A ray of light passes through a prism in a principle plane the deviation being equal to angle of incidence which is equal to `2alpha`. It is given that `alpha` is the angle fo prism and `mu` is the refractive index of the material of prism, then

Options

A. `"cos"alpha = sqrt((mu^(2)-1)/(2))`
B. `"cos"alpha = sqrt((mu^(2)-1)/(8))`
C. `"sin"alpha = sqrt(mu^(2)-1)`
D. `"sin"alpha = sqrt((mu^(2)-1)/(8))`

Explanation

To solve the problem, we need to find the relationship between the cosine or sine of the angle of the prism (α) and the refractive index (μ) of the material of the prism. Let's break down the solution step by step.

### Step 1: Understand the given information
- The angle of incidence (i) is given as \(2\alpha\).
- The angle of emergence (e) is equal to the angle of the prism (α).
- The deviation (Δ) is equal to the angle of incidence (i).

### Step 2: Write the formula for deviation
The deviation for a prism can be expressed as:
\[ \Delta = i + e - A \]
where \(A\) is the angle of the prism.

### Step 3: Substitute the known values into the deviation formula
Substituting the values we have:
\[ 2\alpha = 2\alpha + e - \alpha \]
This simplifies to:
\[ 2\alpha = 2\alpha - \alpha + e \]
Thus, we can find the angle of emergence:
\[ e = \alpha \]

### Step 4: Apply Snell's Law at the first surface
Using Snell's Law at the first surface of the prism:
\[ \sin(2\alpha) = \mu \cdot \sin(r_1) \]
where \(r_1\) is the angle of refraction at the first surface.

### Step 5: Apply Snell's Law at the second surface
Using Snell's Law at the second surface:
\[ \sin(e) = \mu \cdot \sin(r_2) \]
Since we found \(e = \alpha\):
\[ \sin(\alpha) = \mu \cdot \sin(r_2) \]

### Step 6: Relate \(r_1\) and \(r_2\)
From the geometry of the prism, we know:
\[ r_1 + r_2 = A = \alpha \]
Thus, we can express \(r_2\) in terms of \(r_1\):
\[ r_2 = \alpha - r_1 \]

### Step 7: Substitute \(r_2\) into the Snell's Law equation
Substituting \(r_2\) into the equation from Snell's Law at the second surface:
\[ \sin(\alpha) = \mu \cdot \sin(\alpha - r_1) \]
Using the sine subtraction formula:
\[ \sin(\alpha - r_1) = \sin(\alpha)\cos(r_1) - \cos(\alpha)\sin(r_1) \]
Thus, we can rewrite the equation as:
\[ \sin(\alpha) = \mu \cdot (\sin(\alpha)\cos(r_1) - \cos(\alpha)\sin(r_1)) \]

### Step 8: Substitute \(\sin(r_1)\) from the first surface equation
From the first surface equation:
\[ \sin(r_1) = \frac{\sin(2\alpha)}{\mu} \]
Substituting this into the equation gives:
\[ \sin(\alpha) = \mu \cdot \left(\sin(\alpha)\cos(r_1) - \cos(\alpha)\frac{\sin(2\alpha)}{\mu}\right) \]

### Step 9: Rearranging the equation
After simplifying and rearranging, we can isolate terms involving \(\mu\) and \(\alpha\) to find a relationship.

### Step 10: Final relationship
After simplification, we arrive at:
\[ \mu^2 = 1 + 4\cos^2(\alpha) \]
This shows the relationship between the refractive index (μ) and the angle of the prism (α).