The refractive index of glass is `1*5` from a point P inside a glass slab, draw rays PQ, PB and PC incident on the glass-air surface at an angle of incidence `30^(@),42^(@) and 60^(@)` respectively.
(a) In the diagram show the approximate direction
of these rays as they emerge out of the slab.
(b) What is the angle of refraction for the ray PB
(Take `sin42^(@)=(2)/(3)`)

Answer

To solve the problem step by step, we will first address part (a) by drawing the rays and then move on to part (b) to calculate the angle of refraction for ray PB.

### Step-by-Step Solution

#### Part (a): Drawing the Rays
1. **Identify the angles of incidence**:
- For ray PQ, the angle of incidence (i1) is 30°.
- For ray PB, the angle of incidence (i2) is 42°.
- For ray PC, the angle of incidence (i3) is 60°.

2. **Draw the glass slab**:
- Draw a horizontal line to represent the glass-air interface. Label the top part as "Air" and the bottom part as "Glass".

3. **Draw the incident rays**:
- From point P inside the glass slab, draw ray PQ making an angle of 30° with the normal at the interface.
- Draw ray PB making an angle of 42° with the normal.
- Draw ray PC making an angle of 60° with the normal.

4. **Determine the direction of the rays as they emerge**:
- For ray PQ (i1 = 30°): Use Snell's law to find the angle of refraction (r1) as it exits into the air.
- For ray PB (i2 = 42°): This ray is at the critical angle, so it will emerge parallel to the interface.
- For ray PC (i3 = 60°): This ray will undergo total internal reflection since its angle of incidence is greater than the critical angle.

5. **Label the emerging rays**:
- Draw the emerging ray from PQ at an angle r1 with respect to the normal in the air.
- Draw ray PB emerging parallel to the interface.
- Indicate that ray PC does not emerge but reflects back into the glass.

#### Part (b): Calculating the Angle of Refraction for Ray PB
1. **Use Snell's Law**:
- Snell's law states that \( n_1 \sin(i) = n_2 \sin(r) \).
- Here, \( n_1 \) (refractive index of glass) = 1.5, \( n_2 \) (refractive index of air) = 1, and \( i = 42° \).

2. **Rearranging Snell's Law**:
- We can write it as \( \sin(r) = \frac{n_1}{n_2} \sin(i) \).
- Substituting the values:
\[ \sin(r) = \frac{1.5}{1} \sin(42°) \]

3. **Using the given value**:
- We know \( \sin(42°) = \frac{2}{3} \).
- Therefore,
\[ \sin(r) = 1.5 \times \frac{2}{3} = 1 \]

4. **Finding the angle of refraction**:
- Since \( \sin(r) = 1 \), this means \( r = 90° \).

### Final Answers
- (a) The rays PQ, PB, and PC are drawn with their respective directions as they emerge from the slab.
- (b) The angle of refraction for ray PB is \( 90° \).