Optical components - lens and mirror

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SUMMARY

The discussion focuses on optical components, specifically diverging lenses and mirrors, and their effects on image formation and magnification. The user presents calculations for determining object distances and image positions using formulas related to cross-magnification (Z). Key equations include Z = -a'/a and a = f(Z+1)/Z, with specific values derived for various scenarios, such as achieving a magnification of 1/2 and 1/4. The user seeks validation of their solutions and understanding of the principles involved.

PREREQUISITES
  • Understanding of optical principles, specifically lens and mirror equations.
  • Familiarity with magnification concepts in optics.
  • Knowledge of the sign conventions used in optics.
  • Ability to manipulate algebraic equations related to optics.
NEXT STEPS
  • Study the derivation and application of the lens maker's equation.
  • Learn about ray diagrams for diverging lenses and mirrors.
  • Explore the concept of virtual images and their characteristics.
  • Investigate the effects of multiple optical components on image formation.
USEFUL FOR

Students and educators in physics, particularly those focusing on optics, as well as anyone interested in understanding the behavior of lenses and mirrors in image formation.

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Homework Statement


a) Despite the diverging lens with focal length f we observed object which is perpendicular to the optical axis (Figure 6 in the article ,,5. Rozptylka a zrcátko"). To what distance from the diverging lens we have to put the subject to be scaled twice? (cross-magnification Z = 1 / 2). Determine also the position of the image of subject.

b) NextLOK to the diverging lens we placed a mirror perpendicular to the optical axis of the distance d from the lens. The rays pass through the lens, reflected from the mirror and then again pass through the lens. Where we will find the image of object and what will be its cross-magnification?

c) How we have to choose a distance d that the resulting image will be four times smaller than the object? Where we will see the final image?

1. The attempt at a solution
a) I use this equations:
Z = -a'/a = -(a'-f/f) = -(f/(a-f) = 1/2 (minus only for decisions - the image upright / inverted)
a = f(Z+1)/Z = f*1,5/0,5 = 3f
a' = f(Z-1) = f*(0,5-1) = -0,5f
Is right?

b) Ok, I can draw it, but numerically... Hehe.
Can I use the typical formula Z=y'/y = - a'/a = -(a'-f)/f = -f/(a-f) ? I think NO, but I try it.
Z = -f/(d-f); a' = Z*d = -fd*(d-f)

c)
Hmmm... Also the formula.
Z = -f/(a-f)
d=f(Z+1)/Z = f*(0,25+1)/0,25=5f
a, = -Z*d = -0,25*5f = -3,75f

Ok, this is my solutions. What do you think about it? What is true, what is false? Please, give me a advice, thank you.
 
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