A lens (positioned at x=1.5) and a lit object in the shape of an upright arrow i

Question

A lens (positioned at x=1.5) and a lit object in the shape of an upright arrow is to the left by the lens. The object is positioned (x = 0.5) such that it results into a virtual image. The position of the virtual image is not given. A converging lens of focal length 0.7 can be added at x=2.5. The resulting image is at x=3.97.

a) Knowing that the focal length (df) is related to the object distance and the image distance (di) by this expression:
1/df = 1/do + 1/di
Find the do, the object distance for this second lens.

b) Now find the xo, the position of the object for this second lens.

Explanation / Answer

Given that,

do1 = 0.5 ; f2 = 0.7 ; at x = 2.5 ; di2 = 3.97

(a)Let do2 be the object distance from the seconf lens. We know from lens eqn that,

1/f = 1/di + 1/do

do = (di - f)/ di x f

do2 = (di2 - f2) / di2 x f2 = (3.97 - 0.7) / 3.97 x 0.7 = 3.27 / 2.78 = 1.18

Hence, do2 = 1.18.

(b)position of the object for second lens will be:

x = 2.5 - 1.18 = 1.32

Hence, x = 1.32

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