Critical angle optics problem

1. Nov 24, 2007

Winzer

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. The attempt at a solution
Not sure where to start

2. Nov 24, 2007

mgb_phys

Lookup 'critical angle'

3. Nov 24, 2007

Winzer

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?

4. Nov 24, 2007

mgb_phys

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. Nov 24, 2007

Winzer

oops.
that $$r=hcos(\theta_c)$$ should be a $$r=htan(\theta_c)$$

6. Nov 24, 2007

Winzer

I get 90.48 m^2 but it is incorrect

7. Nov 24, 2007

mgb_phys

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. Nov 24, 2007

Winzer

Funny calc mistake,
but i got 62.36 m^2 and that is he correct answer

9. Nov 24, 2007

mgb_phys

oops - typed it out wrong! Always check your arithmatic!