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(C) Let the position of the lens be at $x = L$. The object is at $x_o = -40 \, cm$ and the image is at $x_i = 50 \, cm$. The object distance is $u = -

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$x = -10 \, cm$

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(C) Let the position of the lens be at $x = L$. The object is at $x_o = -40 \, cm$ and the image is at $x_i = 50 \, cm$. The object distance is $u = -(L - x_o) = -(L + 40)$ and the image distance is $v = (x_i - L) = (50 - L)$.The magnification $m$ is given by $m = \frac{h_i}{h_o} = \frac{-2 \, cm}{1 \, cm} = -2$.Using the magnification formula $m = \frac{v}{u}$,we have $-2 = \frac{50 - L}{-(L + 40)}$.This simplifies to $2 = \frac{50 - L}{L + 40}$.$2(L + 40) = 50 - L \implies 2L + 80 = 50 - L$.$3L = -30 \implies L = -10 \, cm$.Thus,the lens is located at $x = -10 \, cm$.

The optical axis of a thin equiconvex lens is the $x$-axis. The coordinates of the object and its image are $(-40 \, cm, 1 \, cm)$ and $(50 \, cm, -2 \, cm)$ respectively. Find the position of the lens.

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