Question

The power of a thin lens is +5 D. When it is immersed in a liquid, it behaves like a concave lens with a focal length of 100 cm. Calculate the refractive index of the liquid. Given the refractive index of glass = 1.5.

Answer 1

Given, the convergent lens,

The refractive index of air, µ_a = 1

The refractive index of glass, µ_g = 1.5

Power of lens, P = +5 D

So, its focal length,

`f = 1/P`

`f = 100/5`

f = 20 cm

Using lens maker's formula,

`1/f = ((mu_g - mu_a))/mu_a xx (1/R_1 - 1/R_2)`

Substituting the values given in the problem,

⇒ `1/20 = (1.5 - 1) xx (1/R_1 - 1/R_2)` ...........(i)

Now, when the convergent lens is immersed in a liquid of refractive index µ₁, the convergent lens acts as a diverging lens of focal length -100 cm.

By lens maker's formula,

`1/(-100) = ((1.5 - mu_1))/mu_1(1/R_1 - 1/R_2)` ...........(ii)

Dividing (i) by (ii), we get,

`-5 = (1/2)/((1.5/mu_1 - 1))`

⇒ `1 - 1.5/mu_1 = 1/10`

⇒ `1.5/mu_1 = (1 - 1)/10`

⇒ `3/(2mu_1) = 9/10`

`mu_1 = 5/3`