# Homework Help: Setup help needed:apparent depth vs. actual depth

1. Apr 25, 2007

### RoseCrye

there's a light source, S, sitting 2.60 metres below the surface of a pool 1.00 metres away from one side. i'm supposed to find the angle at which the light left the water and the difference between the apparent and actual depths of the light source.

i attempted using Snell's Law with n of water = 1.33 and the incedent angle = 90 degrees. 1.33(sin90)=1.00(sin_theta) I got theta=1.33. that can't be right.

I also tried trig. Theta is the angle of a rt triangle with the light source. 2.60 is the opposite, 1.00 is adjacent. Using the cot_theta, my answer was 69 degrees. neither of these answers was accepted by WebAssign.

HELP!!!

2. Apr 25, 2007

### RoseCrye

Also, there is a question as to how to find the critical angle of a "special glass" (the index of refraction of which I am not given... ) when submerged in water. Hereis the question:

"The critical angle for a special glass in air is 43°. What is the critical angle if the glass is immersed in water?"

I tried to use sin_theta=1/n, but I couldn't figure out how to set up the problem. Someone have an idea?

3. Apr 25, 2007

### mgb_phys

It depends on where you are looking from.
If you are directly above the light then, as you said i=90deg and the light goes straight out - the apparent depth is then just real_depth/n.

If you are standing on the side of the pool then just draw a triangle, and the real depth is the hypotenuse.

4. Apr 25, 2007

### RoseCrye

but if you go from the standpoint that the observer is @ the side of the pool, how do you find the apparent depth?

5. Apr 25, 2007

### RoseCrye

By the way, i worked it out, the index of refraction for the glass is 1.466, if that helps anyone... i just need at setup for this, I'm not asking for a solution.

6. Apr 25, 2007

### RoseCrye

WAITAMINNIT....
sorry, that didn't register... thnx.

7. Apr 25, 2007

### RoseCrye

this didn't work.

what i did:
Pythag. theorem => hyp=2.78m
D(apparant)=D(real)/n
D(apparant)=2.78/1.33
D(apparant)=2.09
Difference = .69m

Say that ang.B=rt ang. Ang.A=observer's eyes to S (the angle that the light meets the water/air boundary to the normal inside the water). Ang.C=S to ang.B

tanA=2.6/1
tanA=2.6
tan-1A= 68.96

Should i input this angle into Snell's Law?

8. Apr 25, 2007

### hage567

Now that you have n for the glass, just try applying the same reasoning now that it is in the water instead of the air, and solve for theta.

9. Apr 25, 2007

### RoseCrye

whatcha mean? Put it in Snell's Law?

10. Apr 25, 2007

### hage567

Well yes, for the special case of a critical angle.

11. Apr 25, 2007

### RoseCrye

Tried pitting it in Snell's, hage, but it didn't work... I came out with 38.22 degrees, and used 43 degrees as the incedent angle.

how do i find the crit angle in the water? if i use the glass' n, i'd only get 43 degrees again. i think i need the index of refraction at the water/glass boundary. but i am so confused....

12. Apr 25, 2007

### hage567

You know that n=1.466 for the glass, and n=1.33 for the water (looking it up).

13. Apr 25, 2007

### RoseCrye

How do you set up snell's law for a critical angle?! i thought snell's law didn't work with critical angles...

14. Apr 25, 2007

### hage567

15. Apr 25, 2007

### RoseCrye

theta1 with which n, glass or water?

16. Apr 25, 2007

### RoseCrye

GOT IT! Thanks, hage!

17. Apr 25, 2007

### RoseCrye

got some insight into the other problem of mine?

18. Apr 25, 2007

### RoseCrye

to find the apparent depth, real depth over the n of what? tried it with the n of the water, and it's telling me it's wrong.