A light ray is incident at `45^(@)` on a glass slab. The slab is `3cm` thick, and the refractive index of the glass is `1.5`. What will the displacement of the ray be as a result of its passage through the slab? At what angle will the ray emerge formn the slab?

Explanation

To solve the problem of a light ray incident at an angle of \( 45^\circ \) on a glass slab that is \( 3 \, \text{cm} \) thick with a refractive index of \( 1.5 \), we can follow these steps:

### Step 1: Determine the angle of refraction using Snell's Law
Snell's Law states that:
\[ n_1 \sin i = n_2 \sin r \]
Where:
- \( n_1 = 1 \) (refractive index of air)
- \( n_2 = 1.5 \) (refractive index of glass)
- \( i = 45^\circ \) (angle of incidence)
- \( r \) is the angle of refraction

Substituting the values:
\[ 1 \cdot \sin(45^\circ) = 1.5 \cdot \sin(r) \]
\[ \sin(45^\circ) = \frac{\sqrt{2}}{2} \]
Thus,
\[ \frac{\sqrt{2}}{2} = 1.5 \cdot \sin(r) \]
\[ \sin(r) = \frac{\sqrt{2}}{3} \]
Now, we can find the angle \( r \) using the inverse sine function.

### Step 2: Calculate the displacement of the ray
The displacement \( d \) of the ray as it passes through the slab can be calculated using the formula:
\[ d = t \cdot \sin(i - r) \]
Where:
- \( t = 3 \, \text{cm} \) (thickness of the slab)
- \( i = 45^\circ \)
- \( r \) is the angle of refraction calculated in Step 1.

### Step 3: Calculate the angle of emergence
Since the ray emerges from the slab, it will again follow Snell's Law. The angle of emergence will be equal to the angle of incidence at the first surface due to the symmetry of the situation:
\[ \text{Angle of emergence} = i = 45^\circ \]

### Final Calculation
1. **Calculate \( r \)**:
\[ r = \sin^{-1}\left(\frac{\sqrt{2}}{3}\right) \]
(Use a calculator to find the value of \( r \))

2. **Calculate the displacement \( d \)**:
\[ d = 3 \cdot \sin(45^\circ - r) \]
(Substitute the value of \( r \) to find \( d \))

### Summary of Results
- The displacement \( d \) can be calculated using the above formula.
- The angle of emergence is \( 45^\circ \).