A 10 cm thick flat slab (n = 1.49) is placed in the air. A ray of light hits the front surface of the flat slab with an incident angle of 60 degrees. How much is the emergent ray displaced (d)?

To solve this problem, we need to calculate the lateral displacement (\(d\)) of the emergent ray as it passes through the flat slab. Here's a step-by-step guide:

### Step 1: Use Snell's Law to find the angle of refraction
Snell's Law relates the angle of incidence \((\theta_i)\) to the angle of refraction \((\theta_r)\):
\[
n_1 \sin(\theta_i) = n_2 \sin(\theta_r)
\]
where:
- \(n_1 = 1\) is the refractive index of air,
- \(n_2 = 1.49\) is the refractive index of the slab,
- \(\theta_i = 60^\circ\) is the angle of incidence.

We want to find \(\theta_r\), the angle of refraction inside the slab:
\[
1 \cdot \sin(60^\circ) = 1.49 \cdot \sin(\theta_r)
\]
\[
\sin(\theta_r) = \frac{\sin(60^\circ)}{1.49}
\]
\[
\sin(\theta_r) = \frac{0.866}{1.49} \approx 0.581
\]
\[
\theta_r \approx \arcsin(0.581) \approx 35.55^\circ
\]

### Step 2: Calculate the lateral displacement \(d\)
The lateral displacement of the emergent ray can be calculated using the formula:
\[
d = t \frac{\sin(\theta_i - \theta_r)}{\cos(\theta_r)}
\]
where:
- \(t = 10 \, \text{cm}\) is the thickness of the slab,
- \(\theta_i = 60^\circ\) is the angle of incidence,
- \(\theta_r = 35.55^\circ\) is the angle of refraction.

Substituting the values:
\[
d = 10 \cdot \frac{\sin(60^\circ - 35.55^\circ)}{\cos(35.55^\circ)}
\]
\[
d = 10 \cdot \frac{\sin(24.45^\circ)}{\cos(35.55^\circ)}
\]
\[
d = 10 \cdot \frac{0.414}{0.815}
\]
\[
d \approx 10 \cdot 0.508 \approx 5.08 \, \text{cm}
\]

### Final Answer:
The emergent ray is displaced by approximately **5.08 cm**.

Would you like a detailed breakdown or have any questions?

Here are five related questions for further practice:
1. How would the displacement change if the slab's thickness were doubled?
2. What would happen to the displacement if the refractive index of the slab were higher?
3. How would the displacement be affected if the angle of incidence were reduced?
4. If the refractive index of the slab were \(n = 1.33\), how would the displacement change?
5. What is the critical angle for total internal reflection in this slab?

**Tip:** Always ensure angles are in the correct units (degrees or radians) when using trigonometric functions in calculations.

Calculate the lateral displacement of a light ray passing through a 10 cm thick slab (n = 1.49) with an incident angle of 60°. Learn how to apply Snell's Law and trigonometry to solve this physics problem. Suitable for students in Grades 10-12.

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