Optics, involving index of refraction

[thezac11]

Homework Statement

At what distance below the surface of a lake will a fish have a 180 degree field view above the water surface (if the fish is looking up toward the surface of the lake) ?

-Assume the index of refraction for water to be 1.33 and that of air to be 1.00

Homework Equations

I'm not given any equations for this problem. I'm assuming it has something to do with Snell's Law: (n1)sin(i)=(n2)sin(r) , where n=index of refraction, i=angle of incidence, and r=angle of refraction

The Attempt at a Solution

I have no clue where to start this problem. I feel like I need more information. If anyone could point me in the right direction it would be greatly appreciated.

Thanks,
-Zac

 

[ehild]

Start it with drawing a picture of the problem.

ehild

 

[thezac11]

I have a drawing that shows the viewing area from a point under the water going towards the surface at an unknown angle, then the viewing area widening once it breaks the surface. But I don't understand how to find the depth at which the viewing area is 180 degrees once it breaks the surface.

 

[ehild]

I do not really understand the question. The fish sees everything above water in an angle of view twice the critical angle -no matter at what depth it is. You can calculate the angle from Snell's law. Only the diameter of the circle within everything above water can be seen would change with depth.

ehild

 

[NRock]

This is possibly a weird way to solve the problem, but here it is: I assuming what the problem means is at what distance will the fish see a panoramic 180 degree field of vision (I wasn't quite sure). If your interface is oriented along the y-axis (the surface of the water is parallel to the y-axis) then the widest rays this fish will see will be coming in at theta = 90 and negative 90 degrees. These rays will bend following snell's law at the surface of the water and intersect at a specific point. You're fish needs to be at that point.