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Lenses
Lenses  
I am having difficulty with this question. Any help is appreciated.

A 2-cm tall object is located 5 cm in front of a diverging lens of unknown focal length . At a distance of 32 cm behind the first lens, there is a converging lens with . Knowing that the final image is located 45 cm behind the second lens and upside down, determine the focal length of the diverging lens.
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Answer
Two Lenses
Two Lenses  
Treat each lens independently!
Analyzing the second lens first, the object, image and focal lengths are:
di=45cm, f=20cm
1/do+1/di=1/f  becomes  1/do+1/45=1/20  solving for do=20*45/(45-20)=36cm
That object distance places the object 4 centimeters to the left of the first lens:
36-32=4cm
So the image for the first lens is found 4 centimeters to the left of the first lens and is, therefore a virtual image of the first lens.
Applying the lens equation to the first lens where the values are:
do=5.0cm, di=-4.0cm  the focus becomes 1/5.0-1/4.0=1/f  solving for f=1/-0.05=-20cm

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James J. Kovalcin

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I am teaching or have taught AP physics B and C [calculus based mechanics & electricity and magnetism] as well as Lab Physics for college bound students. I have a BS in Physics from the University of Pittsburgh and a Master of Arts in Teaching from same. I have been teaching physics for 34 years. I am constantly updating my skills and have a particular interest in modern physics topics.

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