Double-convex lenses are to be manufactured from a glass of refractive index 1.55,

Question:

Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?

Solution:

Refractive index of glass, $\mu=1.55$

Focal length of the double-convex lens, f = 20 cm

Radius of curvature of one face of the lens = R1

Radius of curvature of the other face of the lens = R2

Radius of curvature of the double-convex lens = R

$\therefore R_{1}=R$ and $R_{2}=-R$

The value of R can be calculated as:

$\frac{1}{f}=(\mu-1)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]$

$\frac{1}{20}=(1.55-1)\left[\frac{1}{R}+\frac{1}{R}\right]$

$\frac{1}{20}=0.55 \times \frac{2}{R}$

$\therefore R=0.55 \times 2 \times 20=22 \mathrm{~cm}$

Hence, the radius of curvature of the double-convex lens is 22 cm.
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