Part A: Using Snell's Law to Compute the Angle of Refraction

Snell's Law relates the angles of incidence and refraction when light passes between two media with different refractive indices. The equation is:

$n_1 \sin(\theta_1) = n_2 \sin(\theta_2)$

Where:

• $n_1$ is the refractive index of the first medium (air in this case, $n_1 = 1.0$).
• $\theta_1$ is the angle of incidence.
• $n_2$ is the refractive index of the second medium (glass, $n_2 = 1.4$).
• $\theta_2$ is the angle of refraction.

Given:

• $\theta_1 = 25^\circ$
• $n_1 = 1.0$
• $n_2 = 1.4$

We need to find $\theta_2$, the angle of refraction. Using Snell's law:

$\sin(\theta_2) = \frac{n_1 \sin(\theta_1)}{n_2}$

Let’s calculate $\theta_2$.

Part B: Geometry of Viewing Through Glass

This part requires calculating how far the spot on the wall appears to shift due to refraction as you look through the glass. To solve this, we need to consider the geometry of the problem, including the distances involved, and create a diagram that shows how the light bends as it passes through the glass.

Given:

• You are 8 feet from the wall.
• The glass is 3 inches (0.25 feet) thick and positioned 2 feet in front of you.
• The refractive index of the glass is 1.4.

We need to find how much lower the spot appears when looking through the glass, which involves calculating the apparent displacement due to refraction.

I’ll calculate the refraction angle from Part A first, and then proceed with the geometry calculations for Part B.### Part A Result: Using Snell's Law, the angle of refraction ($\theta_2$) when light enters the glass is approximately $17.57^\circ$.

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Part B: Geometry of Viewing Through Glass

To calculate how far below your eye level the spot appears on the wall, we need to take into account:

1. The actual distance between your position and the glass (2 feet).
2. The thickness of the glass (3 inches = 0.25 feet).
3. The distance from the glass to the wall (8 - 2 = 6 feet).

Steps:

1. The light enters the glass at an angle of $\theta_1 = 25^\circ$ and refracts to $\theta_2 = 17.57^\circ$.
2. We'll calculate the lateral shift caused by refraction, which results in the spot appearing lower than it would without the glass.

Let me compute the vertical displacement next.It seems like I can’t do more advanced data analysis right now. Please try again later.

If you'd like, I can walk you through the process step-by-step so you can complete the calculation manually, or you can upload another question.

Would you like me to continue with the explanation or assist with something else?

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In the meantime, here are some follow-up questions based on the concepts:

1. How would the angle of refraction change if the refractive index were higher?
2. What would happen to the lateral shift if the glass were thinner?
3. How does Snell's Law apply when light enters a denser medium from a less dense one?
4. What is total internal reflection, and when does it occur?
5. How would the shift change if the angle of incidence were increased?

Tip: Snell's Law is only applicable until the critical angle is reached. After that, total internal reflection occurs.