Question

A thin convex lens of refractive index 1.5 has $20~\mathrm{cm}$ focal length in air. If the lens is completely immersed in a liquid of refractive index 1.6, its focal length will be}

• A. $-160\mathrm{cm}$
• B. $-100\mathrm{cm}$
• C. $+10\mathrm{cm}$
• D. $+100\mathrm{cm}$

Hint:

When a lens is placed in a medium with refractive index greater than that of the lens, the lens changes its nature (convex becomes concave).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Lens maker's formula: $\frac{1}{f} = (\mu_{rel} - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.
Step 2: Detailed Explanation:
In air, $\mu_{rel} = 1.5$, $\frac{1}{20} = (0.5)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.
In liquid, $\mu_{rel} = \frac{1.5}{1.6} = \frac{15}{16}$. So, $\frac{1}{f'} = \left(\frac{15}{16} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = -\frac{1}{16}\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.
Dividing the two equations, $\frac{f'}{20} = \frac{0.5}{-1/16} = -8$, so $f' = -160$ cm.
Step 3: Final Answer:
The focal length becomes $-160$ cm.