Question

An optometrist prescribes a corrective lens with a power of +1.5 diopters. The lens maker will start with glass blank with an index of refraction of 1.6 and a convex front surface whose radius of curvature is 20 cm. To what radius of curvature should the other surface be ground?

Answer 1

Given:

The power of the lens is,

$P=+1.5\text{ D}$

The refractive index of the lens is,

$n=1.6$

The radius of curvature of the convex surface is,

$\begin{gathered} R_1=20\text{ cm} \\ =0.20\text{ m} \end{gathered}$

To find:

The radius of curvature of the other surface

Explanation:

If the radius of curvature of the other surface is

$R_2$

we can write,

$\frac{1}{f}=P=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})$

Now, substituting the values we get,

$\begin{gathered} 1.5=(1.6-1)(\frac{1}{0.20}-\frac{1}{R_2}) \\ \frac{1.5}{0.6}=\frac{1}{0.20}-\frac{1}{R_2} \\ \frac{1}{R_2}=\frac{1}{0.20}-\frac{1.5}{0.6} \\ \frac{1}{R_2}=2.5 \\ R_2=0.40\text{ m} \\ R_2=40\text{ cm} \end{gathered}$

Hence, the radius of curvature of the other surface is 40 cm.