Answer :
Let's solve this step-by-step.
### Step 1: Understand the Given Data
We are given:
1. Incident angle ([tex]\(\theta_1\)[/tex]) with the normal: 60 degrees.
2. Emergence angle ([tex]\(\theta_2\)[/tex]) with the normal when the ray emerges back into the air: 42 degrees.
3. Refractive index of air ([tex]\(n_1\)[/tex]) is typically 1.
### Step 2: Snell's Law Application
According to Snell's Law:
[tex]\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \][/tex]
where:
- [tex]\( n_1 \)[/tex] is the refractive index of the first medium (air, in this case).
- [tex]\( \theta_1 \)[/tex] is the angle of incidence.
- [tex]\( n_2 \)[/tex] is the refractive index of the second medium (glass, in this case).
- [tex]\( \theta_2 \)[/tex] is the angle of refraction.
### Step 3: Calculate the Sine of the Angles
We can convert the angles from degrees to radians to find the sine values.
[tex]\[ \theta_1 = 60^\circ \][/tex]
[tex]\[ \theta_2 = 42^\circ \][/tex]
The sine values for these angles are:
- [tex]\( \sin(60^\circ) \approx 0.86603 \)[/tex]
- [tex]\( \sin(42^\circ) \approx 0.66913 \)[/tex]
### Step 4: Solving for the Refractive Index of Glass
We need to find [tex]\( n_2 \)[/tex] (refractive index of glass). Rearrange Snell's Law to solve for [tex]\( n_2 \)[/tex]:
[tex]\[ n_2 = \frac{n_1 \sin(\theta_1)}{\sin(\theta_2)} \][/tex]
Given [tex]\( n_1 \)[/tex] is 1, we have:
[tex]\[ n_2 = \frac{1 \cdot 0.86603}{0.66913} \approx 1.29425 \][/tex]
### Step 5: Diagram and Additional Angles
To draw the path of the light ray through the slab:
1. Draw the normal to the glass surface at the point where the light ray enters.
2. Draw the incident ray at an angle of 60 degrees with the normal.
3. At the point where the light ray reaches the other face of the slab, draw the normal again.
4. Draw the refracted ray at an angle of refraction such that it eventually emerges out at an angle of 42 degrees with the normal when entering air.
Here's a simple, labeled diagram of the path of the light ray:
```
Air (n1)
Normal \ Normal
| \ |
| Incident \ / Emergent
| 60° θâ\ / 42°
| / | \ /
Glass / | \
(n2) / θâ | \
|Ray| \
Key:
θâ = Refracted angle inside glass
θâ = Emerged refracted angle (42 degrees)
```
### Final Summary
- The angle of refraction [tex]\(\theta_1\)[/tex] inside the glass is such that the sine of this angle is approximately 0.86603.
- The angle of emergence [tex]\(\theta_2\)[/tex] with the normal in air is 42 degrees, having a sine value of 0.66913.
- The refractive index of the glass [tex]\(n_2\)[/tex] is approximately 1.29425.
### Step 1: Understand the Given Data
We are given:
1. Incident angle ([tex]\(\theta_1\)[/tex]) with the normal: 60 degrees.
2. Emergence angle ([tex]\(\theta_2\)[/tex]) with the normal when the ray emerges back into the air: 42 degrees.
3. Refractive index of air ([tex]\(n_1\)[/tex]) is typically 1.
### Step 2: Snell's Law Application
According to Snell's Law:
[tex]\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \][/tex]
where:
- [tex]\( n_1 \)[/tex] is the refractive index of the first medium (air, in this case).
- [tex]\( \theta_1 \)[/tex] is the angle of incidence.
- [tex]\( n_2 \)[/tex] is the refractive index of the second medium (glass, in this case).
- [tex]\( \theta_2 \)[/tex] is the angle of refraction.
### Step 3: Calculate the Sine of the Angles
We can convert the angles from degrees to radians to find the sine values.
[tex]\[ \theta_1 = 60^\circ \][/tex]
[tex]\[ \theta_2 = 42^\circ \][/tex]
The sine values for these angles are:
- [tex]\( \sin(60^\circ) \approx 0.86603 \)[/tex]
- [tex]\( \sin(42^\circ) \approx 0.66913 \)[/tex]
### Step 4: Solving for the Refractive Index of Glass
We need to find [tex]\( n_2 \)[/tex] (refractive index of glass). Rearrange Snell's Law to solve for [tex]\( n_2 \)[/tex]:
[tex]\[ n_2 = \frac{n_1 \sin(\theta_1)}{\sin(\theta_2)} \][/tex]
Given [tex]\( n_1 \)[/tex] is 1, we have:
[tex]\[ n_2 = \frac{1 \cdot 0.86603}{0.66913} \approx 1.29425 \][/tex]
### Step 5: Diagram and Additional Angles
To draw the path of the light ray through the slab:
1. Draw the normal to the glass surface at the point where the light ray enters.
2. Draw the incident ray at an angle of 60 degrees with the normal.
3. At the point where the light ray reaches the other face of the slab, draw the normal again.
4. Draw the refracted ray at an angle of refraction such that it eventually emerges out at an angle of 42 degrees with the normal when entering air.
Here's a simple, labeled diagram of the path of the light ray:
```
Air (n1)
Normal \ Normal
| \ |
| Incident \ / Emergent
| 60° θâ\ / 42°
| / | \ /
Glass / | \
(n2) / θâ | \
|Ray| \
Key:
θâ = Refracted angle inside glass
θâ = Emerged refracted angle (42 degrees)
```
### Final Summary
- The angle of refraction [tex]\(\theta_1\)[/tex] inside the glass is such that the sine of this angle is approximately 0.86603.
- The angle of emergence [tex]\(\theta_2\)[/tex] with the normal in air is 42 degrees, having a sine value of 0.66913.
- The refractive index of the glass [tex]\(n_2\)[/tex] is approximately 1.29425.
