Lenses - magnification, position, focal length help

  • Thread starter alicia113
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  • #1
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Lenses - magnification, position, focal length help!!!

ok so here is the quesiton..

An object is loacted 3.5 cm from the optical centre of a lens. The lense produces an image of magnification that is +0.667 the size of the object

a) is this lens a diverging or converging lens? explain

b) Find the position of the image and the focal length of the lens using the magnification equation and the thin lens equation

c) make a diagram of the situation, showing the axis of symmetry and the prinicple axis of the lens, its optical centre itrs principle focal points and the positions of the object and image



ok so you said it was a diverging lens and i have NO IDEA how to do b.... i am going back and back to my notes... but i just dont understand :confused: ... someone please help me !!! this is due tomorrow too :cry:
 
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  • #2
berkeman
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ok so here is the quesiton..

An object is loacted 3.5 cm from the optical centre of a lens. The lense produces an image of magnification that is +0.667 the size of the object

a) is this lens a diverging or converging lens? explain

b) Find the position of the image and the focal length of the lens using the magnification equation and the thin lens equation

c) make a diagram of the situation, showing the axis of symmetry and the prinicple axis of the lens, its optical centre itrs principle focal points and the positions of the object and image



ok so you said it was a diverging lens and i have NO IDEA how to do b.... i am going back and back to my notes... but i just dont understand :confused: ... someone please help me !!! this is due tomorrow too :cry:

Welcome to the PF.

Per the PF rules, you need to show some effort before we can be of tutorial help.

What do you mean "you said it was diverging"? Who is you?

A good place to start is to list the lens equations, and show how you calculate magnification. Then try to make the sketch they are asking for -- that should get you pretty far along... Please show us your work.
 
  • #3
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So magnification is


M= hi/ho = -di/do
 
  • #4
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So hi is +3.5cm. But what's the ho?

M=3.5/? = +0.667

So it would end up being 3.5/0.667 = 5.24? Am I correct?
 
  • #5
berkeman
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Good so far. Sketch?
 
  • #6
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Is there an equation for focal length.?
 
  • #7
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But I thought the magnification was +0.667?
 
  • #8
berkeman
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Is there an equation for focal length.?

Yes, the lens equation.
 
  • #9
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Is it

1/Di=1/f - 1/do
 
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  • #10
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Is it

1/Di=1/f - 1/do

But for that I am not sure where the numbers fit in. I do not have focal length. All I have is magnification and where the object is located
 
  • #11
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But for that I am not sure where the numbers fit in. I do not have focal length. All I have is magnification and where the object is located


I have 1/Di = 1/f - 1/3.5cm


I am missing the focal length


Wait that's the thin lens equation wow ok hold on.
 
  • #12
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No I need the thin lens equation .. I am so confused right now !!
 
  • #14
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Ok so I have my focal length as +4.3 magnification as 5.24 and distance of image is -18.3
 
  • #15
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How do I find position of image
 

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