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We place an object in the left part of a divergent lens ...

We place an object in the left part of a divergent lens with focal length f = -30 cm, such that its image is five times smaller than the object. Find out where the object is located compared with the lens?


Solution


As the diverging lens always forms virtual images for real objects we have:


$β = \frac{1}{5}$

$Because \ β = \frac{x_{2}}{x_{1}} => x_{2} = βx_{1}$


Using thin lens formula we get:

$x_{1} = f \frac{(1 - β)}{β} =  - 120 \ cm$


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