# Homework Help: Optics : How many images can be seen in inclined mirrors?

1. Jul 8, 2017

### Jahnavi

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

I tried to draw a rough figure depicting the situation.

Here M is the mosquito and I1, I2, I3 are its images .

Mosquito can clearly see it's image I1 .Now whether it can see I2and I3 is something I am not too sure .

For that I first considered image I2 .M can only see I2 if ray PM is such that it has been originated at M and undergone two reflections at mirrors R1 and R2.Since the mosquito is very close to R1 and mirrors are inclined at an obtuse angle, we cannot draw such multiple reflected rays .

Similar reasoning lies for image I3 . So I think option B) is correct.

Is my reasoning correct ?

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Last edited: Jul 8, 2017
2. Jul 8, 2017

### Staff: Mentor

I think your reasoning is OK, but there's an easier way to consider images with plane mirrors. Think of each mirror as a window into a parallel world where the images lie. Those images are exactly behind those mirrors, along a line normal to the surface. So, assuming I2 is the image of M in mirror R2, then ask: Can M look through that "window" and see I2? No.

Similarly, if I3 is the reflection of I1 in mirror R2, then can M see it by looking through the "window" of mirror R2? Again, no.

3. Jul 8, 2017

### Jahnavi

Why ? Can M see only along a normal ? Is extension of mirror not considered part of window ?

Sorry . I am not understanding this window concept .

Can you explain how does mosquito see the images through the window?

Last edited: Jul 8, 2017
4. Jul 8, 2017

### CWatters

If the mosquito is looking at images of itself then I think your drawing is wrong. The light must originate from and return to the mosquito.

5. Jul 8, 2017

### Jahnavi

These three are images formed by the mirrors , not images which can be seen by M .

This is what I did to reason out why only one image can be seen .

6. Jul 8, 2017

### SammyS

Staff Emeritus
For the mosquito to see an image of itself, a ray originating at the mosquito must be reflected back to the mosquito. That's along a normal. Right?

7. Jul 8, 2017

### Jahnavi

Normal to what ?

8. Jul 8, 2017

### SammyS

Staff Emeritus
I was explaining why the image the mosquito sees of itself in a (single) plane mirror is along the normal from the mosquito through the mirror.

9. Jul 8, 2017

### Jahnavi

To be more specific , let us work with notations used in the sketch .

I3 is image of M in R2 .I2 is image of I1 in R2 .

Is that ok with you ?

10. Jul 8, 2017

### SammyS

Staff Emeritus

Also helpful is to use the ":Reply" feature to make clear which post you are replying to.

The image, I3, of mosquito, M, in R2 should be along the normal from M through R2, or R2 extended. For your drawing, this normal passes very near to O. For M closer to R1 and/or farther from R2, the mosquito will definitely not be able to view this image. I3 is misplaced.

I2 looks to be reasonably well placed, and is definitely not view-able at M.

11. Jul 8, 2017

### Jahnavi

Could you explain your reasoning ?

12. Jul 8, 2017

### SammyS

Staff Emeritus
You show a ray going to M from the direction of I2, but where does that come from? Not from the Mosquito. While the placement of I2 looks fine, the dashed line to I2 and the arrowhead on the ray pointing to M are both misleading.

The ray would start at M, be reflected twice and exit out the top of the figure. An observer along the ray after the two reflections would see the image at I2. A dashed line from I2 to the point of intersection of the ray and R2 would be correct.

13. Jul 8, 2017

### Jahnavi

Well , this same reasoning I had stated/intended in the OP

I thought , may be you had @Doc Al 's reasoning using the window in your mind .

May be @Doc Al can explain his post .

14. Jul 8, 2017

### Jahnavi

When is M able to see I2 and I3 ?

Since I2 and I3 are images formed in mirror R2 , if I draw a line joining M and I2 , that line should intersect R2( not the extended part ) .Then M can see image I2 .

But if that line doesn't intersect the real visible part of R2 and instead intersects the extended part of mirror R2 , then M cannot see image I2

Is that what you intended in your post ?

Another thing I need to clarify is that , the concerned window should be of that mirror only whose image M wants to see .For example , since I2and I3 are images from R2 , the concerned window should be of R2 only , not R1 .

Not sure if it makes sense ?

Last edited: Jul 9, 2017
15. Jul 9, 2017

### CWatters

Draw a line from the point where the mirrors meet and normal to R2. Can the mosquito be above this line and still be close to R1? Badly worded question?

16. Jul 9, 2017

### Staff: Mentor

Let's go over an example that might help.

Imagine a mirror in the y = 0 plane that extends from x = 0 to x = 10 meters. You stand at point (11, 1). So you are off to the side of the mirror, facing a blank wall. Where is your image in the mirror? At point (11, -1). Can you see it? No! Think of the mirror as a window. Looking through that window, can you see your image? No -- it's behind a wall!

Now consider a friend standing at point (1, 1). His image is at point (1, -1). Can he see his image? Sure, he's right in front of that mirror "window". Can he see your image? Sure. Can you see his image in the mirror? Sure. (Of course it's not really a window, it's a mirror. So you are really only seeing reflections. But it works out as if it were a window.)

If you imagine an image world behind the plane of the mirror, then you can easily determine whether an observer can see any particular image by treating the mirror as a 'window' into that image world. (Any extension of the plane of the mirror is a wall, not a window.)

Let me know if that's any clearer.

17. Jul 9, 2017

### Jahnavi

Ok.

I think my earlier post goes in line with your reasoning .

Please see post 14 and let me know whether you find it reasonable .

18. Jul 9, 2017

### Staff: Mentor

Yes.

Perfect. You've got it now.

(You should play around with some examples of your own making to convince yourself that it always works.)

19. Jul 9, 2017

### Jahnavi

Thanks so much . You have been really nice .