Setup help needed:apparent depth vs. actual depth

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In summary, 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. 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.
  • #1
RoseCrye
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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!
 
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  • #2
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... o_O) 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
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
mgb_phys said:
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.

but if you go from the standpoint that the observer is @ the side of the pool, how do you find the apparent depth?
 
  • #5
RoseCrye said:
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... o_O) 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?

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
mgb_phys said:
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.

WAITAMINNIT...
sorry, that didn't register... thnx.
 
  • #7
mgb_phys said:
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.

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

This answer was deemed incorrect.

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
RoseCrye said:
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.

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
hage567 said:
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.

what are you mean? Put it in Snell's Law?
 
  • #10
Well yes, for the special case of a critical angle.
 
  • #11
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
You know that n=1.466 for the glass, and n=1.33 for the water (looking it up).
 
  • #13
How do you set up snell's law for a critical angle?! i thought snell's law didn't work with critical angles...
 
  • #15
theta1 with which n, glass or water?
 
  • #16
GOT IT! Thanks, hage!
 
  • #17
got some insight into the other problem of mine?
 
  • #18
mgb_phys said:
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.

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.
 

1. What is the difference between apparent depth and actual depth?

Apparent depth refers to how deep an object appears to be when viewed from a certain angle or through a medium, such as water. Actual depth refers to the true depth of an object, regardless of how it appears.

2. How does the angle of viewing affect apparent depth?

The angle of viewing can affect apparent depth because it changes the path of light rays as they pass through a medium, causing objects to appear closer or farther away than they actually are.

3. Can you explain the concept of refraction and its role in apparent depth?

Refraction is the bending of light as it passes through a medium with a different density, such as water or air. This bending of light causes objects to appear to be in a different position than they actually are, which is why apparent depth is affected by refraction.

4. How can we measure the difference between apparent depth and actual depth?

To measure the difference between apparent depth and actual depth, we can use the principles of trigonometry to calculate the angle of refraction and the index of refraction of the medium. These measurements can then be used to determine the actual depth of the object.

5. What are some real-world applications of understanding apparent depth vs. actual depth?

Understanding apparent depth vs. actual depth is important in various scientific fields, such as oceanography and optics. It also has practical applications, such as correcting for the apparent depth of objects when designing and building structures that are partially submerged in water, like bridges or dams.

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