Calculating the Speed of Dolphins Seen Through a Glass Window

AI Thread Summary
The discussion revolves around calculating the apparent speed of a dolphin swimming toward a viewer through a glass window, considering the effects of light refraction. The dolphin's actual speed is 15 m/s, but due to the refraction of light, it appears to move slower. Participants emphasize the importance of understanding apparent depth and how it relates to the viewer's perspective, regardless of whether they are at the same horizontal level as the dolphin. The relevant formula for apparent depth involves the refractive indices of air and water, which helps in determining the apparent speed. Ultimately, the key is to relate the actual depth to the apparent depth to calculate the dolphin's perceived speed accurately.
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Homework Statement


At a marine animal park, Alison is looking through a glass window and watching dolphins swim underwater. If the dolphin is swimming directly toward her at 15 m/s, how fast does the dolphin appear to be moving?


Homework Equations


v=15 m/s
I realize that due to the refraction of light coming off of the dolphin, the dolphin appears to be swimming slower than it actually is.

I am supposed to use the concept of the the refraction of light: Snell's Law, but I'm not really seeing how to approach the problem here.


The Attempt at a Solution

 
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Hint: If the dolphin is a distance X behind the glass, what is the apparent distance behind the glass according to Alison?
 
It's the same? X?
 
sona1177 said:
It's the same? X?
No. If it were the same, then the apparent speed would be the same as the actual speed.

Look up apparent depth.
 
Doc Al said:
No. If it were the same, then the apparent speed would be the same as the actual speed.

Look up apparent depth.

I know that n=1.33 but what does that have to do with apparent depth?
 
sona1177 said:
I know that n=1.33 but what does that have to do with apparent depth?
Did you look it up?
 
Yes. But the girl is not looking at the dolphin from an elevated position. The dolphin is swimming directly at her. I don't see how the depth is changing. I see how depth would be relevant if the girl were on top of a chair looking down, but she is at level with the dolphin.
 
sona1177 said:
Yes. But the girl is not looking at the dolphin from an elevated position. The dolphin is swimming directly at her. I don't see how the depth is changing. I see how depth would be relevant if the girl were on top of a chair looking down, but she is at level with the dolphin.
True, she's at the same horizontal level as the dolphin, but what matters is that the dolphin is behind a depth of water. Whether dolphin-water-girl form a vertical line (if she were looking down from above) or a horizontal line (like described here) doesn't matter. By 'depth' we mean the distance beyond the water surface--that surface could be horizontal or vertical--the light travels the same either way.
 
Doc Al, could you give me a little more to go with here? I need at least a start.
 
  • #10
OK, say she is a distance X behind the glass. If I had the time, I could divide change in X over t, and get velocity. but I know nothing about time, all I know is the speed--15 m/s.
 
  • #11
There is also nothing about apparent depth in my text but doing just from looking at the web, I am getting the formula

Apparent depth= Actual depth(ref index of air/ref index of water)

But we don't know the actual depth here! :(
 
  • #12
sona1177 said:
OK, say she is a distance X behind the glass. If I had the time, I could divide change in X over t, and get velocity. but I know nothing about time, all I know is the speed--15 m/s.
But you know that ΔX/Δt is the actual speed.

sona1177 said:
There is also nothing about apparent depth in my text but doing just from looking at the web, I am getting the formula

Apparent depth= Actual depth(ref index of air/ref index of water)

But we don't know the actual depth here! :(
Good! That formula is all you need.

If the dolphin is a distance X behind the glass, then the actual 'depth' is X. So what's the apparent depth (distance) in terms of X? Then you can compare the actual speed (ΔX/Δt) with the apparent speed.
 
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