Problem with reflection and refraction.

Click For Summary
SUMMARY

The discussion revolves around the optical behavior of a converging lens and a flat plane mirror. A 1 cm tall object is positioned 6 cm to the left of a converging lens with a focal length of 10 cm. The calculated image distance (di) for the lens is -15 cm, indicating that the image is virtual and located on the same side as the object. The subsequent virtual image formed by the flat mirror is located 17 cm behind the mirror, leading to further calculations for the final image position.

PREREQUISITES
  • Understanding of lens formulas, specifically 1/f = 1/do + 1/di
  • Knowledge of magnification calculations, m = hi/ho = -di/do
  • Familiarity with the properties of converging lenses and flat mirrors
  • Basic principles of optics, including image formation and virtual vs. real images
NEXT STEPS
  • Study the principles of ray diagrams for converging lenses
  • Learn about the behavior of light in flat mirrors and their image formation
  • Explore advanced lens systems and their applications in optical devices
  • Investigate the concept of total internal reflection and its implications in optics
USEFUL FOR

Students of physics, optical engineers, and anyone interested in understanding the principles of light behavior through lenses and mirrors.

spikers220
Messages
1
Reaction score
0
1. A 1 cm tall object sits 6cm to the left of a converging lens, f = 10 cm. A flat plane mirror sits 2 cm to the right of the lens. Light passes through the lens, reflects off the mirror, then back through the lens. Where is the final image? Real or Virtual? Upright or inverted? How big is it?



2. 1/f = 1/do+1/di
m = hi/ho = -di/do




3. The first thing that I did was calculate the di for the converging lens and got that to be -15 cm, which means that it will be on the same side as the object. I am kind of unsure on what to do next. Do I then take that image that was formed, make it a real object for the flat mirror, in which a virtual image forms 17cm behind the mirror? Then use 19 cm as the new object distance for my next calculation? It doesn't seem like that is the right thing to do.

I am new to forum so hopefully all this is clear.
 
Physics news on Phys.org
It must work, go ahead.

ehild
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
19
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K