Physics - Geometrical Optics Question with Solution | TestHub

Differentiating the lens formula$\frac{1}{v}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$with respect to time, we get

$-\frac{1}{v^2}·\frac{\text{d}\upsilon }{\text{dt}}+\frac{1}{{\text{u}}^2}·\frac{\text{d}\text{u}}{\text{dt}}=\mathrm{0\; \; \; \; \; \; }\left(\text{as f = constant}\right)$

$\therefore \left(\frac{\text{d}\upsilon }{\text{dt}}\right)=\left(\frac{v^2}{{\text{u}}^2}\right)·\frac{\text{du}}{\text{dt}}$

Further, substituting proper values in lens formula, we
have

$\frac{1}{v}+\frac{1}{0·4}=\frac{1}{0·3}\left(\text{u}=-0·4\text{m}\text{,}\mathit{\text{f}}=0·3\text{m}\right)$

or,$v=1·2\text{m}$

Putting the values in Eq.(i) Magnitude of rate of change of position of image

$=0·09\text{m}/\text{s}$

Lateral magnification,$\mathit{\text{m}}\mathit{=}\frac{v}{\mathit{\text{u}}}$

$\therefore \frac{\mathit{\text{dm}}}{\mathit{\text{dt}}}\mathit{=}\frac{\mathit{\text{u}}\mathit{·}\frac{\mathit{\text{dv}}}{\mathit{\text{dt}}}\mathit{-}v{\displaystyle \frac{du}{dt}}}{{\mathit{\text{u}}}^{\mathit{2}}}=\frac{\left(-0·4\right)\left(0·09\right)-\left(1·2\right)\left(0·01\right)}{{\left(0·4\right)}^2}$

$=-0·3/\text{s}$
$\therefore$Magnitude of rate of change of lateral magnification$=0·3/\text{s}$