# Homework Help: Diverging Lens and Concave Mirror

1. Mar 19, 2004

### dorthod

Can someone get me started on this, I know how to work a diverging lens and a concave mirror, just not together. The problem:

A concave mirror with a radius of curvature of 20.0 cm is placed 25.0 cm from a diverging lens with a focal length of 16.7 cm. An object is placed midway between the lens and the mirror. Image you are looking through the lens and considering only the light that leaves the object and travels first to the mirror:
a. Locate the image formed by the mirror(which becomes a virtual object for the lens) by using a ray diagram, and then verify that image distance with the lens equation.
b. Locate the final image formed by the lens/mirror system, using a ray diagram, and then verify that image distance with the lens equation.
c. Is this image real or virtual? Explain why this is not such a simple answer.

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Here's what I can do:
a)1/10 = 1/Si + 1/12.5(not sure though about 12.5)
50cm = Si

b)1/16.7 = 1/Si - 1/50(assuming part a is correct)
12.52 = Si

c) No clue

Image I cooked up:

Last edited by a moderator: May 1, 2017
2. Mar 19, 2004

### NateTG

Slightly OT but shouldn't the lens be lenticular?

Why don't you try tracing a couple of rays off of the mirror?

3. Mar 22, 2004

### dorthod

Attempted ray-tracing:

Last edited by a moderator: May 1, 2017
4. Mar 23, 2004

### Severian596

I agree with 50 cm being the $$d_{i}$$ from the mirror. In your ray diagram you may want to place the object on the median line that runs through the vertex of both the mirror and the lens.

Secondly you should reexamine your equation (2). You have the object - which in this case the image of the mirror - at 50 cm from the lens. I don't believe this is true. If the image formed by the mirror is 50 cm from the mirror, it's actually located 25 cm PAST the lens, right? That means the image from the mirror has has not converged by the time the light rays meet the diverging lens.

See this image (it's rough but it'll do):
http://copperplug.no-ip.org/homesite/Lens.gif [Broken]

I'm wondering if part C of this question hints at the fact that the diverging lens exactly counteracts the converging of the mirror, thus never actually forming a final image. If all the rays are parallel when they leave this lens this is what will happen.

Last edited by a moderator: May 1, 2017