A convex lens of refractive index 1.55, with both the surfaces of the same radius of curvature has a focal length of 20 cm. The radius of curvature of the surface will be:

22 cm

The correct answer is Option (2) → 22 cm

Given,

$μ$, Refractive index = 1.55

R, (Both the surface have same radius) = $r_1=r_2$

F, focal length = 20 cm

$ \frac{1}{F} = (\mu -1)(\frac{1}{R} - \frac{1}{-R})$

$= \frac{2(\mu -1)}{R}$

$⇒R = 2F (\mu -1)$

$⇒R= 40\times 0.55 cm$

$⇒R= 22cm$