Angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of prism The angle of prism is `(cos 41^(@) = 0.75)`

Options

A. `62^(@)`
B. `41^(@)`
C. `82^(@)`
D. `31^(@)`

Answer

To solve the problem, we need to find the angle of the prism (A) given that the angle of minimum deviation (D) is equal to the angle of the prism and that the refractive index (n) of the prism is 1.5. We also have the information that \( \cos 41^\circ = 0.75 \).

### Step-by-step Solution:

1. **Understanding the Prism Formula**:
The prism formula relates the refractive index (n), the angle of the prism (A), and the angle of minimum deviation (D) as follows:
\[ n = \frac{\sin\left(\frac{A + D}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]
Since \( D = A \), we can substitute \( D \) with \( A \) in the formula.

2. **Substituting D with A**:
The formula becomes:
\[ n = \frac{\sin\left(\frac{A + A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin(A)}{\sin\left(\frac{A}{2}\right)} \]

3. **Using the Given Refractive Index**:
We know that \( n = 1.5 \). Therefore, we can write:
\[ 1.5 = \frac{\sin(A)}{\sin\left(\frac{A}{2}\right)} \]

4. **Using the Double Angle Identity**:
We can express \( \sin(A) \) using the double angle formula:
\[ \sin(A) = 2 \sin\left(\frac{A}{2}\right) \cos\left(\frac{A}{2}\right) \]
Substituting this back into our equation gives:
\[ 1.5 = \frac{2 \sin\left(\frac{A}{2}\right) \cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]
This simplifies to:
\[ 1.5 = 2 \cos\left(\frac{A}{2}\right) \]

5. **Solving for \( \cos\left(\frac{A}{2}\right) \)**:
Rearranging gives:
\[ \cos\left(\frac{A}{2}\right) = \frac{1.5}{2} = 0.75 \]

6. **Finding \( \frac{A}{2} \)**:
From the cosine value, we can find:
\[ \cos\left(\frac{A}{2}\right) = 0.75 \implies \frac{A}{2} = 41^\circ \]

7. **Calculating the Angle of Prism (A)**:
Thus, we find:
\[ A = 2 \times 41^\circ = 82^\circ \]

### Final Answer:
The angle of the prism \( A \) is \( 82^\circ \).

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