The correct relation between the refractive index (`mu`) of the material of prism (A) and angle of minimum deviation (`delta`) is

Options

A. `mu= sin(A/2) /sin (A + delta /2)`
B. `mu= cos(A + delta /2) /sin (A /2)`
C. `mu= cos(A /2) /sin (A+delta /2)`
D. `mu= sin(A + delta /2) /sin (A/2)`

Answer

To find the correct relation between the refractive index (\( \mu \)) of the material of a prism and the angle of minimum deviation (\( \delta \)), we can follow these steps:

### Step 1: Understand the Geometry of the Prism
- A prism has an apex angle \( A \). When light passes through the prism, it is refracted at both surfaces.
- The angle of deviation \( \delta \) is defined as the angle between the original path of the light and the path after it exits the prism.

**Hint:** Visualize the path of light through the prism and how it bends at each interface.

### Step 2: Apply the Condition for Minimum Deviation
- At the angle of minimum deviation, the angles of incidence and emergence are equal (\( I_1 = I_2 \)), and the angles of refraction are also equal (\( R_1 = R_2 \)).
- Let \( R \) be the angle of refraction at each surface. Thus, we can write:
\[ A = R_1 + R_2 \quad \text{(1)} \]
Since \( R_1 = R_2 = R \), we can rewrite this as:
\[ A = 2R \quad \text{(2)} \]

**Hint:** Remember that at minimum deviation, the light path is symmetric.

### Step 3: Relate the Angles Using Snell's Law
- According to Snell's law, we have:
\[ \mu = \frac{\sin I_1}{\sin R_1} \quad \text{(3)} \]
Substituting \( I_1 = R + \frac{\delta}{2} \) and \( R_1 = R \) into Snell's law gives:
\[ \mu = \frac{\sin\left(R + \frac{\delta}{2}\right)}{\sin R} \quad \text{(4)} \]

**Hint:** Use the sine addition formula to simplify the expression.

### Step 4: Substitute for \( R \)
- From equation (2), we know \( R = \frac{A}{2} \). Substitute this into equation (4):
\[ \mu = \frac{\sin\left(\frac{A}{2} + \frac{\delta}{2}\right)}{\sin\left(\frac{A}{2}\right)} \quad \text{(5)} \]

**Hint:** This substitution is crucial for relating the angles directly to the refractive index.

### Step 5: Final Expression
- The final expression that relates the refractive index \( \mu \) of the prism material and the angle of minimum deviation \( \delta \) can be simplified further using trigonometric identities if necessary.
- The final relation is:
\[ \mu = \frac{\sin\left(\frac{A + \delta}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

**Hint:** This equation shows how the refractive index depends on both the prism angle and the angle of minimum deviation.

### Summary
The correct relation between the refractive index \( \mu \) of the material of the prism and the angle of minimum deviation \( \delta \) is given by:
\[ \mu = \frac{\sin\left(\frac{A + \delta}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]