The refracting angle of a glass prism is `30^@.` A ray is incident onto one of the faces perpendicular to it. Find the angle `delta` between the incident ray and the ray that leaves the prism. The refractive index of glass is `mu=1.5.`

Explanation

To solve the problem step by step, we will follow the given information and apply the principles of refraction.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Refracting angle of the prism, \( A = 30^\circ \)
- Refractive index of glass, \( \mu = 1.5 \)
- The ray is incident perpendicular to one of the faces of the prism.

2. **Determine the Incident Angle:**
- Since the ray is incident perpendicular to the face, the angle of incidence \( I_1 = 0^\circ \).

3. **Determine the Angle of Refraction at the First Face:**
- According to Snell's law, when light enters the prism, the angle of refraction \( R_1 \) is also \( 0^\circ \) because the incident angle \( I_1 = 0^\circ \).

4. **Calculate the Angle of Refraction at the Second Face:**
- The relationship between the angles in the prism is given by:
\[ R_1 + R_2 = A \]
- Since \( R_1 = 0^\circ \), we can substitute:
\[ 0 + R_2 = 30^\circ \implies R_2 = 30^\circ \]

5. **Apply Snell's Law at the Second Face:**
- Using Snell's law:
\[ \mu_1 \sin I_2 = \mu_2 \sin R_2 \]
- Here, \( \mu_1 = 1 \) (for air), \( \mu_2 = 1.5 \), and \( R_2 = 30^\circ \):
\[ 1 \cdot \sin I_2 = 1.5 \cdot \sin 30^\circ \]
- Since \( \sin 30^\circ = 0.5 \):
\[ \sin I_2 = 1.5 \cdot 0.5 = 0.75 \]

6. **Calculate the Angle of Refraction \( I_2 \):**
- Now, find \( I_2 \):
\[ I_2 = \sin^{-1}(0.75) \approx 48.6^\circ \]

7. **Calculate the Total Deviation \( \delta \):**
- The total deviation \( \delta \) is given by:
\[ \delta = I_2 + I_1 - A \]
- Substitute the values:
\[ \delta = 48.6^\circ + 0^\circ - 30^\circ = 18.6^\circ \]

### Final Answer:
The angle \( \delta \) between the incident ray and the ray that leaves the prism is \( 18.6^\circ \).

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