Light in Underwater Pool Refraction Problem

Click For Summary
The problem involves calculating the radius of the circle of light on the surface of a swimming pool from a small underwater light positioned 1 meter below the surface. The key to solving this is applying Snell's Law to determine the critical angle for total internal reflection, which is found to be 48.6 degrees. Using this angle, the radius of the light circle is calculated as 1.13 meters. The discussion emphasizes the importance of visualizing the scenario through a diagram to aid understanding. Overall, the solution hinges on understanding light behavior at the water's surface.
JSGandora
Messages
92
Reaction score
0

Homework Statement


A small underwater pool light is 1 meter below the surface of a swimming pool. What is the radius of the circle of light on the surface, from which light emerges from the water?

Homework Equations


Snell's Law
Fermat's Principle

The Attempt at a Solution

I have no idea about how to approach the first part because we're not given the radius of the light.
 
Last edited:
Physics news on Phys.org
JSGandora said:
A small underwater pool light is 1 meter below the surface of a swimming pool. I also have no idea how to approach the first part because we're not given the radius of the light.

It is small. Consider it a point source of light. It will illuminate the surface: the light rays cross the water surface and arrive to the eyes of the observer. But not all rays can escape from the water: Find the angle where the rays incident on the surface of the water totally reflect. Make a drawing!

ehild
 
Ohh, thank you so much. It's just the critical angle, or 48.6 degrees, making the radius to be 1/tan(48.6)=1.13m.
 
It is correct.

ehild
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
3K
Replies
14
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
5K
Replies
3
Views
2K
Replies
3
Views
2K