Question

A double convex lens of + 5 D is made of glass of refractive index 1.55 with both faces of equal radii of curvature. Find the value of its radius of curvature.

Answer 1

From lens maker formula, we have

\[P = (\mu - 1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right)\]

\[\text { where} \]

\[ P =\text { Power of lens } = + 5 D\]

\[\mu = \text { Refractive index of the lens } = 1 . 55\]

\[ R_1 =\text {  Radius of curvature of first face } \left( + ve \right)\]

\[ R_2 = \text { Radius of curvature of second face } \left( - ve \right)\]

\[\text {Given }: \]

\[ R_1 = R_2 = R\]

\[ \Rightarrow P = (\mu - 1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right)\]

\[ \Rightarrow 5 = (1 . 55 - 1)\left( \frac{1}{R} - \frac{1}{- R} \right)\]

\[ \Rightarrow 5 = (1 . 55 - 1)\left( \frac{2}{R} \right)\]

\[ \Rightarrow 5 = 0 . 55\left( \frac{2}{R} \right)\]

\[ \Rightarrow R = \frac{0 . 55 \times 2}{5}\]

\[ \Rightarrow R = 0 . 22 m\]

The radius of curvature of the lens is 22 cm.