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^(@)`

Answer

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

### Step 1: Identify the critical angle
The critical angle \( c \) can be calculated using the formula:
\[ \sin c = \frac{1}{\mu} \]
where \( \mu \) is the refractive index of the prism. Given \( \mu = 1.5 \):
\[ \sin c = \frac{1}{1.5} = \frac{2}{3} \]

### Step 2: Calculate the critical angle
Using the inverse sine function:
\[ c = \sin^{-1}\left(\frac{2}{3}\right) \]
From the problem, we know that \( \sin^{-1}\left(\frac{2}{3}\right) = 42^\circ \). Thus, the critical angle \( c = 42^\circ \).

### Step 3: Analyze the prism
An equilateral prism has angles of \( 60^\circ \). When a ray of light is incident normally on one face of the prism, the angle of incidence \( i \) is \( 90^\circ \).

### Step 4: Determine the angle of refraction at the first face
Using Snell's law:
\[ n_1 \sin i = n_2 \sin r \]
Here, \( n_1 = 1 \) (air), \( i = 90^\circ \), and \( n_2 = 1.5 \). Thus:
\[ 1 \cdot \sin(90^\circ) = 1.5 \cdot \sin r \]
This simplifies to:
\[ 1 = 1.5 \cdot \sin r \implies \sin r = \frac{1}{1.5} = \frac{2}{3} \]
So, \( r = 42^\circ \).

### Step 5: Calculate the angle of incidence at the second face
At the second face of the prism, the angle of incidence \( i_2 \) is given by:
\[ i_2 = 60^\circ - r = 60^\circ - 42^\circ = 18^\circ \]

### Step 6: Determine the angle of refraction at the second face
Using Snell's law again:
\[ 1.5 \cdot \sin(18^\circ) = 1 \cdot \sin r' \]
Calculating \( \sin(18^\circ) \):
\[ \sin(18^\circ) \approx 0.309 \]
Thus:
\[ 1.5 \cdot 0.309 = \sin r' \implies \sin r' \approx 0.464 \implies r' \approx 27.7^\circ \]

### Step 7: Calculate the angle of deviation
The angle of deviation \( \delta \) is given by:
\[ \delta = i - r' = 60^\circ + 60^\circ - 180^\circ = 60^\circ \]

### Conclusion
Thus, the angle of deviation \( \delta \) is:
\[ \delta = 60^\circ \]