- #1
SaltMiner
- 6
- 3
- Homework Statement
- You are spearfishing in waist-deep water when you spot a fish that appears to be 45° below the water and about 50cm underneath the surface. You recognize that the light coming from the fish to your eye has been refracted and you must therefore aim at some depth below where the fish appears to be. How far below where the fish appears to be should you aim?
- Relevant Equations
- n air = 1
n water = 1.35
angle 1 = 45 degrees
assume a flat interface
This is a question on a past exam at university. The answer was provided (for revision purposes and exam preparation) but I never understood it and it continues to frustrate me because even if I can't come up with the right answer to a problem, I'll at least 'get' (understand) the proper answer when it is given. But not this one.
There were three steps to solving this problem as presented:
1. First we find the angle of refraction using Snell's law which turns out to be 31.58 degrees. But then we take from 90 degrees this value to get 58.42 degrees.
2. We then find what the y value is by using some simple trig (tan(degrees) = O/A) and then solving for A which is our y value in the diagram. This turns out to be 50cm.
3. Lastly we use the same formula as in step 2 to find what the x value is. x = 50tan(58.42degrees).
Now I get each part of this answer, I just don't see how it is correct. This is because the y value here doesn't have to do with the actual fish's position, only its apparent position.
If we knew either how far down or away the fish actually was, then it would be easy to find the missing value because we also know the angle of refraction. But we know neither. In my university textbook, it even uses a similar diagram of a fish underwater showing that the fish is further AWAY and DOWN from its apparent position.
So even though its true that at 50cm away and at an angle of 58.42 degrees, the x value would be 81.3cm, this doesn't actually have to do with the fish's real position. What am I just not getting here? I hope this has made some sense.
There were three steps to solving this problem as presented:
1. First we find the angle of refraction using Snell's law which turns out to be 31.58 degrees. But then we take from 90 degrees this value to get 58.42 degrees.
2. We then find what the y value is by using some simple trig (tan(degrees) = O/A) and then solving for A which is our y value in the diagram. This turns out to be 50cm.
3. Lastly we use the same formula as in step 2 to find what the x value is. x = 50tan(58.42degrees).
Now I get each part of this answer, I just don't see how it is correct. This is because the y value here doesn't have to do with the actual fish's position, only its apparent position.
If we knew either how far down or away the fish actually was, then it would be easy to find the missing value because we also know the angle of refraction. But we know neither. In my university textbook, it even uses a similar diagram of a fish underwater showing that the fish is further AWAY and DOWN from its apparent position.
So even though its true that at 50cm away and at an angle of 58.42 degrees, the x value would be 81.3cm, this doesn't actually have to do with the fish's real position. What am I just not getting here? I hope this has made some sense.