Physics - Geometrical Optics Question with Solution | TestHub

Let $L$ is the distance between a real object and its real image formed by a convex lens, then as $L=(|u|+|v|)$ $=(\sqrt{u}-\sqrt{v}{)}^2+2\sqrt{uv}$
L will be minimum, when $(\sqrt{u}-\sqrt{v}{)}^2=0$ $\text{ i.e., }u=v$ Putting, $u=-u$ and $v=+u$ in lens formula, $\frac{(0}{v}-\frac{@}{u}=\frac{\text{ @) }}{F}$ $\frac{(0}{u}-\frac{\stackrel{◯}{R}}{-u}=\frac{\text{ ๑) }}{F}$ $u=2F$ $\therefore (L{)}_{\min }=2\sqrt{2-\times 2}=4$ $(U\mathrm{sing}(\mathrm{i}))$