How Many Reflections Appear Between Angled Mirrors?

  • Thread starter Thread starter fliptomato
  • Start date Start date
  • Tags Tags
    Images Mirrors
AI Thread Summary
The discussion revolves around determining how many images are visible when standing between two angled mirrors. The key factors influencing the number of reflections include the angle between the mirrors (\phi) and the distance between them (D). A recursive formula, d(\theta, n), has been developed to calculate the distance from the original position to the nth image based on the angle of reflection. To find the total number of visible images, one must identify the value of n that makes d(\theta, n) equal to the length of the mirrors (L). This approach helps clarify the relationship between the angle, the number of reflections, and the visibility of images.
fliptomato
Messages
78
Reaction score
0
Hi everyone--I've been mulling over a homework question from earlier in the term:

Suppose you are standinig in the middle of two mirrors of length L separated by a distance D. ((i.e. if you are standing at the origin, the centers of the two mirrors are at x=\pm \L/2)) The mirrors are at a relative angle \phi. ((i.e. if the mirror behind you was in the yz plane, then the mirror in front of you will be rotated by \phi in the xy plane))

The question is: how many images of yourself do you see in the mirror?

This seems like a purely geometry problem, given the constraints of L and D, how many admissible angles \theta (in the xy plane) will produce a ray that bounces around a certain number of times and ends up back at the starting point (with the added trivial subtlety that it must hit the front of your face, not the back of your head).

Anyway, I think I've got it down to a formula for d(\theta, n) that is recursive in n, where \theta is the angle at which you are looking at n is the number of times (forward and back) that the ray is reflected. The problem is that this isn't particularly helpful to answer the question (how many images)!

The assigned problem gave the numerical values L, D = 1 meter, and \phi = 1 degree.

I have a copy of the solution for this case, but I can't even make heads or tails of it! (Indeed I'm skeptical if this is even the correct solution for this problem.)

I've posted this at:
http://www.stanford.edu/~flipt/upload/number3.pdf

Any feedback about my approach or a rough summary of what's going on in that solution set would be appreciated! (I'll take down the solution promptly after some discussion since it's not a good idea to leave such solutions online for next year's class!)

-Flip
 
Last edited by a moderator:
Physics news on Phys.org


Hi Flip,

First of all, great job on working through this problem and coming up with a recursive formula for d(\theta, n)! It shows that you have a good understanding of the geometry involved in this scenario.

To answer the question of how many images you would see in the mirror, let's take a step back and think about the concept of reflections. When you stand between two mirrors, you are essentially creating multiple reflections of yourself. The first reflection is in the front mirror, and the subsequent reflections are between the two mirrors. Each reflection creates a new image of yourself, and the number of images you see will depend on the angle between the two mirrors (\phi) and the number of times the ray bounces between the mirrors (n).

Now, let's consider the recursive formula you came up with, d(\theta, n). This formula represents the distance between the original point (where you are standing) and the nth image. So, if we know the value of d(\theta, n) for a certain angle \theta and number of reflections n, we can calculate the distance between each image and the original point.

To answer the question of how many images you see, we need to find the value of n that makes d(\theta, n) equal to the length of the mirror (L). This is because when the distance between the original point and the nth image is equal to the length of the mirror, the nth image will overlap with the original point and you won't be able to see it.

So, in summary, the number of images you see will depend on the angle between the mirrors (\phi) and the number of times the ray bounces between the mirrors (n). To find the value of n, you can use your recursive formula and set d(\theta, n) equal to the length of the mirror (L).

I hope this helps clarify things for you. Keep up the good work!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top