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Old May 2nd 2009, 12:11 PM   #1
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trouble with optics

A diverging lens (f = -10.0cm) is located 20.0cm to the left of a converging lens (f = 30.0cm). An object stands to the left of the diverging lens, exactly at its focal point. Determine the distance (in cm) of the final image relative to the converging lens
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Old May 3rd 2009, 09:23 AM   #2
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The question has to be solved in two steps.
First let us find the position of the image formed by the diverging lens. For this-
u = -10 cm (as the object is placed at the focus of diverging lens)
f = -10 cm
1/v - 1/u = 1/f
1/v - 1/(-10) = 1/(-10)
This gives v = -5 cm. This means the image is formed at a distance of 5 cm and at the left side of the diverging lens.

This image will act as the object for the converging lens. The distance of this image from the converging lens is 5 +20 = 25 cm left of the converging lens. Hence for this lens-

u = -25 cm
f = 30 cm
1/v - 1/(-25) = 1/30
this gives v = -150cm. It means the final image is 150 cm from and left of the converging lens.
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