A prism having refractive index `sqrt2` and refractive angle `30^@` has one of the refractive surfaces polished. A beam of light incident on the other surfaces will trace its path if the angle of incidence is

Options

A. `0^@`
B. `30^@`
C. `45^@`
D. `60^@`

Explanation

To solve the problem step by step, we will follow the reasoning and calculations presented in the video transcript.

### Step 1: Understand the Problem
We have a prism with a refractive index of \( \sqrt{2} \) and a refractive angle of \( 30^\circ \). One of the refractive surfaces is polished, and we need to find the angle of incidence for a beam of light incident on the other surface.

### Step 2: Draw the Diagram
1. Draw the prism with the refractive angle \( A = 30^\circ \).
2. Identify the two surfaces of the prism. Since one surface is polished, we will consider the other surface for the incident light ray.
3. Draw a perpendicular line to the polished surface.

### Step 3: Identify Angles
1. Let the angle of incidence be \( I \).
2. The angle of refraction \( r \) will occur at the interface between air (refractive index \( n_1 = 1 \)) and the prism (refractive index \( n_2 = \sqrt{2} \)).
3. The angle between the refracted ray and the perpendicular to the surface is \( 30^\circ \) (the refractive angle).

### Step 4: Use Geometry to Find Angles
1. The total angle between the refracted ray and the perpendicular line is \( 90^\circ \).
2. Since the angle at the prism is \( 30^\circ \), we can find the angle \( r \) using:
\[ r = 90^\circ - 30^\circ = 60^\circ \]

### Step 5: Apply Snell's Law
Using Snell's Law:
\[ n_1 \sin I = n_2 \sin r \]
Substituting the known values:
\[ 1 \cdot \sin I = \sqrt{2} \cdot \sin(60^\circ) \]

### Step 6: Calculate \( \sin(60^\circ) \)
We know that:
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \]
So, substituting this into the equation gives:
\[ \sin I = \sqrt{2} \cdot \frac{\sqrt{3}}{2} \]

### Step 7: Simplify the Equation
Now we simplify:
\[ \sin I = \frac{\sqrt{6}}{2} \]

### Step 8: Find the Angle of Incidence
To find \( I \):
\[ I = \sin^{-1}\left(\frac{\sqrt{6}}{2}\right) \]
This value corresponds to an angle of \( 45^\circ \).

### Final Answer
Thus, the angle of incidence \( I \) is:
\[ \boxed{45^\circ} \]