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: Understanding the Concept
In air, \(\dfrac1f=(\mu-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)\). In a liquid, the relative index is \(\dfrac{\mu}{\mu_0}\).

Step 2: Key Formula or Approach
\[ \frac1{f_0}=\left(\frac{\mu}{\mu_0}-1\right)\left(\frac1{R_1}-\frac1{R_2}\right) \]

Step 3: Detailed Explanation
Divide the two equations:
\[ \frac{f_0}{f}=\frac{\mu-1}{\frac{\mu}{\mu_0}-1}=\frac{\mu_0(\mu-1)}{\mu-\mu_0} \]
\[ f_0=\frac{\mu_0(\mu-1)f}{\mu-\mu_0} \]

Final Answer:
The focal length in the liquid is \(\frac{\mu_0(\mu-1)f}{\mu-\mu_0}\), option (D). \[ \boxed{\dfrac{\mu_0(\mu-1)f}{\mu-\mu_0}\ \text{(D)}} \]