Question

A convex lens of refractive index $1.5$ has a focal length of $20\text{ cm}$ in air. Find its nature and focal length when it is immersed in a transparent liquid of refractive index $1.25$.

Hint:

General ratio formula: $\frac{f_l}{f_a} = \frac{\mu_g - 1}{\frac{\mu_g}{\mu_l} - 1}$.
Here: $\frac{f_l}{20} = \frac{1.5 - 1}{\frac{1.5}{1.25} - 1} = \frac{0.5}{0.2} = 2.5 \implies f_l = 20 \times 2.5 = 50\text{ cm}$.

Answer 1

Concept:
• Lens Maker's Formula in air: $\frac{1}{f_a} = (\mu_g - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.

• Lens Maker's Formula in a liquid of refractive index $\mu_l$: $\frac{1}{f_l} = \left(\frac{\mu_g}{\mu_l} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.

Step 1: Express curvature factor using air focal length
Given focal length in air $f_a = +20\text{ cm}$, refractive index of glass lens $\mu_g = 1.5$.
Using Lens Maker's formula in air:
\[ \frac{1}{f_a} = (\mu_g - 1) K \]
where $K = \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ is the geometric curvature factor.
\[ \frac{1}{20} = (1.5 - 1) K = 0.5 K \]
\[ K = \frac{1}{20 \times 0.5} = \frac{1}{10}\text{ cm}^{-1} \]

Step 2: Calculate focal length in liquid
Given refractive index of liquid $\mu_l = 1.25$.
Relative refractive index of lens with respect to liquid is $\mu_{rel} = \frac{\mu_g}{\mu_l} = \frac{1.5}{1.25} = 1.2$.
Using Lens Maker's formula in liquid:
\[ \frac{1}{f_l} = (\mu_{rel} - 1) K = (1.2 - 1) K \]
\[ \frac{1}{f_l} = 0.2 K \]
Substitute $K = \frac{1}{10}$:
\[ \frac{1}{f_l} = 0.2 \times \frac{1}{10} = \frac{0.2}{10} = \frac{1}{50}\text{ cm}^{-1} \]
\[ f_l = +50\text{ cm} \]

Step 3: Determine nature of lens in liquid
Since $f_l = +50\text{ cm}$ remains positive ($\mu_g > \mu_l$), the lens retains its original focal sign.
Hence, it continues to act as a converging (convex) lens.

Step 4: Conclusion
The focal length of the lens in liquid is $+50\text{ cm}$ and its nature remains converging.