What is the Critical Angle for Light Passing Through a Contaminated Lake?

In summary, the problem involves finding the area of the largest circle on the surface of Lake Ontario through which light can emerge from a point source located 3.56 m below the surface. By using the critical angle formula and trigonometry, the radius of the circle is found to be tan(theta) * depth, and the area is calculated to be approximately 62.36 m^2.
  • #1
Winzer
598
0

Homework Statement



A point source of light is located 3.56 m below the surface of a large lake of clear, but contaminated, water (Lake Ontario, where n = 1.28). Find the area of the largest circle on the lake's surface through which light coming from the source can emerge into the air.


Homework Equations





The Attempt at a Solution


Not sure where to start
 
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  • #2
Lookup 'critical angle'
 
  • #3
ok:
[tex] sin(\theta)_c=\frac{n_air}{n_water}[/tex]
[tex] \theta_c= 51.37[/tex]
I use some trig to find the radius which is:
[tex] r=hcos(\theta_c)[/tex]
And I can find the area from there, correct?
 
  • #4
Yes - but draw a sketch just to make sure you have the angle the right way round - it's always tricky when it's near 45deg.
 
  • #5
oops.
that [tex] r=hcos(\theta_c)[/tex] should be a [tex] r=htan(\theta_c)[/tex]
 
  • #6
I get 90.48 m^2 but it is incorrect
 
  • #7
Theta is the angle between the ray that would just exit and the normal to the surface.
So by similair triangles it is also the internal angle between the ray and a line straight up from the surface to the source.

So the radius of the patch on the surface is tan(theta) * depth. The area is then of course pi r^2.

(I get 65.5m^2)
 
  • #8
Funny calc mistake,
but i got 62.36 m^2 and that is he correct answer
 
  • #9
oops - typed it out wrong! Always check your arithmatic!
 

Related to What is the Critical Angle for Light Passing Through a Contaminated Lake?

1. What is the critical angle in optics?

The critical angle in optics refers to the angle at which light passing through a boundary between two different materials is no longer refracted, but instead is reflected back into the original material. This angle is dependent on the refractive indices of the two materials and can be calculated using Snell's law.

2. How is the critical angle used in practical applications?

The critical angle is used in a variety of practical applications, such as in fiber optics, where it is important to understand the angle at which light will be reflected back into the fiber. It is also used in the design of lenses and prisms, as well as in determining the maximum angle at which total internal reflection can occur in optical devices.

3. What factors affect the critical angle in optics?

The critical angle is affected by several factors, including the refractive indices of the materials involved, the angle of incidence of the light, and the polarization of the light. It is also influenced by external factors such as temperature and pressure.

4. How is the critical angle related to total internal reflection?

The critical angle is directly related to total internal reflection, as it is the angle at which light will be completely reflected back into the original material. Total internal reflection occurs when the angle of incidence is greater than the critical angle, resulting in no light passing through the boundary between the two materials.

5. Can the critical angle be manipulated or controlled?

Yes, the critical angle can be manipulated or controlled by changing the refractive indices of the materials, the angle of incidence, or the polarization of the light. This can be useful in practical applications where the angle of refraction needs to be controlled, such as in the design of lenses and prisms.

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