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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.

Would you like further explanation on these problems?

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.