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Lens system |
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| May21-07, 08:01 AM | #1 |
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Lens system
1. The problem statement, all variables and given/known data
A combination of two thin convex lenses of equal focal lengths,is kept separated along the optic axes by a distance of 20 cm between them.The combination behaves as a lens system of infinite focal length.If an object is kept at 10 cm from the first lens,its image will be formed on the other side at a distance x from the second lens.Find x. 2. Relevant equations 1/f=1/v + 1/u where f is focal length,u is object distance and v is image distance. 3. The attempt at a solution Can the above equation be applied for the lens system as a whole,instead of just a single lens at a time? If it can be applied,then I think the problem can be solved in the following way: Since f of the lens system is infinity,this means that rays initially travelling parallel to the axis before the 1st lens,will continue to do so after passing through 2nd lens.The lens system behaves symmetrically.If we apply the above equation to the system as a whole,we get v= -10 cm and the image is virtual.Therefore x= 1o cm.Is this answer and reasoning correct? |
| May23-07, 10:41 AM | #2 |
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Someone please help........
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| May23-07, 11:51 AM | #3 |
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Hello,
I have tried to prove some relation that you want by some simple assumptions: (1)two identical lenses (2)the curvatures of two surface of one lens are the same (3) lenses are thin. But no such a general relation is satisfied. More intuitively, you can consider some special examples with the lenses of the problem: Example1. an point object is in front of the first lens with [tex]p_1=25\text{(cm)}[/tex] (i guess that you can calculate about the image problems with Gauss' formula) [tex]\frac{1}{25}+\frac{1}{q_1}=\frac{1}{10}\quad\Rightarrow\quad q_1=\frac{50}{3}\text{(cm)}[/tex] [tex]p_2=20-\frac{50}{3}=\frac{10}{3}[/tex] [tex]\frac{1}{10/3}+\frac{1}{q_2}=\frac{1}{10}\quad\Rightarrow\quad q_2=-5\text{(cm)}[/tex] Therefore the image is a virtual one and between the two lenses with a distance 5(cm) from the second lens. If one calculates again with your point of view, one always gets the same image distance with the distance between the object and the first lens (due to the infinite forcus length). Example2. an point object is in front of the first lens with [tex]p_1=20\text{(cm)}[/tex]. You can try this example without calculations. Therefore, the problem is just a coincidence. Hope these helpful. |
| May24-07, 03:49 AM | #4 |
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Lens system
I don't quite understand.
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