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In a convex lens of focal length $F$, the minimum distance between an object and its real image must be
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Verified Answer
The correct answer is:
$4 F$
Let $L$ is the distance between a real object and its real image formed by a convex lens, then as
$L=(|u|+|v|)$
$=(\sqrt{u}-\sqrt{v})^2+2 \sqrt{u v}$
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,
$\frac{(0}{v}-\frac{@}{u}=\frac{\text { @) }}{F}$
$\frac{(0}{u}-\frac{®}{-u}=\frac{\text { ๑) }}{F}$
$u=2 F$
$\therefore \quad(L)_{\min }=2 \sqrt{2-\times 2}=4$
$(U \operatorname{sing}(\mathrm{i}))$
$L=(|u|+|v|)$
$=(\sqrt{u}-\sqrt{v})^2+2 \sqrt{u v}$
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,
$\frac{(0}{v}-\frac{@}{u}=\frac{\text { @) }}{F}$
$\frac{(0}{u}-\frac{®}{-u}=\frac{\text { ๑) }}{F}$
$u=2 F$
$\therefore \quad(L)_{\min }=2 \sqrt{2-\times 2}=4$
$(U \operatorname{sing}(\mathrm{i}))$
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