Question

A convex lens of refractive index 3/2 has a power of 2.5 D in air. If it is places in a liquid of refractive index 2 then the new power of lens is?

Answer 1

Answer:

Focal length of a convex lens having power 2.5 D, = 1/2.5 m Also focal length of a lens in a medium of refractive index μ is given by

Explanation:https://www.sarthaks.com/?qa=blob&qa_blobid=7442159782517558825

Answer 2

Answer:

Explanation:

Solution,

Here, we have

Convex lens of refractive index 3/2 has a power of 2.5 D in air.

Places in a liquid of refractive index 2.

Here, we know that,

According to the Lens Maker's formula,

Power of lens, P = (μ(r) - 1) (1/R₁ - 1/R₂)

where μ(r) is the refractive index of the material of lens

And its relative to the surrounding medium = μ (material)/μ (medium)

Here, we have

For convex lens in air, μ(r) = 3/2/1 = 3/2

For convex lens in liquid, μ(r) = 3/2/2 = 3/4

Here, we get

⇒ P₂/P₁ = μ(r)₂ - 1/μ(r)₁ - 1

⇒ (3/2 - 1)/(3/2 - 1) = - 1/2

⇒ P₂ = - 1/2 × 2.5 D

⇒ P₂ = - 1.25 D.

Hence, the new power of the lens is - 1.25 D.