Question

A biconvex lens of refractive index 1.5 has a focal length of 20 cm in air. Its focal length when immersed in a liquid of refractive index 1.6 will be:

• A. -160 cm
• B. 160 cm
• C. 16 cm
• D. -16 cm

Answer 1

The Correct Option is A

To determine the focal length of a biconvex lens immersed in a liquid, we need to apply the lens maker's formula, which takes into account the change in refractive index of the surrounding medium. 

The lens maker's formula is given by:

\[\frac{1}{f} = \left( \frac{n_{\text{lens}}}{n_{\text{medium}}} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\]

Let's break down the steps:

1. Initially, the lens is in air. The refractive index of the lens, \(n_{\text{lens}}\), is 1.5 and that of air, \(n_{\text{medium}}\), is 1.0. Substituting these values into the lens formula when the lens is in air gives us the initial focal length, \(f_{\text{air}}\), of 20 cm. This is expressed as:
2. Re-arranging this to solve for \(\frac{1}{R_1} - \frac{1}{R_2}\):
3. Now, when the lens is immersed in a liquid with a refractive index, \(n_{\text{medium}} = 1.6\), we use the same lens maker's formula:
4. Substitute \(\frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{10 \text{ cm}}\):
5. Simplify:
6. Hence, invert to find the focal length:

Therefore, the focal length of the lens when immersed in a liquid of refractive index 1.6 is -160 cm.

Answer 2

Step 1: Given Data: - Refractive index of the lens μ_l = 1.5 - Refractive index of the medium (liquid) μ_m = 1.6 - Focal length in air f_a = 20 cm

Step 2: Use the Lens Formula in Different Mediums: - The relationship between the focal length in air f_a and the focal length in the medium f_m is given by:

\( \frac{f_m}{f_a} = \frac{\mu_l - 1}{\mu_l - \mu_m} \)

Step 3: Substitute the Values:

\( \frac{f_m}{20} = \frac{(1.5 - 1)}{(1.5 - 1.6)} \)

\( \frac{f_m}{20} = \frac{0.5}{-0.1} \)

\( f_m = 20 \times -5 = -160 \, \text{cm} \)

So, the correct answer is: -160 cm