A ray of light is incident normally on one face of an equilateral prism of refractive index `1.5`. The angle of deviation is: (Given `sin^-1.(2)/(3)=42^@`)

Options

A. `30^@`
B. `45^@`
C. `60^@`
D. `75^@`

Explanation

To find the angle of deviation for a ray of light incident normally on one face of an equilateral prism with a refractive index of 1.5, we can follow these steps:

### Step 1: Understand the Geometry of the Prism
An equilateral prism has three equal angles of 60 degrees each. When light enters normally (perpendicularly) to one face, it does not bend at the first interface.

### Step 2: Calculate the Angle of Refraction
Since the light is incident normally, the angle of incidence (i) is 0 degrees. According to Snell's law:
\[ n_1 \sin(i) = n_2 \sin(r) \]
Where:
- \( n_1 = 1 \) (refractive index of air)
- \( n_2 = 1.5 \) (refractive index of the prism)
- \( i = 0^\circ \)
- \( r \) is the angle of refraction.

Since \( \sin(0) = 0 \), the equation simplifies to:
\[ 1 \cdot 0 = 1.5 \cdot \sin(r) \]
This means that the light will pass straight through the prism without bending at the first interface.

### Step 3: Determine the Angle of Deviation
The angle of deviation (D) can be calculated using the formula:
\[ D = i + e - A \]
Where:
- \( i \) is the angle of incidence,
- \( e \) is the angle of emergence,
- \( A \) is the angle of the prism.

For an equilateral prism, \( A = 60^\circ \).

Since the light enters normally:
- \( i = 0^\circ \)
- The angle of emergence \( e \) can be determined from the geometry of the prism. The light will emerge at an angle of \( 60^\circ \) from the second face of the prism.

Thus:
\[ D = 0^\circ + 60^\circ - 60^\circ \]
\[ D = 0^\circ \]

### Step 4: Conclusion
The angle of deviation for the ray of light incident normally on one face of the equilateral prism is:
\[ D = 0^\circ \]