Images from Two Parallel Mirrors problem

[jiten827]

Homework Statement

Two parallel mirrors that face each other are placed along and perpendicular to the x-axis at x = +11.00 m and x = -11.00 m. Assume that a point source of light is placed on the axis at x = +2.70 m. In principle, there will be an infinite number of images generated. Consider the locations of the four images of the point source with the smallest absolute values of image distance i. Going from left to right (i.e from negative to positive), what is the x-coordinate of the left most image?

2. What is the x-coordinate of the next image?

3. What is the x-coordinate of the next image?

4. What is the x-coordinate of the right most image?


I really do not know where to start, so any help appreciated

 

[learningphysics]

Draw a sketch... first draw the reflections of the point source in the two mirrors... what are the coordinates of these two images. I'll call the I1 (reflection in mirror 1) and I2 (reflection in mirror 2). Now I1 and I2 also have reflections... I1 has a reflection in mirror 2... call it I12. you can get the coordinates of this image because you have the coordinates of I1, and therefore you know its distance from mirror 2... so I12 is located on the other side of mirror 2 at the same distance.

And I2 has a reflection in mirror 1... call it I21. You can get I21's coordinates in a similar manner as you got I12's coordinates.

that's 4 images... I1, I2, I12 and I21.

 

[ogb p]

I would approach this problem by first understanding the concept of multiple reflections between parallel mirrors. This means that light from the source will bounce back and forth between the two mirrors, creating an infinite number of images.

To answer the questions, I would use the formula for image distance, which states that the distance between the source and an image is equal to the distance between the two mirrors divided by the number of reflections (n). In this case, n can be any positive integer, as there is no limit to the number of reflections.

1. The leftmost image will have the smallest absolute value of image distance, which means it will be the closest to the source. Using the formula, we can calculate the image distance as (11-2.7)/n. Since n can be any positive integer, we can assume n=1, giving us an image distance of 8.3m. Since the source is at x=2.7m, the leftmost image will have an x-coordinate of 2.7-8.3 = -5.6m.

2. The next image will have an image distance of (11-2.7)/n = 8.3m. Assuming n=2, the x-coordinate of this image will be 2.7-2*8.3 = -13.3m. However, this is not a real image as it is outside the two mirrors. The next real image will have an image distance of (11-2.7)/n = 8.3m, and assuming n=3, the x-coordinate will be 2.7-3*8.3 = -21.3m.

3. The next image will have an image distance of (11-2.7)/n = 8.3m. Assuming n=4, the x-coordinate of this image will be 2.7-4*8.3 = -29.3m.

4. The rightmost image will have an image distance of (11-2.7)/n = 8.3m. Assuming n=0, the x-coordinate of this image will be 2.7-0*8.3 = 2.7m. However, this is not a real image as it is outside the two mirrors. The next real image will have an image distance of (11-2.7)/n = 8.3