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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.
The graph in the link is not to scale.

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:
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


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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