Physics - Geometrical Optics Question with Solution | TestHub

Mirror formula$\frac{1}{\text{v}}+\frac{1}{\text{u}}=\frac{1}{\text{f}}$

Differentiating w.r.t.$\mathrm{t}$,

$0=-\frac{1}{{\text{v}}^2}\frac{\text{dv}}{\text{dt}}-\frac{1}{{\text{u}}^2}\frac{\text{du}}{\text{dt}}$

$\frac{\text{dv}}{\text{dt}}=-\frac{{\text{v}}^2}{{\text{u}}^2}\frac{\text{du}}{\text{dt}}$

$\text{But}\frac{\text{du}}{\text{dt}}=\text{15 m/s}$

$\text{Now}\frac{1}{\text{v}}=(\frac{1}{\text{20}}+\frac{1}{\text{600}})\therefore \text{v}=\frac{\text{600}}{\text{31}} \text{cm}$

$\text{and}\frac{\text{v}}{\text{u}}=\frac{\text{600/31}}{\text{600}}=\frac{1}{\text{31}}$

$\text{Speed of image}=\frac{\text{d}\text{v}}{\text{dt}}=-(\frac{1}{\text{31}}{)}^2\times \text{15}$

$=\text{0.016} \mathrm{m}/\mathrm{s}$