A ray of light incident at an angle `theta` on a refracting face of a prism emerges from the other face normally. If the angle of the prism is `5^@` and the prism is made of a material of refractive index `1.5`, the angle of incidence is.

Options

A. `7.5^@`
B. `5^@`
C. `15^@`
D. `2.5^@`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Geometry of the Prism
We have a prism with an angle \( A = 5^\circ \). A ray of light is incident at an angle \( \theta \) on the refracting face of the prism and emerges normally from the other face.

### Step 2: Identify Angles of Refraction
When the light ray enters the prism, it will refract at the first face. We denote the angle of refraction at the first face as \( R_1 \). Since the ray emerges normally from the second face, the angle of refraction at the second face, \( R_2 \), is \( 0^\circ \).

### Step 3: Apply the Prism Formula
For a prism, the relationship between the angles is given by:
\[ A = R_1 + R_2 \]
Since \( R_2 = 0^\circ \), we have:
\[ R_1 = A = 5^\circ \]

### Step 4: Use Snell's Law
Now we apply Snell's Law at the first interface (air to prism):
\[ \mu_1 \sin(\theta) = \mu_2 \sin(R_1) \]
Where:
- \( \mu_1 = 1 \) (refractive index of air)
- \( \mu_2 = 1.5 \) (refractive index of the prism)
- \( R_1 = 5^\circ \)

Substituting the values, we have:
\[ 1 \cdot \sin(\theta) = 1.5 \cdot \sin(5^\circ) \]

### Step 5: Calculate \( \sin(5^\circ) \)
Using a calculator or trigonometric tables:
\[ \sin(5^\circ) \approx 0.0872 \]

### Step 6: Substitute and Solve for \( \sin(\theta) \)
Substituting \( \sin(5^\circ) \) into the equation:
\[ \sin(\theta) = 1.5 \cdot 0.0872 \]
Calculating the right side:
\[ \sin(\theta) = 0.1308 \]

### Step 7: Find \( \theta \)
Now, to find \( \theta \):
\[ \theta = \sin^{-1}(0.1308) \]
Using a calculator:
\[ \theta \approx 7.5^\circ \]

### Conclusion
The angle of incidence \( \theta \) is approximately \( 7.5^\circ \).

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