Physics - Geometrical Optics Question with Solution | TestHub

From the lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\Rightarrow v=f-\frac{f^2}{u+f}$

 now, if we plot a graph between $u \text{and} v$

When $u\to \infty ; v=f-\frac{f^2}{\infty }=f$

and for $u\to 0; v=0$

Given that $u=v$is a reference line, therefore, $\frac{dv}{du}=0+\frac{f^2}{{\left(u+f\right)}^2}$

${\overline{)\frac{dv}{du}}}_{u=0}=0+\frac{f^2}{{\left(0+f\right)}^2}=1$

The equation of tangent of the curve at the origin, $v-0=1\left(u-0\right)\Rightarrow v=u$