The focal lengths of a lens are in the ratio 8:3 when it is immersed in two different liquids refractive indices 1.6 and 1.2 respectively. The refractive index of the material of the lens i

Options

A. 1.25
B. 1.5
C. 1.8
D. 2

Explanation

To solve the problem, we need to find the refractive index of the lens material (μ) given the focal lengths of the lens in two different liquids and their respective refractive indices.

### Step-by-Step Solution:

1. **Understanding the Problem:**
We are given two different liquids with refractive indices:
- Liquid 1: μ₁ = 1.6 (focal length f₁)
- Liquid 2: μ₂ = 1.2 (focal length f₂)
The ratio of the focal lengths is given as:
\[ \frac{f₁}{f₂} = \frac{8}{3} \]

2. **Using the Lens Maker's Formula:**
The formula for the focal length of a lens in a medium is given by:
\[ \frac{1}{f} = \frac{μ_{lens}}{μ_{medium} - 1} \cdot \frac{1}{R} \]
For our case, we can express the focal lengths as:
\[ \frac{1}{f₁} = \frac{μ}{1.6 - 1} \cdot \frac{1}{R} = \frac{μ}{0.6} \cdot \frac{1}{R} \]
\[ \frac{1}{f₂} = \frac{μ}{1.2 - 1} \cdot \frac{1}{R} = \frac{μ}{0.2} \cdot \frac{1}{R} \]

3. **Setting Up the Ratio:**
From the given ratio of focal lengths:
\[ \frac{f₁}{f₂} = \frac{8}{3} \]
This implies:
\[ \frac{1/f₁}{1/f₂} = \frac{8}{3} \]
Substituting the expressions for \(1/f₁\) and \(1/f₂\):
\[ \frac{\frac{μ}{0.6}}{\frac{μ}{0.2}} = \frac{8}{3} \]
This simplifies to:
\[ \frac{0.2}{0.6} = \frac{8}{3} \]
or:
\[ \frac{1}{3} = \frac{8}{3} \]

4. **Cross Multiplying:**
Cross multiplying gives us:
\[ 8 \cdot 0.6 = 3 \cdot 0.2 \]
Simplifying:
\[ 4.8 = 0.6μ - 0.2μ \]
This leads to:
\[ 4.8 = 0.4μ \]

5. **Solving for μ:**
Dividing both sides by 0.4:
\[ μ = \frac{4.8}{0.4} = 12 \]

6. **Conclusion:**
The refractive index of the lens material is:
\[ μ = 12 \]