Question

A lens of refractive index '\(μ\)' has focal length '\(f\)'. When the lens is immersed in a liquid of refractive index '\(μ_0\)', its focal length becomes '\(f_0\)'. Then '\(f_0\)' is given by

• A. \(\frac{(μ_0-μ)f}{μ(μ_0-1)}\)
• B. \(\frac{μ(μ_0-1)f}{(μ_0-μ)}\)
• C. \(\frac{(μ-μ_0)f}{μ_0(μ-1)}\)
• D. \(\frac{μ_0(μ-1)f}{(μ-μ_0)}\)

Hint:

Use the lens maker's formula in air and in liquid and divide.

Answer 1

The Correct Option is D

Step 1: Write the ratio of powers
$\dfrac{P_{liquid}}{P_{air}}=\dfrac{\mu/\mu_0-1}{\mu-1}$, and focal length is the inverse of power.

Step 2: Simplify
$f_0=f\cdot\dfrac{\mu-1}{(\mu-\mu_0)/\mu_0}=\dfrac{\mu_0(\mu-1)f}{\mu-\mu_0}$, option (D).

Final Answer:
Option (D) gives $f_0$. \[ \boxed{\dfrac{\mu_0(\mu-1)f}{\mu-\mu_0}} \]