Question

Fill in the Blank : A thin lens of refractive index 1.5 has a focal length of 15 cm in air. When the lens is placed in a medium of refractive index 4/3, its focal length will become . . . . . . cm.

Answer 1

Answer:

The answer is 22.8 cm.

Explanation:

Case 1 :-

Refractive index of lens in air, μ₁ = 1.5

focal length, f₁ = 15 cm

∴(μ₁-1)(1/R₁+1/R₂) = 15 ----------(1)

Case 2 :-

Refractive index, μ₂ = 4/3 = 1.33

focal length,f₂ = (μ₂-1)(1/R₁+1/R₂)

∴(μ₂-1)(1/R₁+1/R₂) = f₂ ----------(2)

Dividing equation (1) by (2), we get

(μ₁-1)/(μ₂-1) = 15/f₂

or, (1.5-1)/(1.33-1) = 15/f₂

or, 0.5/0.33 = 15/f₂

or, 1.52 = 15/f₂

∴f₂ = 22.8 cm