Question

A convex lens of refractive index \(1.5\) and focal length \(f=18\) cm is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is \( \alpha \times f \). Find the value of \( \alpha \). (Given: refractive index of water \(=\dfrac{4}{3}\))

Hint:

When a lens is immersed in a medium, always use the {relative refractive index} in the lens maker formula.

Answer 1

Correct Answer: 3

Step 1: Understanding the Concept:
The focal length of a lens depends on the refractive index of its material relative to the surrounding medium. This relationship is described by the Lens Maker's Formula. When a lens is immersed in a liquid like water, its focal length increases because the relative refractive index decreases.
Step 2: Key Formula or Approach:
Lens Maker's Formula: \[ \frac{1}{f} = \left( \frac{\mu_l}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] where \(\mu_l\) is the refractive index of the lens and \(\mu_m\) is the refractive index of the medium.
Step 3: Detailed Explanation:
Let \(f_a\) be the focal length in air (\(\mu_m = 1\)) and \(f_w\) be the focal length in water (\(\mu_m = 4/3\)).
In air: \[ \frac{1}{f_a} = (1.5 - 1) K = 0.5 K \quad \dots(1) \] where \(K = \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\).
In water: \[ \frac{1}{f_w} = \left( \frac{1.5}{4/3} - 1 \right) K = \left( \frac{4.5}{4} - 1 \right) K = \left( \frac{9}{8} - 1 \right) K = \frac{1}{8} K \quad \dots(2) \] Dividing equation (1) by equation (2): \[ \frac{f_w}{f_a} = \frac{0.5 K}{\frac{1}{8} K} = 0.5 \times 8 = 4 \] So, \(f_w = 4 f_a\).
The difference in focal lengths is: \[ \Delta f = f_w - f_a = 4f_a - f_a = 3f_a \] Given \(\Delta f = \alpha \times f\), where \(f = f_a\): \[ 3f = \alpha f \implies \alpha = 3 \] Step 4: Final Answer:
The value of \(\alpha\) is 3.