A ray of light is travelling from medium ‘A’ into a rarer medium B. The angle of incidence ^siS 45° and the angle of deviation is 15° . The refractive index of medium A w.r.to B is

Options

A. `sqrt(3/2)`
B. `sqrt(3)/2`
C. `1/sqrt(2)`
D. `sqrt(2/3)`

Explanation

To solve the problem, we need to find the refractive index of medium A with respect to medium B, given the angle of incidence and the angle of deviation. Here's the step-by-step solution:

### Step 1: Understand the Given Information
- Angle of incidence (I) = 45°
- Angle of deviation (Δ) = 15°
- Since light is traveling from a denser medium A to a rarer medium B, we know that the refractive index of medium A (μA) is greater than that of medium B (μB).

### Step 2: Determine the Angle of Refraction
The angle of refraction (R) can be calculated using the relationship:
\[ R = I + Δ \]
Substituting the values:
\[ R = 45° + 15° = 60° \]

### Step 3: Apply Snell's Law
Snell's Law states:
\[ μA \sin(I) = μB \sin(R) \]
We can rearrange this to find the ratio of the refractive indices:
\[ \frac{μA}{μB} = \frac{\sin(R)}{\sin(I)} \]

### Step 4: Calculate the Sine Values
Now we need to find the sine values for the angles:
- \( \sin(45°) = \frac{1}{\sqrt{2}} \)
- \( \sin(60°) = \frac{\sqrt{3}}{2} \)

### Step 5: Substitute the Sine Values into Snell's Law
Substituting the sine values into the rearranged Snell's Law:
\[ \frac{μA}{μB} = \frac{\sin(60°)}{\sin(45°)} \]
\[ \frac{μA}{μB} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}} \]

### Step 6: Simplify the Expression
To simplify:
\[ \frac{μA}{μB} = \frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{1} \]
\[ \frac{μA}{μB} = \frac{\sqrt{3} \cdot \sqrt{2}}{2} = \frac{\sqrt{6}}{2} \]

### Final Answer
The refractive index of medium A with respect to medium B is:
\[ μA = \frac{\sqrt{6}}{2} \]