A prism `(mu=1.5)` has the refracting angle of `30^(@)` The deviation of a monochromatic ray incident normally on its one surface will be `(sin 48^(@) 36 = 0.75)`

Options

A. `18^(@)` 36
B. `20^(@)` 30
C. `18^(@)`
D. `22^(@)1'`

Answer

To solve the problem, we need to calculate the deviation of a monochromatic ray incident normally on one surface of a prism with a refractive index (μ) of 1.5 and a refracting angle of 30 degrees.

### Step-by-Step Solution:

1. **Understand the Geometry of the Prism**:
- The prism has a refracting angle \( A = 30^\circ \).
- The ray of light is incident normally on one of the surfaces of the prism.

2. **Identify the Angles**:
- When the ray hits the surface normally, it enters the prism without any deviation at the first surface. Therefore, the angle of incidence \( i = 0^\circ \).
- The angle of refraction \( r_1 = 0^\circ \) (since it enters normally).

3. **Calculate the Angle of Deviation**:
- The angle of deviation \( \delta \) can be calculated using the formula:
\[ \delta = i + e - A \]
where \( e \) is the angle of emergence.

4. **Using Snell's Law**:
- At the second surface of the prism, we can apply Snell's law:
\[ \mu = \frac{\sin i}{\sin r} \]
- Here, \( i \) is the angle of incidence at the second surface, which is equal to the angle of refraction at the first surface, \( r_1 = 0^\circ \). Therefore, \( i = A - r_2 \) where \( r_2 \) is the angle of refraction at the second surface.

5. **Calculate the Angle of Refraction**:
- Using the refractive index:
\[ \mu = \frac{\sin(30^\circ)}{\sin(e)} \]
- Substituting the values:
\[ 1.5 = \frac{0.5}{\sin(e)} \]
- Rearranging gives:
\[ \sin(e) = \frac{0.5}{1.5} = \frac{1}{3} \]

6. **Finding the Angle \( e \)**:
- Using the inverse sine function:
\[ e = \sin^{-1}(0.75) \]
- This gives:
\[ e = 48^\circ 36' \]

7. **Calculate the Deviation \( \delta \)**:
- Now substituting back into the deviation formula:
\[ \delta = 0^\circ + 48^\circ 36' - 30^\circ \]
- Simplifying gives:
\[ \delta = 18^\circ 36' \]

### Final Answer:
The deviation of the monochromatic ray is \( \delta = 18^\circ 36' \).

---