 Solution

Use positive real object and image distances $u,v$ for a thin lens. The required absolute magnification is $M=1000/15=200/3$, and the specified LCD-to-screen separation gives $u+v=2000\mathrm{mm}$, not $v=2000\mathrm{mm}$. Since $v/u=M$,The thin lens equation yieldsPlace the projection lens $29.56\mathrm{mm}$ beyond the LCD towards the screen. Its real image is inverted; the signed magnification is $-200/3$.