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:
Answer & explanation
Correct answer: option 2
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})$
$= \frac{2(\mu -1)}{R}$
$⇒R = 2F (\mu -1)$
$⇒R= 40\times 0.55 cm$
$⇒R= 22cm$