Deviation `delta` produced by a prism of angle A, which is assumed to be small, made of a material of refractive index `mu` is given by

Options

A. `delta=(mu-1)A`
B. `delta=(mu+1)A`
C. `((sin"A+delta)/2)/(sin(A)/(2))`
D. `delta=(mu-1)/(mu+1)`

Answer

To solve the problem of finding the deviation \( \delta \) produced by a thin prism of angle \( A \) and refractive index \( \mu \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Prism Formula**:
The deviation \( \delta \) for a prism can be derived using the formula:
\[ \mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]
where \( \delta_m \) is the minimum deviation and \( A \) is the angle of the prism.

2. **Assuming Small Angles**:
Since we are dealing with a thin prism, we can assume that the angle \( A \) is small. For small angles, we can use the approximation:
\[ \sin x \approx x \quad \text{(in radians)} \]
Thus, we can rewrite the sine functions:
\[ \sin\left(\frac{A + \delta_m}{2}\right) \approx \frac{A + \delta_m}{2} \quad \text{and} \quad \sin\left(\frac{A}{2}\right) \approx \frac{A}{2} \]

3. **Substituting the Approximations**:
Substituting these approximations into the prism formula gives:
\[ \mu = \frac{\frac{A + \delta_m}{2}}{\frac{A}{2}} = \frac{A + \delta_m}{A} \]

4. **Rearranging the Equation**:
Rearranging the equation leads to:
\[ \mu = 1 + \frac{\delta_m}{A} \]

5. **Solving for Minimum Deviation**:
From the above equation, we can express the minimum deviation \( \delta_m \):
\[ \delta_m = (\mu - 1) \cdot A \]

6. **Conclusion**:
Therefore, the deviation \( \delta \) produced by a thin prism of angle \( A \) and refractive index \( \mu \) is given by:
\[ \delta = (\mu - 1) \cdot A \]

### Final Answer:
The deviation \( \delta \) produced by a thin prism of angle \( A \) and refractive index \( \mu \) is:
\[ \delta = (\mu - 1) \cdot A \]