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

Options

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

Answer

To solve the problem, we will use the lens maker's formula and the information given about the focal lengths in two different liquids.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a lens with focal lengths in the ratio of 8:3 when immersed in two different liquids with refractive indices \( n_1 = 1.6 \) and \( n_2 = 1.2 \). We need to find the refractive index \( \mu_l \) of the lens material.

2. **Using the Lens Maker's Formula**:
The lens maker's formula is given by:
\[ \frac{1}{f} = \left( \mu_l - n \right) \frac{1}{R} \]
where \( f \) is the focal length, \( \mu_l \) is the refractive index of the lens, \( n \) is the refractive index of the medium, and \( R \) is the radius of curvature.

3. **Setting Up the Equations**:
For the first liquid (refractive index \( n_1 = 1.6 \)):
\[ \frac{1}{f_1} = \left( \mu_l - 1.6 \right) \frac{2}{R} \]
For the second liquid (refractive index \( n_2 = 1.2 \)):
\[ \frac{1}{f_2} = \left( \mu_l - 1.2 \right) \frac{2}{R} \]

4. **Expressing Focal Lengths**:
Rearranging the equations gives us:
\[ f_1 = \frac{2R}{\mu_l - 1.6} \]
\[ f_2 = \frac{2R}{\mu_l - 1.2} \]

5. **Using the Ratio of Focal Lengths**:
We know that:
\[ \frac{f_1}{f_2} = \frac{8}{3} \]
Substituting the expressions for \( f_1 \) and \( f_2 \):
\[ \frac{\frac{2R}{\mu_l - 1.6}}{\frac{2R}{\mu_l - 1.2}} = \frac{8}{3} \]
The \( 2R \) cancels out:
\[ \frac{\mu_l - 1.2}{\mu_l - 1.6} = \frac{8}{3} \]

6. **Cross-Multiplying**:
Cross-multiplying gives:
\[ 3(\mu_l - 1.2) = 8(\mu_l - 1.6) \]
Expanding both sides:
\[ 3\mu_l - 3.6 = 8\mu_l - 12.8 \]

7. **Rearranging the Equation**:
Rearranging the equation to isolate \( \mu_l \):
\[ 12.8 - 3.6 = 8\mu_l - 3\mu_l \]
\[ 9.2 = 5\mu_l \]
Dividing both sides by 5:
\[ \mu_l = \frac{9.2}{5} = 1.84 \]

8. **Final Answer**:
The refractive index of the material of the lens is \( \mu_l = 1.84 \).