# Converging lenses

A coin is located 20.0 cm to the left of a converging lens (f=16.0 cm). A second identical lens is placed to the right of the first lens, such that the image formed by the combination has the same size and orientation as the original coin. Find the separation between the lenses.

For the first converging lens

$1/x1+1/x2 =1/f$ where

$x1$ is the distance to the object positive is the object is to the left of the lens

$x2$ is the distance to the image positive is the image is to the right of the lens

$f$ is positive for converging lens

$1/x2 =1/x1 – 1/f = 1/16-1/20$

$x2 =80 cm$

the magnification of the first lens is $Y2/Y1 =-X2/X1 =-80/20 = -4$

(image inverted and bigger)

For the second converging lens $1/x1′ -1/x2′ =1/f$

with the same meaning of x1′ and x2′ as for the ffirst lens

the magnification of the second lens is $Y2’/Y1′ = -X2’/x1′ = -1/4$ (image re-inverted and smaller)

$(y1’=y2)$

$x2’/x1′ =1/4 x1’=4*x2’$

$1/(4*x2′)+1/x2′ =1/f$

$5/(4×2′) =1/f$

$x2′ = 5f/4 =20 cm$

$x1′ = 4*20 =80 cm$

The distance between the two lens is $D = x2+x1′ =80+80 =160 cm$

The diagram should be symetrical.