Question

A biconcave lens of power P vertically splits into two identical plano
concave parts. The power of each part will be
(a) 2P
(b) P/2
(c) P
(d) PW2​

Answer 1

answer : option (b) P/2

it is given that a biconcave lens of power P vertically splits into two identical plano concave parts.

Let radius of curvature of biconcave lens is R and refractive index of lens is μ.

so, 1/f = (μ - 1)[1/R + 1/R] = 2(μ - 1)/R

⇒f = R/2(μ - 1) ..........(1)

now when biconcave lens vertically splits into two identical plano concave parts.

let focal length of each part = f'

1/f' = (μ - 1)[1/R - 1/∞] = (μ - 1)/R

⇒f' = R/(μ - 1)...........(2)

from equations (1) and (2) we get,

f' = 2f

as we know power of lens is inversely proportional to focal length.

so, P/P' = (1/f)/(1/f')

⇒P/P' = (1/f)/(1/2f) = 2

⇒P' = P/2

therefore, power of each part is P/2.

Answer 2

Given:

The power of lens = P

To find:

Power on each part of concave parts.

Solution:

There is relation between the focal length and the thickness.

Focal length is inversely proportional to the thickness.

f ∝ 1/t

And the focal length is also inversely proportion to the power

P ∝ 1/f

When the lens splits into two parts that are equal

The focal length be half

There power will be half.

P∝1/2

The option B is correct the power of each part be P/2