How do you calculate the location of the third virtual image?

AI Thread Summary
The discussion centers on calculating the location of virtual images created by two mirrors meeting at a corner, with a red ball positioned at (-1m, -2m). It is established that there are at least two images: one at (-1m, 2m) and another at (1m, -2m). The third image is identified at (1m, 2m), although the method for determining this is unclear to the original poster. The conversation emphasizes the importance of understanding angles of incidence and the concept of extending mirrors to visualize reflections. Overall, the topic highlights the complexities of image formation in reflective surfaces.
aleksbooker
Messages
22
Reaction score
0

Homework Statement



Two 3.0m wide mirrors meet at a corner. Taking the corner as the origin of the x/y axis, A red ball is placed at point A (-1m, -2m).

1) How many images are seen by an observer at point O? [Point O is not given coordinates, but looks to be at approximately (-3m, -3m)].

2) What are the (x,y) coordinates of each image?

Homework Equations



The Attempt at a Solution



I figured there were at least two virtual images, one behind each mirror at a point perpendicular to the red ball. So there's an image directly north of the ball (-1m, 2m) and directly east of the ball (1m, -2m).

Apparently, the third image is located at (1m, 2m) and I have no idea why or how I would have found that. I know that light from the ball strikes both mirrors before reaching the observer, and that the angles of incidence are all the same, but I don't know how to calculate *what* the angle of incidence is! :(
 
Physics news on Phys.org
Each mirror reflects everything in front of it, i.e. everything that you would see if you were placed in the mirror looking out. In that position, how many images would you see?
 
Should I then mentally extend the top mirror beyond the corner so that it "sees" the reflection from the bottom right mirror?
 
aleksbooker said:
Should I then mentally extend the top mirror beyond the corner so that it "sees" the reflection from the bottom right mirror?
Yes, but that's usual. An object placed in front of a mirror, but not directly in front (i.e. off to the side a bit) still produces an image in the mirror. It's just that you have to stand off to the other side to see it.
 
Cool. Thanks for confirming/clarifying. :)
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top