Help with critical angle calculation

In summary, the problem involves determining the maximum distance from the edge of the pool that the penny can be and still be visible to the observer, taking into account the refractive index of water. The solution involves calculating the apparent depth of the penny and using Snell's law to find the critical angle. By considering the light ray coming from the observer's eye, the distance can be determined and then used to solve the next part of the problem.
  • #1
shar_p
14
0

Homework Statement


A penny sits at the bottom of a pool of water (n=1.33) at a depth of 3.0m. If the observer 1.8m tall stands 30cm away from the ledge, how close to the side can the penny be and still be visible to the observer. Suppose there is another penny 10 times farther away than the 1st one, will a light ray going from this new penny to the top edge of the pool emerge from the water ? And if yes, what is the angle made by the light ray?


Homework Equations


To still be visible, the angle should be less than critical angle, is that correct? but I can't figure this out .. please help


The Attempt at a Solution


real depth = 3.0 so apparent depth can be calculated to be 3/1.33 = 2.25 m.
I can also calculate the critical angle with Snell's sin(theta) = 1.33 (since sin 90 = 1). So theta = 48.75 . Which means that the angle has to be less that 48.75 for the penny to be visible.
Now how do I get the distance from this?

Please help I have a test on Wednesday. Thanks
 
Last edited:
Physics news on Phys.org
  • #2
Why not turn the problem around? Imagine a light ray coming from the observer's eye, grazing the edge of the pool, and going down into the water. Where would it reach the bottom?

Now you've got that distance sorted out you'll be able to attack the next part of the problem.
 
  • #3

To calculate the critical angle, you can use the formula sin(theta) = 1/n, where n is the refractive index of the medium (in this case, water with a refractive index of 1.33). This will give you a critical angle of approximately 48.75 degrees, as you have correctly calculated.

To determine the distance from the edge of the pool that the penny can be and still be visible, you can use the trigonometric relationship tan(theta) = opposite/adjacent. In this case, the adjacent side is the distance from the edge of the pool to the observer (30cm), and the opposite side is the distance from the edge of the pool to the penny (x). So, you can rearrange the equation to solve for x: x = tan(theta)*30cm. Plugging in the critical angle of 48.75 degrees, you get x = tan(48.75)*30cm = 34.28cm. This means that the penny can be up to 34.28cm away from the edge of the pool and still be visible to the observer.

As for the second part of the question, yes, the light ray from the second penny will emerge from the water. This is because the light ray will be traveling at an angle less than the critical angle, so it will not undergo total internal reflection and will emerge from the water. To calculate the angle made by the light ray, you can use the same formula as before, but with the new distance from the edge of the pool (10 times farther away than the first penny). So, the angle made by the light ray will be tan^-1(10*tan(48.75)) = 84.96 degrees.

I hope this helps! Good luck on your test.
 

Related to Help with critical angle calculation

1. What is the critical angle?

The critical angle is the angle of incidence at which the refracted ray of light travels along the surface of the medium, instead of passing through it.

2. How is the critical angle calculated?

The critical angle can be calculated using Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the refractive indices of the two media.

3. What factors affect the critical angle?

The critical angle is affected by the refractive indices of the two media, as well as the angle of incidence and the wavelength of the incident light.

4. Why is the critical angle important?

The critical angle is important because it determines whether light will be refracted or totally internally reflected when passing through different media. This phenomenon is used in many applications, such as fiber optics and prism-based devices.

5. Can the critical angle be greater than 90 degrees?

Yes, the critical angle can be greater than 90 degrees in certain situations where the refractive index of the second medium is greater than the first medium. This results in total internal reflection, where light is completely reflected back into the first medium.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
913
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
5
Replies
157
Views
7K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
Back
Top