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Old Sep 8th 2009, 05:41 PM   #1
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Lens Problems

1.) How far apart are an object and an image formed by an 95 -focal-length converging lens if the image is 3.50 larger than the object and is real?

do + di = ?

I know to use 1/f = 1/do + 1/di and magnification is m = -di/do but i cant figure out how to apply the two together to find the distance between the object and image.

2.) A 48.0 -focal-length converging lens is 28.0 behind a diverging lens. Parallel light strikes the diverging lens. After passing through the converging lens, the light is again parallel.

What is the focal length of the diverging lens?


i set up a ray diagram but i cant figure out how to do the ray tracing when the focal point is 20cm past the second lens and then what the rays will do when they pass through the second lens because you are not starting from the object you are starting from the projected image of the first lens.

Any help is greatly appreciated,
Thank you,
Dank420
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Old Sep 9th 2009, 12:01 AM   #2
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1) Well you have everything in place.

From $\displaystyle m = -\frac{d_o}{d_i} $ you get


$\displaystyle d_o = -m d_i$. Use this value of $\displaystyle d_o$ in


$\displaystyle \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} $.

Find do, di and then do + di.


2) When parallel rays are incident on one side of a diverging lens, when viewed from the other side,
they appear to diverge from its focus. If you use the standard 1/f equation with the understanding that parallel rays imply that the source or object is at infinitely large distance from the lens you will get this result.
For parallel rays to emerge from a converging lens, the source should be at its focus by the same logic.
Thus the point behind the diverging lens from where the rays appear to diverge is actually at the focus of the converging lens giving 48 = 28 + f where f is the focal length of the diverging lens or
f = 20 cm.
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Old Sep 9th 2009, 12:15 AM   #3
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Oops missed the ray diagram bit. You seem to have got the value 20 already.
Let the parallel rays strike the divergent lens, diverge , then strike the convergent lens and emerge parallel from it.
Extend the divergent rays backward ( behind the divergent lens i.e. the side from where the parallel rays first come) till they meet at a point. Use dotted lines for this portion and mark the distance from this point to the div lens as f its focal length.
The projected image as you call it ,is the object for the converging lens and is at its focus.
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Old Sep 9th 2009, 04:44 PM   #4
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Thank you for the help, but i still cant seem to figure out the first one.


I solved do=-mdi and got do=-3.5di

i then set 1/95 = -1/3.5di + 1/di which gives me 1/95 = -1/2.5di

When i solve that out i get di = -38

when i go back and put 1/95= 1/38 + 1/x i get do to be 27.14.

When i add these two together i get -10.86 but that is not the correct answer.

I also tried adding the abs. values of do and di and i got 65.14 which is also incorrect.


What am i doing wrong?
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Old Sep 9th 2009, 10:29 PM   #5
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I may be way off here, but are the values as follows?

do = 122.14 , di = 427.5 , do + di = 549.63 ?

If answers dont match, it would be helpful if you quote the correct ones.
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