Question

The radius of curvature of each face of a biconcave lens made of glass of refractive index is 1.5 is 30 cm. calculate the focal length of the lens in air

Answer 1

Answer: The focal length of the lens in air is 30 cm.

Explanation:

Given that,

Refractive index n = 1.5

Radius of curvature R = 30 cm

We know that,

The relation between focal length, radius of curvature and refractive index is

$\dfrac{1}{f}=(n-1)(\dfrac{1}{R_{1}}-\dfrac{1}{R_{2}})$

$\dfrac{1}{f}= (1.5-1)(\dfrac{1}{30}-\dfrac{1}{-30})$

$\dfrac{1}{f}=\dfrac{1}{30}$

$f = 30 cm$

Hence, The focal length of the lens in air is 30 cm.

Answer 2

Answer:

answer is -30cm

Explanation:

1/f=[n-1][1/r1 -1 /r2]