You place an object 22.4 cm in front of a diverging lens which has a focal lengt

Question

You place an object 22.4 cm in front of a diverging lens which has a focal length with a magnitude of 14.6 cm. Determine how far in front of the lens the object should be placed in order to produce a new image that is 3.30 times smaller than the original image.

I asked this before, and the answer was wrong. its not 33.58.

help text from the question:

Can you write the thin lens equation that gives a relationship between the focal length of a lens, the object distance, and the image distance?
Can you write the thin lens equation that gives a relationship between the object distance, image distance, and magnification?

Explanation / Answer

lens is diverging means focal length is negative

equation used will be

1/u+1/v=1/f

where u=object distance

v=image distance

f =focal length

for u=22.4

1/22.4 +1/v=-1/14.6

v=-8.84 cm

let object is x cm long

image heigth=(8.84/22.4)*x=0.395 x

image height for second position=0.395x/3.3=0.1196 x

object height = x (will remain same)

-v/u=0.1196x/x=0.1196

v=-0.1196u

putting in equation

1/u-1/0.1196u=-1/14.6

u=107.47 cm

so object should be kept at 107.47 cm to the left of lens

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