Calculate the apparent distance between the image of the fish and the front wall

Question

The back wall of a home aquarium is a mirror that is a distance of 40.0 away from the front wall. The walls of the tank are negligibly thin. A fish, swimming midway between the front and back walls, is being viewed by a person looking through the front wall. The index of refraction of air is = 1.000 and that of water is = 1.333. (a) Calculate the apparent distance between the fish and the front wall.
(b) Calculate the apparent distance between the image of the fish and the front wall.
I understand and got the answer to part A- 15cm, but I don't know how to solve part b. I'm not sure to what image it is referring to either. Is it the refracted image from when looking through the aquarium, or is it the image made by the mirror of the back wall? My professors answer is 45cm. Please explain the steps to take.

Answer 1

Hi Olivia!

 

I think Part b is referring to the image of the fish in the mirror that is the back wall of the tank.   From outside the tank, that can be considered another "object."

 

You have already correctly shown that if an object is a certain distance away from the wall of the tank on the inside, it will appear, looking directly at it (along the normal line) from the outside, to be closer to the wall by a factor of the ratio of n_out/n_in = n_air/n_water = 1/(4/3) = 3/4.  So, if you multiply the actual distance of an object inside the tank by that factor, you get its apparent distance from the front wall, as seen by someone looking at it directly from the outside.

 

The image of the fish in the back wall mirror is just another object (making the assumption that someone looking in directly along the normal line can see it around the actual fish in the tank, which is in between the viewer and the mirror image; we can assume the viewer is just enough off-line to be able to see the image around the fish without changing this result).  The main question is, what is the actual distance from the mirror image to the front wall?  If we know that, we just multiply it by the 3/4 factor you already solved for (or just repeat whichever technique you used to correctly solve Part (a)) to determine the apparent distance from the front wall to the mirror image of the fish, as viewed directly from the outside.

 

The key to calculating the actual distance of the mirror image of the fish from the front wall is realizing that , even though water is in front of it, the mirror works exactly the same as it would in air.  Assuming the back wall is a plane mirror (since there is no information given about curvature), the only questions to ask are:  what is the distance from the fish to the mirror? and how far "behind" the plane mirror is the virtual image of the fish, given that fish-to-mirror distance?

 

Then

 

[the distance from the mirror image of the fish to the mirror] +

[the distance from the mirror to the actual fish in the tank] +

[the distance from the fish in the tank to the front wall] = [the "actual" distance from the mirror image to the front wall]

 

Then operate on that actual distance the same way you did in correctly solving Part (a) (or just multiply it by the 3/4 factor we talked about above), and you should get your professor's answer.

 

If you have any more questions or want to talk about it further, just let me know!