For a diverging lens with a focal length of f=-12.5cm, where should an object be

Question

For a diverging lens with a focal length of f=-12.5cm, where should an object be placed to form an upright, virtual image that is one-fourth the height of the object?
I posted this question earlier but I could not attach a photo. Asking again, with photo attached. Now lefs apply the thinensequation to a dverging lens, and SET UP AND SOLVE Parn (aj: The behavier of the parallel incident ays indicates hat the focal ngth is negatve -200m We want the lateria using a ray diagram. Suppose that you ane given a thin diverging lens and you ind that a beam of parallelrays 40.0cm -13.3 c Part (bj: Figure 1) stows our peop@cay dagwn for problem Part A·Practice Problem: Express your answer to two signi cant figures and include appropriate units Value Units Figure

Explanation / Answer

given,

focal length = -12.5 cm

height of image = 1/4th of object

so,

lateral magnification of object = 1/4

lateral magnification = -distance of image / distance of object

1 / 4 = -distamce of image / distance of object

distance of image = -distance of object / 4

by lens equation

1 / distance of object + 1 / distance of image = 1 / focal length

1 / distance of object + 1 / (-distance of object / 4) = 1 / focal length

1 / distance of object + 1 / (-distance of object / 4) = 1 / -12.5

distance of object = 37.5 cm

object should be placed 37.5 cm from the lens

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