A ray of light strikes a transparent rectangular slab of refractive index `sqrt(2)` at an angle of incidence of `45^(@)`. The angle between the reflected and refracted rays is

Options

A. `75^(@)`
B. `90^(@)`
C. `105^(@)`
D. `120^(@)`

Explanation

To solve the problem of finding the angle between the reflected and refracted rays when a ray of light strikes a transparent rectangular slab of refractive index \(\sqrt{2}\) at an angle of incidence of \(45^\circ\), we can follow these steps:

### Step 1: Identify the given values
- Refractive index of air, \(n_1 = 1\)
- Refractive index of the slab, \(n_2 = \sqrt{2}\)
- Angle of incidence, \(I = 45^\circ\)

### Step 2: Apply the law of reflection
According to the law of reflection, the angle of reflection \(R\) is equal to the angle of incidence \(I\):
\[ R = I = 45^\circ \]

### Step 3: Apply Snell's Law
Snell's law relates the angles of incidence and refraction to the refractive indices of the two media:
\[ n_1 \sin I = n_2 \sin R \]
Substituting the known values:
\[ 1 \cdot \sin(45^\circ) = \sqrt{2} \cdot \sin(\theta) \]
Where \(\theta\) is the angle of refraction. We know that \(\sin(45^\circ) = \frac{1}{\sqrt{2}}\):
\[ 1 \cdot \frac{1}{\sqrt{2}} = \sqrt{2} \cdot \sin(\theta) \]
This simplifies to:
\[ \frac{1}{\sqrt{2}} = \sqrt{2} \cdot \sin(\theta) \]

### Step 4: Solve for \(\sin(\theta)\)
Rearranging the equation gives:
\[ \sin(\theta) = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2} \]
Thus,
\[ \theta = \sin^{-1}\left(\frac{1}{2}\right) = 30^\circ \]

### Step 5: Calculate the angle between the reflected and refracted rays
The angle between the reflected ray and the refracted ray is given by:
\[ \text{Angle between reflected and refracted rays} = R + \theta \]
Substituting the values we found:
\[ \text{Angle} = 45^\circ + 30^\circ = 75^\circ \]

### Step 6: Find the angle between the rays
Since the total angle around the normal is \(180^\circ\), we can also express this as:
\[ \text{Angle} = 180^\circ - (R + \theta) = 180^\circ - 75^\circ = 105^\circ \]

### Final Answer
The angle between the reflected and refracted rays is \(105^\circ\).

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