Question

Find the radius of curvature of the convex surface of a plano convex lens whose focal length is 0.3 and the refractive index of a lens material is 1.5

Answer 1

Solution :-

[ Refer to attachment ]

From figure, using sign convention

Radius of curvature of convex surface surface of the lens R₁ = - R

Radius of curvature of plane surface of the lens R₂ = ∞

Focal length of the lens = - 0.3 units

Refractive index of the lens n = 1.5

Using lens maker's formula

$\sf \dfrac{1}{f} = (n - 1) \bigg( \dfrac{1}{R_1} - \dfrac{1}{ R_2} \bigg)$

By substituting the values

$\implies \sf \dfrac{1}{ - 0.3} = (1.5- 1) \bigg( \dfrac{1}{ - R } - \dfrac{1}{ \infty } \bigg)$

$\implies \sf - \dfrac{1}{0.3} = (0.5) \bigg( - \dfrac{1}{ R } - 0 \bigg)$

[ Since 1/∞ = 0 ]

$\implies \sf - \dfrac{1}{0.3} = (0.5) \bigg( - \dfrac{1}{ R } \bigg)$

$\implies \sf - \dfrac{1}{0.3 \times 0.5} = - \dfrac{1}{ R }$

$\implies \sf - \dfrac{1}{0.15} = - \dfrac{1}{ R }$

$\implies \sf \dfrac{1}{R} = \dfrac{1}{0.15}$

$\implies \sf R = 0.15$

Therefore the radius of curvature of the convex surface is 0.15 units.