A convex lens is held directly above a con lying on a table and forms an image of it. After the lens has been moved vertically a distance equal to its focal length, it forms another image of the coin equal in size with the previous image. If the diameter of the coin is 16 mm, what is the diameter in milimeters of the image?

A convex lens is held directly above a con lying on a table and forms

1.	A convex lens is held directly above a con lying on a table and forms an image of it. After the lens has been moved vertically a distance equal to its focal length, it forms another image of the coin equal in size with the previous image. If the diameter of the coin is 16 mm, what is the diameter in milimeters of the image?
Answer» Magnitude of magnification remain unchanged `m=\|(f)/(f+(-x_(0)))\|=\|(f)/(f+(-x_(0)-f))\|=(1)/(f-x_(0))=+-(1)/(x_(0))` `+-x_(0)=f-x_(0)` either `f=0` or `x_(0)=(f)/(2)` `m=(f)/(f-f//2)=2(d_("image"))/(d_("coin"))` `rArrd_("image")=2xxd_("coin")=2xx16=32mm`

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