Question

Focal length of a convex lens in air is f = 18 cm. It is immersed in a liquid of refractive index 4/3. If change in focal length of lens is \(\Delta f\) = nf, Find n. [Given refractive index of lens is 1.5] :

Hint:

For a glass lens (\(\mu=1.5\)) in water (\(\mu=1.33\)), the focal length roughly quadruples (\(4f\)). Thus, the change is \(3f\).

Answer 1

Correct Answer: 3

Step 1: Understanding the Concept:
The focal length of a lens depends on the relative refractive index between the lens material and the surrounding medium.
When immersed in a liquid, the relative refractive index decreases, so the focal length increases.
Step 2: Key Formula or Approach:
Lens Maker's Formula:
\[ \frac{1}{f} = \left( \frac{\mu_l}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Step 3: Detailed Explanation:
Let \(f_a = 18 \text{ cm}\) be the focal length in air (\(\mu_a = 1\)).
\[ \frac{1}{18} = (1.5 - 1)K \implies 0.5K = \frac{1}{18} \implies K = \frac{1}{9} \]
Where \(K = (\frac{1}{R_1} - \frac{1}{R_2})\).
In liquid (\(\mu_l = 4/3\)):
\[ \frac{1}{f_l} = \left( \frac{1.5}{4/3} - 1 \right) K = \left( \frac{4.5}{4} - 1 \right) \frac{1}{9} = \left( \frac{1.125 - 1}{9} \right) = \frac{0.125}{9} = \frac{1/8}{9} = \frac{1}{72} \]
So, \(f_l = 72 \text{ cm}\).
The change in focal length \(\Delta f = f_l - f_a = 72 - 18 = 54 \text{ cm}\).
Given \(\Delta f = nf = n(18)\).
\[ 54 = 18n \implies n = 3 \]
Step 4: Final Answer:
The value of \(n\) is 3.