# Homework Help: Reflecting glass and a convex mirror.

1. Jul 2, 2007

### Alpha Russ Omega

Hello!
I'm stuck on this problem...

A thin flat plate of partially reflecting glass is a distance (b) from a convex mirror. A point source of light (S) is placed a distance (a) in front of the plate so that its image in the partially reflecting plate coincides with its image in the mirror. If b = 5.20 cm., and the focal length of the mirror is f = - 51.0 cm., find "a" in centimeters.

http://img254.imageshack.us/img254/8801/221ch3031p04lt9.png [Broken]

I tried solving for p but my answer came out incorrect.

1/p + 1/q = 1/f
p = 1/[(1/f)-(1/q)]
p = 1/[(1/-51.0 cm)-(1/5.20cm)] = ...

Last edited by a moderator: May 2, 2017
2. Jul 4, 2007

### Staff: Mentor

The object distance from the convex mirror is a+b, not b. (What must the image distance be?)

3. Jul 5, 2007

### Alpha Russ Omega

Ah... So the image distance must be b-a. Thus: [1/(a+b)]+[1/(b-a)]=1/f
Then I would probably just solve for a.

4. Jul 5, 2007

### Alpha Russ Omega

Converging lens, object, and mirror

Sorry for double posting, but I'm stuck on a similar one. This one is even more confusing to me.

"An object is placed 1.00 m. in front of a converging lens, of focal length 0.54 m., which is 2.00 m. in front of a plane mirror. How far from the lens is the final image located (in meters), if viewed by looking toward the mirror through the lens?"

I couldn't find a graphic for this problem so I drew something that could look like what they are describing.

I'm not even sure how to apply the previous equation here. Hopefully I drew the picture correctly.

Can that a+b or b-a method apply here as well?

Thanks in advance for any help!

Last edited by a moderator: May 3, 2017
5. Jul 5, 2007

### Staff: Mentor

Do it step by step. The light is first focused by the lens: find the image location. That image becomes the object for the plane mirror: find its image. Now that image becomes the object for the second pass through the lens (in the opposite direction): find that image, which is the final one.

Give it a shot.

6. Jul 7, 2007

### Alpha Russ Omega

Alrighty, I got it figured out. Thanks! I did the step-by-step thing and that worked out nicely.

7. Feb 1, 2012

### AbbeFaria

I'm stuck on the second question as well. Applying the method suggested by Doc Al, I calculated the image distance for the image initially formed by the lens using the thin lens equation as follows:

1/1 + 1/i = 1/0.5 (in my version of the question, f = 0.5m, not 0.54m)

Therefore, i = 1. Thus, the image is formed 1m between the converging lens and the mirror.

Applying the equation i = -p for the mirror, you get an image distance of -1 with respect to the mirror.

Then, applying the thin lens equation again for this image, with the light now travelling in the opposite direction (compared to when the equation was first used), I got:

-1/3 + 1/i = 1/0.5

Therefore, i = 3/7. Thus, the image is 3/7m from the lens, on the side of it closest to the mirror.

My textbook gives the answer as 0.6m, however, with the image being on the other side of the lens.

I think I must have misunderstood the sign conventions for the thin lens equation. I have p = -3 and f = 0.5 the second time I'm applying the equation because this time, p is 3m on the other side of the lens, while f remains unchanged. It seems that to get an answer of 0.6m, however, p and i have to have the same sign.

Any help with this would be much appreciated.

8. Feb 2, 2012

### Staff: Mentor

Since the light (on its second pass) is moving from right to left, that's the orientation you should use. The object distance for that pass should be positive. (You only use a negative object distance for virtual objects.)