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

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Homework Help Overview

The discussion revolves around determining the critical angle for light emerging from a contaminated lake, specifically focusing on the geometry involved in calculating the area of the largest circle on the lake's surface through which light can escape.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the concept of critical angle and its calculation, with some attempting to apply trigonometric relationships to find the radius and area of the light-emitting circle. Questions arise regarding the correct application of trigonometric functions and the interpretation of angles in the context of the problem.

Discussion Status

There is an ongoing exploration of different approaches to the problem, with participants providing calculations and corrections to each other's methods. Some guidance is offered regarding the importance of sketching the scenario to clarify the angles involved, indicating a productive direction in the discussion.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information available and the methods they can use. There is a focus on ensuring that calculations are accurate and that assumptions about angles are correctly understood.

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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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Lookup 'critical angle'
 
ok:
sin(\theta)_c=\frac{n_air}{n_water}
\theta_c= 51.37
I use some trig to find the radius which is:
r=hcos(\theta_c)
And I can find the area from there, correct?
 
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.
 
oops.
that r=hcos(\theta_c) should be a r=htan(\theta_c)
 
I get 90.48 m^2 but it is incorrect
 
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)
 
Funny calc mistake,
but i got 62.36 m^2 and that is he correct answer
 
oops - typed it out wrong! Always check your arithmatic!
 

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