Find the minimum and maximum angle of deviation for a prism with angle `A = 60^(@)` and `mu = 1.5`

Options

A. `37^(@)`
B. `49^(@)`
C. `60^(@)`
D. `90^(@)`

Explanation

To find the minimum and maximum angle of deviation for a prism with an angle \( A = 60^\circ \) and refractive index \( \mu = 1.5 \), we can follow these steps:

### Step 1: Understand the relationship between angles
For a prism, the angle of deviation \( \delta \) is related to the angle of incidence \( i \) and the angle of emergence \( e \) by the formula:
\[ \delta = i + e - A \]

### Step 2: Find the minimum angle of deviation
The minimum angle of deviation occurs when the angle of incidence \( i \) equals the angle of emergence \( e \). In this case, we can denote both angles as \( i \).

### Step 3: Use the relationship between angles in the prism
The relationship between the angles of incidence and refraction can be expressed using Snell's law:
\[ \mu_1 \sin(i) = \mu_2 \sin(R) \]
where \( R \) is the angle of refraction inside the prism. Since we have a prism with angle \( A \), we know that:
\[ R_1 + R_2 = A \]
If \( R_1 = R_2 = R \), then:
\[ 2R = A \implies R = \frac{A}{2} = \frac{60^\circ}{2} = 30^\circ \]

### Step 4: Calculate the angle of incidence for minimum deviation
Using Snell's law at the first surface:
\[ \mu_1 \sin(i) = \mu_2 \sin(R) \]
Substituting the values:
\[ 1 \cdot \sin(i) = 1.5 \cdot \sin(30^\circ) \]
Since \( \sin(30^\circ) = \frac{1}{2} \):
\[ \sin(i) = 1.5 \cdot \frac{1}{2} = 0.75 \]
Thus,
\[ i = \sin^{-1}(0.75) \approx 48.6^\circ \]

### Step 5: Calculate the minimum angle of deviation
Now, substituting \( i \) back into the deviation formula:
\[ \delta_{min} = 2i - A \]
\[ \delta_{min} = 2 \cdot 48.6^\circ - 60^\circ \approx 37.2^\circ \]

### Step 6: Find the maximum angle of deviation
The maximum angle of deviation occurs when the angle of incidence \( i \) is \( 90^\circ \). In this case, we will find the corresponding angle of refraction \( R \).

### Step 7: Calculate the angle of refraction for maximum deviation
Using Snell's law again:
\[ \mu_1 \sin(90^\circ) = \mu_2 \sin(R) \]
This simplifies to:
\[ 1 = 1.5 \sin(R) \implies \sin(R) = \frac{1}{1.5} = \frac{2}{3} \]
Thus,
\[ R \approx \sin^{-1}\left(\frac{2}{3}\right) \approx 42^\circ \]

### Step 8: Calculate the angle of emergence
Using the prism angle:
\[ R_1 + R_2 = A \implies R_1 + 42^\circ = 60^\circ \implies R_1 = 18^\circ \]

### Step 9: Calculate the angle of emergence using Snell's law
Using Snell's law at the second surface:
\[ \mu_2 \sin(R_2) = \mu_1 \sin(e) \]
Substituting the values:
\[ 1.5 \sin(18^\circ) = 1 \cdot \sin(e) \]
Calculating \( \sin(18^\circ) \approx 0.309 \):
\[ \sin(e) = 1.5 \cdot 0.309 \approx 0.464 \]
Thus,
\[ e \approx \sin^{-1}(0.464) \approx 27.6^\circ \]

### Step 10: Calculate the maximum angle of deviation
Substituting \( i \) and \( e \) back into the deviation formula:
\[ \delta_{max} = i + e - A \]
\[ \delta_{max} = 90^\circ + 27.6^\circ - 60^\circ \approx 57.6^\circ \]

### Final Answers
- Minimum angle of deviation \( \delta_{min} \approx 37.2^\circ \)
- Maximum angle of deviation \( \delta_{max} \approx 57.6^\circ \)