An object is 20.0 cm from a converging lens, and the image falls on a screen. Wh

Question

An object is 20.0 cm from a converging lens, and the image falls on a screen. When the object is moved 3.10 cm closer to the lens, the screen must be moved 2.20 cm farther away from the lens to register a sharp image. Determine the focal length of the lens.


HINT: Let be the distance between the object and the screen. Since the object and image are on opposite sides of the screen, we have . Eliminate by combining this equation with the thin lens equation and solve for . You'll need to use the quadratic formula (see appendix E).

the answer is not 9.769 cm

Explanation / Answer

1/f=1/20+1/v----------eqn1 1/f=1/16.9+1/v+2.1-----eqn2 so 1/20+1/v=1/16.9+1/v+2.1 v(v+2.1)=228.96 v^2+2.1V-228.96=0 so v=14.11 substituting v in eqn 1 we get focal length=8.27m

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