Preparing for Optics Exam: Solving the Refraction of Light in Water Problem

[ChEJosh]

Hello, first time poster. I'm in an optics class because I thought it would be fun, but I'm having a terrible time. And I didn't do so well on the first test. The second test is tomorrow and I want to do well on it. We were given a practice exam with 3 questions. I got two of them, but the third is still eluding me. Thank you for any guidance.

Homework Statement

A small fish, 40cm below the surface of a lake, is viewed through a thin converging lens with focal length 20cm. the lens is located 30cm above the water. The refractive index of water is 1.33 and air is 1. Assume that the fish lies on the optical axis of the lens.

Homework Equations

The only equation that I know of is the various forms of (1/so)+(1/si)=(1/f)

The Attempt at a Solution

I followed the advice of someone else to draw a ray diagram and use Snell's law at the interface of air/water, but that got me no where. I don't know what the angles of the rays are to use Snell's law and even if I did, I'm not sure how that would help me.

 

[Doc Al]

I presume the problem is to locate the image of the fish. First thing to do is to ignore the lens (for the moment) and determine the apparent position (depth) of the fish below the water surface. Do that by analyzing a few rays from the fish as they refract at the surface.

Once you find the apparent position of the fish due to refraction at the water surface, then you can plug into the thin lens equation.

 

[ChEJosh]

Thank you for the reply. It did ask to locate the image of the fish, sorry.

I don't understand how to determine the apparent depth. We were never given anything like this problem in lecture or in homework. Could you explain that a little more in depth please?

 

[Doc Al]

Sure. Make yourself a diagram of the fish at a distance D under the water surface. Draw a vertical line from the fish upward. Draw a ray of light from the fish (assume it's a point source) that makes some angle with the vertical (not just straight up). Show how the ray refracts as it traverses the boundary. Use Snell's law to find its new angle in the air. Then trace back the refracted ray until it hits the vertical line. Use a bit of trig to figure out at what distance below the surface it hits the line. That's the apparent depth of the fish as seen from the air. (If you did this with several rays at different angles, they would all hit that vertical line at the same point.)

 

[ChEJosh]

Thank you again for the reply. A couple clarifications to make sure I understand.
So the ray I draw from the fish is arbitrary and I just choose a random angle, and use that angle in Snell's law to calculate the refracted angle?
Also, which line is the apparent depth? The entire line made from the fish to the point where the refracted ray crosses the vertical? If that's the case then it takes into account the actual distance of 40cm in water and the 30cm in air (in other words the entire length from the lens)?
And then, I plug that in for "so" in the (1/so) term of the lens equation?

 

[Doc Al]

So the ray I draw from the fish is arbitrary and I just choose a random angle, and use that angle in Snell's law to calculate the refracted angle?

Yes.

Also, which line is the apparent depth? The entire line made from the fish to the point where the refracted ray crosses?

The actual depth is D, the real distance of fish to surface. The apparent position of the fish is where the refracted ray crosses the vertical line. So the apparent depth is the distance from that crossing point and the surface. (Just like all the rays from the fish emanate from the same point; all the refracted rays trace back to the same point.)

If that's the case then it takes into account the actual distance of 40cm in water and the 30cm in air?

No. We're just finding the apparent position of the fish, as seen by us in the air. Once we find that point, we can ignore the water and pretend the fish is really at its apparent position (at least for the purpose of locating its final image). Solve for the apparent depth (D') in terms of D.

And then, I plug that in for "so" in the (1/so) term of the lens equation?

"so" would be the total distance between the apparent location of the fish and the lens, not just the apparent depth.

If you are familiar with the equations for refraction at curved surface (which is the starting point for deriving the lens equation) you can also treat this problem by treating the water surface as a refracting surface of infinite radius. First we find the image of the fish created by the water/air boundary. Then we use that image as the source for analyzing what the lens does.

 

[ChEJosh]

Are you referring to (n1/so)+(n2/si)=(n2-n1)/R?
My initial thought about this problem was to use that equation, but I wasn't sure what to do about R. I didn't think about setting it to infinity.

Thank you.

 

[Doc Al]

Are you referring to (n1/so)+(n2/si)=(n2-n1)/R?

Exactly!

My initial thought about this problem was to use that equation, but I wasn't sure what to do about R. I didn't think about setting it to infinity.

If you're comfortable using that equation, that's definitely the easy way to account for the refraction at the surface.

But just for the heck of it, try the other method I outlined above. You'll get the same answer. (It's a lot easier than it sounds, once you draw the diagram.)

 

[ChEJosh]

I did indeed get the same answer; although, it took a few tries with that apparent depth method.
Thank you very much.