Two small angled prisms A and B deviate the blue rays by 7 0 and 9 0 and the red rays by 5 0 and 7 0 respectively. Which prism has a greater . p dispersive power ?

Options

A. Prism A
B. Prism B
C. same for both Prism A & B
D. none of these

Explanation

To determine which prism has a greater dispersive power, we can follow these steps:

### Step 1: Understand the Concept of Dispersive Power
Dispersive power (ω) of a prism is defined as the ability of the prism to separate different colors of light. It can be calculated using the formula:
\[ \omega = \frac{\Delta V - \Delta R}{\Delta M} \]
where:
- \( \Delta V \) = deviation for violet (or blue) light,
- \( \Delta R \) = deviation for red light,
- \( \Delta M \) = mean deviation (usually taken as the average deviation of the middle color, which can be approximated as the average of the deviations of the extreme colors).

### Step 2: Gather Given Data
From the problem, we have the following deviations:
- For Prism A:
- Deviation for blue rays (\( \Delta V_A \)) = 7°
- Deviation for red rays (\( \Delta R_A \)) = 5°

- For Prism B:
- Deviation for blue rays (\( \Delta V_B \)) = 9°
- Deviation for red rays (\( \Delta R_B \)) = 7°

### Step 3: Calculate the Mean Deviation for Each Prism
Mean deviation (\( \Delta M \)) can be calculated as the average of the deviations for blue and red light.

For Prism A:
\[ \Delta M_A = \frac{\Delta V_A + \Delta R_A}{2} = \frac{7 + 5}{2} = \frac{12}{2} = 6° \]

For Prism B:
\[ \Delta M_B = \frac{\Delta V_B + \Delta R_B}{2} = \frac{9 + 7}{2} = \frac{16}{2} = 8° \]

### Step 4: Calculate the Dispersive Power for Each Prism
Now we can calculate the dispersive power for both prisms.

For Prism A:
\[ \omega_A = \frac{\Delta V_A - \Delta R_A}{\Delta M_A} = \frac{7 - 5}{6} = \frac{2}{6} = \frac{1}{3} \]

For Prism B:
\[ \omega_B = \frac{\Delta V_B - \Delta R_B}{\Delta M_B} = \frac{9 - 7}{8} = \frac{2}{8} = \frac{1}{4} \]

### Step 5: Compare the Dispersive Powers
Now we compare the dispersive powers:
- \( \omega_A = \frac{1}{3} \)
- \( \omega_B = \frac{1}{4} \)

Since \( \frac{1}{3} > \frac{1}{4} \), we conclude that:
\[ \omega_A > \omega_B \]

### Conclusion
Thus, Prism A has a greater dispersive power than Prism B.

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