Angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of prism of given prism. Then, the angle is prism is….
`(sin 48^(@)36'=0.75)`

Options

A. `80^(@)`
B. `41^(@)24'`
C. `60^(@)`
D. `82^(@)48'`

Explanation

To solve the problem, we will use the relationship between the refractive index of a prism, the angle of the prism (A), and the angle of minimum deviation (Δm). The formula we will use is:

\[ \mu = \frac{\sin\left(\frac{A + \Delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

Given that the angle of minimum deviation (Δm) is equal to the angle of the prism (A), we can set Δm = A. Therefore, the equation simplifies to:

\[ \mu = \frac{\sin\left(\frac{A + A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin(A)}{\sin\left(\frac{A}{2}\right)} \]

### Step-by-step Solution:

1. **Identify the given values:**
- Refractive index, \( \mu = 1.5 \)
- Angle of minimum deviation, \( \Delta_m = A \)

2. **Substitute into the formula:**
\[ 1.5 = \frac{\sin(A)}{\sin\left(\frac{A}{2}\right)} \]

3. **Use the sine double angle identity:**
The sine of an angle can be expressed as:
\[ \sin(A) = 2 \sin\left(\frac{A}{2}\right) \cos\left(\frac{A}{2}\right) \]

4. **Substituting this into the equation:**
\[ 1.5 = \frac{2 \sin\left(\frac{A}{2}\right) \cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

5. **Cancel \(\sin\left(\frac{A}{2}\right)\) from numerator and denominator:**
\[ 1.5 = 2 \cos\left(\frac{A}{2}\right) \]

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

7. **Finding the angle:**
We know that \( \cos\left(\frac{A}{2}\right) = 0.75 \). To find the angle, we take the inverse cosine:
\[ \frac{A}{2} = \cos^{-1}(0.75) \]

8. **Calculating \(\cos^{-1}(0.75)\):**
Using a calculator or trigonometric tables, we find:
\[ \frac{A}{2} \approx 41^\circ 34' \]

9. **Multiplying by 2 to find A:**
\[ A = 2 \times 41^\circ 34' = 82^\circ 48' \]

### Final Answer:
The angle of the prism \( A \) is approximately \( 82^\circ 48' \).

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