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Optical components - lens and mirror

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1. 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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