Physics - Geometrical Optics Question with Solution | TestHub

We know:Let the seperation between object and image by $x$, $$x=u-v$$ Using equation (1), on rewriting, $$\begin{array}{rl}& x=u-\frac{xf}{u}\\ & \Rightarrow x=\frac{u^2}{(u+f)}---(1)\end{array}$$ we first take log of equation (1) $$\Rightarrow \ln (x)=2\ln (u)-\ln (u+f)$$ Subsequently, to minimize $x$ w.r.t. $u$, take the derivative of $x$ w.r.t. $u$ : $$\begin{array}{rl}& \Rightarrow \frac{1}{x}\frac{dx}{du}=\frac{2}{u}-\frac{1}{(u+f)}\\ & \Rightarrow \frac{dx}{du}=\frac{x(u+2f)}{u(u+f)}---(2)\end{array}$$ Condition for minimum $x$ requires, $\frac{dx}{du}=0$ and $\frac{d^{2}x}{\partial u^2}<0$ $$\Rightarrow \frac{dx}{du}=\frac{x(u+2f)}{u(u+f)}=0$$ So, $x=0$ \{is the trivial solution $\}$ and the real solution occurs for $u=-2f=-R$. Lets check the minimum value using equation (1), $$\Rightarrow x_{\min }=\frac{u^2}{(u+f)}=\frac{(-2f{)}^2}{(-2f+f)}=4f$$