Department of Energy Argonne National Laboratory Office of Science NEWTON's Homepage NEWTON's Homepage
NEWTON, Ask A Scientist!
NEWTON Home Page NEWTON Teachers Visit Our Archives Ask A Question How To Ask A Question Question of the Week Our Expert Scientists Volunteer at NEWTON! Frequently Asked Questions Referencing NEWTON About NEWTON About Ask A Scientist Education At Argonne Fish Size in Water
Name: Laura L.
Status: other
Age: 40s
Location: N/A
Country: N/A
Date: 7/17/2003


Question:
I was kayaking the other day and saw a rather large fish in no more than 2 1/2 ft of water. My question is would the water magnify the size of the fish or are things only magnified underwater while wearing a mask. If the water does magnify things at that depth how much larger would they appear. I do not want to be telling any untrue fish tales!


Replies:
Interesting question: the usual question is how deep does an object underwater appear. (the answer is 3/4 of its actual depth; an object four feet underwater appears to be 3 feet underwater, using the index of refraction of water to be 4/3).

To answer your question, though, look straight down at one end of the fish from a height h above the water and at an angle a with a line perpendicular to the surface of the water at the other end of the fish, which is at a depth d underwater. A ray of light refracts at the water's surface to an angle b with the perpendicular where sin a = n sin b, where n=4/3 is the index of refraction of the water. If you know a little trigonometry and draw a diagram as I described, I am sure you can obtain:

w = (h + d) tan a,
s = h tan a + d tan b,
m = w/s = ((h+d) tan a)/(h tan a + d tan b)

Here w is the apparent length of the fish, s is the actual length and m is the magnification. Using the approximation sin a = a (accurate at small angles, but good to 2% even at 20 degrees):

m = n(h+d)/(nh+d)

Notice that if the fish is at the surface (d=0), the magnification is 1 and the fish is seen unmagnified.

At d = h, m = 2n/(n+1).

Using n = 4/3, this gives m=8/7.

But wait!!! The fish looks closer to the surface than it is and so you observe the fish to be 3/4 * 8/7 = 6/7 of its actual size if you correct for it seeming to be closer and therefore larger than it actually is! At any rate, it looks somewhere between 6/7 and 8/7 of its actual size, so you are not exaggerating that much (unless you are truly exaggerating!)

I hope this is clear, but I certainly find it confusing... If you would like more words from me, feel free to write again.

Best, Dick Plano, Professor of Physics emeritus, Rutgers University



Click here to return to the Physics Archives

NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.

For assistance with NEWTON contact a System Operator (help@newton.dep.anl.gov), or at Argonne's Educational Programs

NEWTON AND ASK A SCIENTIST
Educational Programs
Building 360
9700 S. Cass Ave.
Argonne, Illinois
60439-4845, USA
Update: June 2012
Weclome To Newton

Argonne National Laboratory