Two prisms A and B are in contact with each other have angular dispersions of `2^(@)` and `4^(@)` respectively. The dispersive power of A is 0.002. If the combination produces dispersion without deviation, the dispersive power of B is

Options

A. 0.001
B. 0.004
C. 0.002
D. 0.006

Answer

To find the dispersive power of prism B, we can follow these steps:

### Step 1: Understand the given information
We have two prisms, A and B:
- Angular dispersion of prism A, \( \theta_A = 2^\circ \)
- Angular dispersion of prism B, \( \theta_B = 4^\circ \)
- Dispersive power of prism A, \( \omega_A = 0.002 \)

We need to find the dispersive power of prism B, \( \omega_B \), under the condition that the combination produces dispersion without deviation.

### Step 2: Use the condition for dispersion without deviation
The condition for dispersion without deviation for two prisms in contact is given by:
\[ \frac{\theta_A}{\omega_A} + \frac{\theta_B}{\omega_B} = 0 \]
This implies:
\[ \frac{\theta_A}{\omega_A} = -\frac{\theta_B}{\omega_B} \]

### Step 3: Substitute the known values
Substituting the known values into the equation:
\[ \frac{2}{0.002} = \frac{4}{\omega_B} \]

### Step 4: Solve for \( \omega_B \)
Cross-multiplying gives:
\[ 2 \cdot \omega_B = 4 \cdot 0.002 \]
\[ 2 \cdot \omega_B = 0.008 \]
Now, divide both sides by 2:
\[ \omega_B = \frac{0.008}{2} = 0.004 \]

### Step 5: Conclusion
The dispersive power of prism B is:
\[ \omega_B = 0.004 \]

### Final Answer
The dispersive power of prism B is \( 0.004 \).

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