# Homework Help: Plane Mirror confusion!

1. Mar 13, 2006

### mr_coffee

Hello everyone! We just had our first class today and it seemed to make sense but now the homework is confusing me on what is what. Here is the problem:
You look through a camera toward an image of a hummingbird in a plane mirror. The camera is 4.30 m in front of the mirror. The bird is at camera level, 4.90 m to your right and 3.10 m from the mirror. What is the distance between the camera and the apparent position of the bird's image in the mirror?

Am i suppose to use the mirrior equations to find this?
I don't knwo what is what though. Like Object Distance/image distance, he said cammera, then he said bird. WHich one is the object? I know the image would be what u see through the mirrior. But he mentinos a distance of a cammera which can be an object, and also a bird which can be an object. Here is my work:
http://suprfile.com/src/1/68pm9w/lastscan.jpg [Broken]

You probably don't even have to draw a picture but i attempted still doesn't make much senes to me! any help woulld be great! thanks!

Last edited by a moderator: May 2, 2017
2. Mar 13, 2006

### eep

If I remember correctly, for a plane mirror, the distance from an object to a mirror is the same as the "distance" from the mirror to the image. Perhaps that can be of use to you.

3. Mar 13, 2006

### mr_coffee

Thanks eep, thats where I wrote i = -p, where I, is the image distance, and p is the object distance.

So i tried the following:
http://suprfile.com/src/1/68z0ku/lastscan.jpg [Broken]
but that was also wrong. Was it suppoe to be:
if p = 4.30m (distance of cammera from mirror)
Birds distance from mirror is 3.10m, that means the birds image distance from the mirror will also be 3.10m. THe book says,
i = -p;
so would i take, (-4.30m)+3.10m = distance from the cammera to the position of the bird image? I didn't do it this way because i never thought distance was suppose to be negative, and i would come out with a small answer anyways if i took the absolute value! any ideas waht i'm dong wrong?

Last edited by a moderator: May 2, 2017
4. Mar 13, 2006

### eep

Think about what other information the problem gives you. Is the image of the bird directly in front of the camera?

5. Mar 13, 2006

### mr_coffee

hm....no its offset by 4.90m, so I took it as this...I didn't submit yet to see if its right because i only have a few more chances but does this look like i'm doing it right?
http://suprfile.com/src/1/69ar6r/lastscan.jpg [Broken]
Or
if its 4.90m to the right, and the cammer is 4.30 from the mirror, would i take (4.90-4.30)+3.10 to get the distance of the bird image to the cammera?

Last edited by a moderator: May 2, 2017
6. Mar 13, 2006

### eep

you're going to have to use triangles and pythagoras' law of triangles. Also, your distance to the right is wrong. It's telling you that the bird is 4.90 meters to the right of the camera. That is, if the bird was directly next to the camera (which it is not), it would be 4.90 meters to the right of the camera.

Last edited: Mar 13, 2006
7. Mar 13, 2006

### mr_coffee

Thanks for the tip! :)

8. Mar 13, 2006

### eep

A tip for all mirror/lense problems: Light always travels in a straight line. There's a reason why this subject is called "geometric optics" ;)

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