A convex lens forms inverted image of a real object on a fixed...

Since ${m}_{1}{m}_{2} = 1 ⇒ \left(\dfrac{{h}_{1}}{{h}_{0}}\right)\left(\dfrac{{h}_{2}}{{h}_{0}}\right) = 1$

${h}_{0} = \sqrt{{h}_{1}{h}_{2}} = 6cm$

for first position of the lens

$m = - \dfrac{3}{2} = - \dfrac{x + 40}{x}$

$⇒ 3x = 2x + 80$

$⇒ x = 80$

$u = - x$, $y = ( 40 + x )$

$\dfrac{1}{f} = \dfrac{1}{120} - \dfrac{1}{- 80}$

$⇒ f = 48cm$