1. Nov 26, 2011

### John982

Robert is on a ship looking at a school of fish below the lake surface. He estimates the apparent depth of the school to be two meters, and that his observation angle is forty five degrees. Calculate the real depth of the school.
Here is what I think we know:

θr=45°, θi=?, ni=1.00, nr=1.33
Because the index of refraction goes from a lower value to a higher one the light will bend towards the normal which I drew in the diagram.

So if we want to find θi would it be sinθi=((sin45°)(1.33))/(1.00)? I'm a bit lost and would really appreciate some help.

Last edited: Nov 26, 2011
2. Nov 26, 2011

### Staff: Mentor

Close. Snell's law says: $\frac{sin(\theta 1)}{sin(\theta 2)} = \frac{n2}{n1}$.

In this case $\theta 1 = 45°~,~n1 = 1.00~,~n2 = 1.33$. It looks like you've got your n1 and n2 swapped.

3. Nov 26, 2011

### John982

Then based on that it would be: sinθ2= (1.00/1.33)(sin45) -> θ2=32.1°. Now that I have the angle I can use tanθ=opp/adj correct? But how do I know what the opposite side length is?

4. Nov 26, 2011

### Staff: Mentor

If you know θ2 (which is θi on your diagram) then you can calculate the angle $\phi$ between the horizontal water surface and the light ray. What triangle side length is the same regardless of the depth?

#### Attached Files:

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5. Nov 26, 2011

### John982

The horizontal side x does not change. Could you find the angle by subtracting θ2 from 90° which would make ϕ 58°?

6. Nov 26, 2011

### Staff: Mentor

That's the idea!

7. Nov 26, 2011

### John982

So could I do the same thing to find x by subtracting 45° from 90° which is 45° and if the length of the apparent depth is 2 meters then x must also be 2 because 45-45-90 triangles have legs of equal length?

8. Nov 26, 2011

### Staff: Mentor

Absolutely!

9. Nov 26, 2011

### John982

our real depth then is 2/Tan32.1=3.19 m. Thank you so much for your help.