For an object placed in front of a converging lens at a distance greater than th

Question

For an object placed in front of a converging lens at a distance greater than the focal length of the lens, a real image wil be created on the opposite side of the lens. We know that if the object distance is close to the focal length,the image distance is very large(approaching infinity),and as the object distance becomes very large, the image distance approaches the focal length. We can define the distance "d"between the object and image and we know that d=o+i, where "o" is the object distance and "i" is the image distance. Prove that the minimum value of d occurs when o=2f and that this minumum value is equal to 4f.

thanks..

Explanation / Answer

object distance -o 1/v+1/o=1/f v=f*o/(o-f) d=o+f*o/(o-f) d=o*(o-f+f)/(o-f) d=o^2/(o-f)...................1 now d=1/((1/0)-(f/o^2) differentiating the denominator and equation to zero gives o=2*f which gives the point to be maxima and hence would be minima for the distance d. putting o in 1 fives d=4f

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