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2 convergent lenses

  • Thread starter DrMcDreamy
  • Start date
  • #1
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Homework Statement



This is a 2 part problem but I figure out the first part. Heres the 1st problem and the solution:
7- Given a convergent lens which has a focal point f. An object is placed at distance p = [tex]\frac{4}{3}[/tex]f to the left of the lens. See the sketch.
Solution: q1 = 4f, and is a real image

8- Now place a convergent lens with a same focal length f at a distance d = f behind the first lens. Determine q2, i.e., the image location measured with respect to the second lens # 2.
I will upload the actual image later.

Given:
p1 = [tex]\frac{4}{3}[/tex]f
q1 = 4f
f1 = f

p2 = ?
q2 = ?
f2 = f

Homework Equations



[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{p}[/tex]+[tex]\frac{1}{q}[/tex]

The Attempt at a Solution



I would show my attempt at the solution but I don't know what is p2. How do I get p2?
 

Answers and Replies

  • #2
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*bump*
 
  • #3
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The image formed by the first lens is now the object for the second lens. Find out where the image forms with respect to the second lens. Without drawing a picture it looks like p2 is -3f. That is:

|--|------q1/p2

where "|" is a lens and "--" is one distance f. As you can hopefully see, q1/p2 is 4f infront of lens one and 3f behind lens two. So do what you did in part one but with p2=-3f.
 
  • #4
68
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Thank you!

My work:

3[tex]\frac{1}{f}[/tex]-[tex]\frac{1}{-3f}[/tex] = [tex]\frac{3-1}{3f}[/tex] = [tex]\frac{4}{3}[/tex]f (inverse) = [tex]\frac{3}{4}[/tex]f
 
  • #5
68
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I have a similar problem heres my work but its wrong cause its not one of the answer choices:

Problem:

Consider the setup of the two-lens system shown in the figure, where the separation of the two lenses is denoted by d = 1.5 a. Their focal lengths are respectively f1 = a and f2 = 2 a. An object is placed at a distance 2 a to the left of lens # 1. Find the location of the final image q2 with respect to lens #2. Take q2 to be positive if it is to the right of lens #2 and negative otherwise.

Formula:

[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{p}[/tex]+[tex]\frac{1}{q}[/tex]

My Work:

p1 = 2a
q1 = ?
f1 = a

p2 = ?
q2 = ?
f2 = 2a

[tex]\frac{1}{q1}[/tex] = [tex]\frac{1}{f}[/tex]-[tex]\frac{1}{p}[/tex] = 2([tex]\frac{1}{a}[/tex]) - [tex]\frac{1}{2a}[/tex] = [tex]\frac{2-1}{2a}[/tex] = [tex]\frac{1}{2a}[/tex] (inverse) = 2a = q1

[tex]\frac{q1}{p2}[/tex] = q1 = [tex]\frac{2a}{p2}[/tex] = 2a [tex]\rightarrow[/tex] [tex]\frac{2a x p2}{2a}[/tex]=[tex]\frac{2a}{2a}[/tex] [tex]\rightarrow[/tex] p2 = 1

[tex]\frac{1}{2a}[/tex]-([tex]\frac{1}{1}[/tex])2 = [tex]\frac{1-2a}{2a}[/tex] = [tex]\frac{a}{2a}[/tex] = [tex]\frac{2a}{a}[/tex] = 2 = q2

What am I doing wrong?
 

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