Question

The radius of curvature of spherical surfaces of a convex lens are 15 cm. and 10 cm.respectively. If Refractive Index of lens is 1.5. then find its focal length.​ Ahy

Answer 1

Focal length of the lens is 100 cm.

Given:

Refractive Index = 1.5

R1 = 15 cm

R2 = 10 cm

To Find:

The focal length of convex lens.

Solution:

We know that :

$\bigstar \: \boxed{\bf \dfrac{1}{f} = \bigg(\dfrac{n_{2}}{n_{1}} - 1 \bigg) \bigg(\dfrac{1}{r_{1}} - \dfrac{1}{r_{2}} \bigg)}$

So,

$\sf \implies \dfrac{1}{f} = (1.5 - 1) \bigg( \dfrac{1}{15} - \dfrac{1}{10} \bigg)$

$\sf \implies \dfrac{1}{f} = (0.5) \bigg( \dfrac{15 - 10}{150} \bigg)$

$\sf \implies \dfrac{1}{f} = (0.5) \bigg( \dfrac{5}{150} \bigg)$

$\sf \implies \dfrac{1}{f} = \dfrac{2.5}{150}$

→ f = 100cm

Therefore,

The focal length is 100 cm.