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### Problem 5
**Problem Statement:**  
A ray of white light (from red to violet) travels from air to glass at an angle of incidence of 70°. The refractive indices of the glass for red and violet light are 1.52 and 1.58, respectively. Find the angle between the paths of red light and violet light in the glass.

**Solution:**  
To solve this, we’ll use Snell’s law:
\[
n_1 \sin \theta_1 = n_2 \sin \theta_2
\]
where:
- \( n_1 \) is the refractive index of air (approximately 1),
- \( \theta_1 \) is the angle of incidence (70°),
- \( n_2 \) is the refractive index of glass for each color of light (1.52 for red, 1.58 for violet),
- \( \theta_2 \) is the angle of refraction in the glass.

#### Step 1: Calculate the Refraction Angle for Red Light
For red light:
\[
1 \cdot \sin(70^\circ) = 1.52 \cdot \sin(\theta_{\text{red}})
\]
\[
\sin(\theta_{\text{red}}) = \frac{\sin(70^\circ)}{1.52}
\]
Using \( \sin(70^\circ) \approx 0.9397 \):
\[
\sin(\theta_{\text{red}}) \approx \frac{0.9397}{1.52} \approx 0.6189
\]
\[
\theta_{\text{red}} \approx \arcsin(0.6189) \approx 38.2^\circ
\]

#### Step 2: Calculate the Refraction Angle for Violet Light
For violet light:
\[
1 \cdot \sin(70^\circ) = 1.58 \cdot \sin(\theta_{\text{violet}})
\]
\[
\sin(\theta_{\text{violet}}) = \frac{\sin(70^\circ)}{1.58}
\]
\[
\sin(\theta_{\text{violet}}) \approx \frac{0.9397}{1.58} \approx 0.5941
\]
\[
\theta_{\text{violet}} \approx \arcsin(0.5941) \approx 36.5^\circ
\]

#### Step 3: Find the Angle Between the Paths of Red and Violet Light
The angle between the paths of red and violet light in the glass is:
\[
\Delta \theta = \theta_{\text{red}} - \theta_{\text{violet}}
\]
\[
\Delta \theta \approx 38.2^\circ - 36.5^\circ = 1.7^\circ
\]

**Answer for Problem 5:** The angle between the paths of red light and violet light in the glass is approximately \( 1.7^\circ \).

---

### Problem 6
**Problem Statement:**  
A diver under the water is looking at a dragonfly that stays still in the air. The apparent position of the dragonfly is _________ than its real position, this is due to the __________.

**Solution:**  
When looking from a denser medium (water) to a less dense medium (air), objects appear **higher** or **closer** than their actual position. This phenomenon is due to **refraction** of light as it passes from water to air, causing the light to bend and the object to appear shifted.

**Answer for Problem 6:**  
The apparent position of the dragonfly is **higher** than its real position, and this is due to **refraction**.

---

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### Related Questions:
1. How does Snell's law apply to refraction in different mediums?
2. What is the significance of different refractive indices for colors in dispersion?
3. Why does light bend towards the normal when entering a denser medium?
4. How is apparent depth different from actual depth in refraction?
5. How would the refraction angle change if the incidence angle were smaller?

**Tip:** Refraction occurs because light changes speed when moving between materials with different optical densities.

A ray of white light (mixture of light, from red to violet) travels from air to glass at an angle of incidence of 70°. Suppose the refractive indices of the glass for red light and violet light are 1.52 and 1.58, respectively. Find the angle between the paths of red light and violet light in the glass.

Solve a refraction problem where a ray of white light passes from air to glass at a 70° incidence angle. Using Snell's law and refractive indices for red and violet light (1.52 and 1.58), find the angle between their paths in glass. Suitable for Grades 10-12.

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