A thin prism `P_(1)` with angle `4degree` and made from glass of refractive index 1.54 is combined with another thin prism `P_(2)` made from glass of refractive index 1.72 to produce dispersion without deviation. The angle of the prism `P_(2)` is

Options

A. `5.33degree`
B. `4degree`
C. `3degree`
D. `2.6degree`

Explanation

To solve the problem, we need to determine the angle of the second prism \( P_2 \) such that the combination of the two prisms produces dispersion without deviation.

### Step-by-step Solution:

1. **Understanding the Deviation Formula**:
The deviation \( \delta \) produced by a prism is given by the formula:
\[ \delta = (\mu - 1) \cdot A \]
where \( \mu \) is the refractive index of the prism material and \( A \) is the angle of the prism.

2. **Setting Up the Equation**:
For two prisms \( P_1 \) and \( P_2 \), the total deviation must be zero for the condition of dispersion without deviation:
\[ \delta_1 + \delta_2 = 0 \]
This implies:
\[ \delta_1 = -\delta_2 \]

3. **Calculating Deviation for Prism \( P_1 \)**:
For prism \( P_1 \) (with \( \mu_1 = 1.54 \) and \( A_1 = 4^\circ \)):
\[ \delta_1 = (1.54 - 1) \cdot 4^\circ = 0.54 \cdot 4^\circ = 2.16^\circ \]

4. **Calculating Deviation for Prism \( P_2 \)**:
Let the angle of prism \( P_2 \) be \( A_2 \) and its refractive index \( \mu_2 = 1.72 \). The deviation for prism \( P_2 \) is:
\[ \delta_2 = (1.72 - 1) \cdot A_2 = 0.72 \cdot A_2 \]

5. **Setting Up the Equation for Zero Deviation**:
From the condition \( \delta_1 + \delta_2 = 0 \):
\[ 2.16^\circ + 0.72 \cdot A_2 = 0 \]
Rearranging gives:
\[ 0.72 \cdot A_2 = -2.16^\circ \]

6. **Solving for \( A_2 \)**:
Dividing both sides by \( 0.72 \):
\[ A_2 = \frac{-2.16^\circ}{0.72} = -3^\circ \]

### Final Result:
The angle of prism \( P_2 \) is \( -3^\circ \).