Focal length of a lens is 0.12 m and refractive index is 1.5. Focal length of the same lens for blue colour is 0.1m. Theh refractive index of the lens for blue colour is

Options

A. 1.51
B. 1.25
C. 1.49
D. 1.6

Explanation

To find the refractive index of the lens for blue color, we will use the Lensmaker's formula, which relates the focal length of a lens to its refractive index and the radii of curvature of its surfaces.

### Step-by-Step Solution:

1. **Write down the Lensmaker's formula**:
\[ \frac{1}{f} = \mu - 1 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
where \( f \) is the focal length, \( \mu \) is the refractive index, and \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

2. **Substitute the known values for the first case**:
Given:
- Focal length \( f = 0.12 \, \text{m} \)
- Refractive index \( \mu = 1.5 \)

Substitute these values into the Lensmaker's formula:
\[ \frac{1}{0.12} = 1.5 - 1 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Simplifying gives:
\[ \frac{1}{0.12} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Therefore:
\[ \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{0.12} \times 2 = \frac{2}{0.12} = \frac{5}{0.03} = 16.67 \]

3. **Set up the equation for blue color**:
Now, for blue color, we have:
- Focal length \( f = 0.1 \, \text{m} \)
- Let the refractive index for blue color be \( \mu_b \).

Using the Lensmaker's formula again:
\[ \frac{1}{0.1} = \mu_b - 1 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substitute \( \frac{1}{R_1} - \frac{1}{R_2} = 16.67 \):
\[ \frac{1}{0.1} = (\mu_b - 1) \cdot 16.67 \]

4. **Solve for \( \mu_b \)**:
Rearranging gives:
\[ \mu_b - 1 = \frac{10}{16.67} \]
Calculate:
\[ \mu_b - 1 = 0.6 \]
Therefore:
\[ \mu_b = 1.6 \]

5. **Conclusion**:
The refractive index of the lens for blue color is \( \mu_b = 1.6 \).