A flat screen is placed 3.80 m to the right of an object. Aconverging lens of fo

Question

A flat screen is placed 3.80 m to the right of an object. Aconverging lens of focal length 0.750 m is placed between theobject and the screen. The thin-lens equation can be used to showthat there are two lens positions that will form sharpimages on the screen.
For the lens position that is closer to theobject, what is the distance between the object and the lens?
The correct answer is 1.0280 m.

How do I find two separate values for the lens positions using thethin lens equation (1/do + 1/di = 1/f)?

Explanation / Answer

L = 3.80 m. f = 0.750 m p = distance between the object and the lens q = distance between the image and thelens q = L - p 1/p + 1/q = 1/f 1/p + 1/(L - p) = 1/f L/[p(L - p)] = 1/f p2 - Lp + Lf = 0 p = [L ± (L2 - 4Lf)]/2 p = 2.772 m, 1.028 m

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