A medium has `n_v = 1.56, n_r=1.44`. Then its dispersive power is:

Options

A. `3//50`
B. `6//25`
C. `0.03`
D. none of these

Explanation

To find the dispersive power of a medium with given refractive indices for violet and red light, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Refractive Indices**:
- Given:
- Refractive index for violet light, \( n_v = 1.56 \)
- Refractive index for red light, \( n_r = 1.44 \)

2. **Calculate the Mean Refractive Index**:
- The mean refractive index \( n_m \) can be calculated as:
\[ n_m = \frac{n_v + n_r}{2} = \frac{1.56 + 1.44}{2} = \frac{3.00}{2} = 1.50 \]

3. **Calculate the Mean Deviation**:
- The mean deviation \( \Delta m \) is given by:
\[ \Delta m = n_m - 1 = 1.50 - 1 = 0.50 \]

4. **Calculate the Angular Dispersion**:
- The angular dispersion \( \Delta \theta \) can be calculated using the formula:
\[ \Delta \theta = (n_v - n_r) \cdot A \]
- Where \( A \) is the angle of the prism. Thus:
\[ \Delta \theta = (1.56 - 1.44) \cdot A = 0.12 \cdot A \]

5. **Calculate the Dispersive Power**:
- The dispersive power \( \omega \) is given by the ratio of angular dispersion to mean deviation:
\[ \omega = \frac{\Delta \theta}{\Delta m} \]
- Substituting the values we calculated:
\[ \omega = \frac{0.12 \cdot A}{0.50} = \frac{0.12}{0.50} \cdot A = \frac{12}{50} \cdot A = \frac{6}{25} \cdot A \]
- Since \( A \) cancels out, the dispersive power is:
\[ \omega = \frac{6}{25} \]

### Final Answer:
The dispersive power of the medium is \( \frac{6}{25} \).

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