Physics - Geometrical Optics Question with Solution | TestHub

$\frac{\text{dv}}{\text{dt}}={\left(\frac{\text{v}}{\text{u}}\right)}^2\times \frac{\text{du}}{\text{dt}}$

${\text{V}}_{\text{image/lens}}={\left(\frac{\text{v}}{\text{u}}\right)}^2\times {\text{V}}_{\text{object/lens}}$

till$u$become$2f$,$m > -1$so$V_{\raisebox{1ex}{$\mathrm{image}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{lens}$}\right.}> V_{\raisebox{1ex}{$\mathrm{object}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{lens}$}\right.}$and the image will move towards left and when

$u$become$> 2f$,$V_{\mathrm{image}/\mathrm{lens}} < V_{\mathrm{object}/\mathrm{lens}}$and the image will move towards the right.