Diverging Lens and Concave Mirror

Click For Summary
SUMMARY

The discussion focuses on the optical interaction between a concave mirror with a radius of curvature of 20.0 cm and a diverging lens with a focal length of 16.7 cm, positioned 25.0 cm apart. Participants analyze the image formation process, using ray diagrams and the lens equation to locate images produced by the mirror and the lens system. The first image formed by the mirror is determined to be located at 50 cm, while the final image from the lens is calculated to be at approximately 12.52 cm. The complexity of determining whether the final image is real or virtual is highlighted, suggesting that the diverging lens counteracts the converging effect of the mirror.

PREREQUISITES
  • Understanding of ray diagrams in optics
  • Familiarity with the lens equation (1/f = 1/di + 1/do)
  • Knowledge of concave mirrors and their properties
  • Basic principles of image formation in optical systems
NEXT STEPS
  • Study the principles of ray tracing for concave mirrors and diverging lenses
  • Learn about the implications of virtual and real images in optical systems
  • Explore the effects of varying object distances on image formation
  • Investigate advanced topics in optical systems, such as aberrations and image quality
USEFUL FOR

Students and educators in physics, optical engineers, and anyone interested in understanding the interactions between lenses and mirrors in optical systems.

dorthod
Messages
5
Reaction score
0
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.

----------------------------
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:
http://home.earthlink.net/~emerson6/lensmirror.gif
 
Last edited by a moderator:
Physics news on Phys.org
Slightly OT but shouldn't the lens be lenticular?

Why don't you try tracing a couple of rays off of the mirror?
 
Attempted ray-tracing:
http://home.earthlink.net/~emerson6/lensmirror2.GIF
 
Last edited by a moderator:
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

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:

Similar threads

  • · Replies 11 ·
Replies
11
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
593
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
2K
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K