex lens,

$$\begin{gathered} {\mathrm{R}}_1=20\mathrm{cm} \\ {\mathrm{R}}_2=-40\mathrm{cm}\left({\mathrm{R}}_2\mathrm{is}\mathrm{negative}\mathrm{as}\mathrm{per}\mathrm{the}\mathrm{sign}\mathrm{conventions}\mathrm{as}\mathrm{it}\mathrm{is}\mathrm{in}\mathrm{the}\mathrm{opposite}\mathrm{direction}\mathrm{of}\mathrm{the}\mathrm{incident}\mathrm{light}\right) \\ \mathrm{f}=20\mathrm{cm} \end{gathered}$$

Step 2: Formula used

Lens maker's formula

$\frac{1}{\mathrm{f}}=\left(\mathrm{\mu }-1\right)\left(\frac{1}{{\mathrm{R}}_1}-\frac{1}{{\mathrm{R}}_2}\right)$

Step 3: Calculating refractive index

Substituting the given values in Lens maker's formula, we get:

$$\begin{gathered} \frac{1}{20}=\left(\mathrm{\mu }-1\right)\left(\frac{1}{20}-\frac{1}{-40}\right) \\ \frac{1}{20}=\left(\mathrm{\mu }-1\right)\left(\frac{1}{20}+\frac{1}{40}\right) \\ \frac{1}{20}=\left(\mathrm{\mu }-1\right)\frac{3}{40} \\ \frac{2}{3}=\mathrm{\mu }-1 \\ \mathrm{\mu }=\frac{5}{3} \end{gathered}$$

Therefore, the refractive index of the material of the lens is $\frac{5}{3}$.