A ray of light is incident normally on one of the faces of a prism of apex angle 30 degree and refractive index sqrt2. The angle of deviation of the ray is…degrees.

Explanation

To solve the problem of finding the angle of deviation of a ray of light incident normally on one of the faces of a prism with an apex angle of 30 degrees and a refractive index of √2, we can follow these steps:

### Step 1: Understand the Geometry of the Prism
The prism has an apex angle (A) of 30 degrees. When light is incident normally on one of its faces, it will not bend at that face but will refract at the second face.

### Step 2: Calculate the Angle of Refraction at the First Face
Since the ray is incident normally, the angle of incidence (i) is 0 degrees. Using Snell's law:
\[ n_1 \sin(i) = n_2 \sin(r) \]
where:
- \( n_1 = 1 \) (refractive index of air),
- \( n_2 = \sqrt{2} \) (refractive index of the prism),
- \( i = 0 \) degrees.

Thus,
\[ 1 \cdot \sin(0) = \sqrt{2} \cdot \sin(r) \]
This simplifies to:
\[ 0 = \sqrt{2} \cdot \sin(r) \]
This implies that \( r = 0 \) degrees.

### Step 3: Determine the Angle of Incidence at the Second Face
The ray travels through the prism and reaches the second face. The angle of incidence (i') at the second face can be calculated as:
\[ i' = A - r \]
Where:
- \( A = 30 \) degrees (apex angle),
- \( r = 0 \) degrees (angle of refraction at the first face).

Thus,
\[ i' = 30 - 0 = 30 \text{ degrees} \]

### Step 4: Calculate the Angle of Refraction at the Second Face
Using Snell's law again at the second face:
\[ n_2 \sin(i') = n_1 \sin(r') \]
Substituting the known values:
\[ \sqrt{2} \cdot \sin(30) = 1 \cdot \sin(r') \]
Since \( \sin(30) = \frac{1}{2} \):
\[ \sqrt{2} \cdot \frac{1}{2} = \sin(r') \]
\[ \sin(r') = \frac{\sqrt{2}}{2} \]
This implies:
\[ r' = 45 \text{ degrees} \]

### Step 5: Calculate the Angle of Deviation
The angle of deviation (D) is given by:
\[ D = i' + r' - A \]
Substituting the values:
\[ D = 30 + 45 - 30 \]
\[ D = 45 \text{ degrees} \]

Thus, the angle of deviation of the ray is **15 degrees**.

### Summary of Solution
The angle of deviation of the ray is **15 degrees**.