A point source S is moving with a speed of $10\text{m/s}$ in XY plane as shown in figure. The radius of curvature of the concave mirror is 4 m. If the speed of the image is $\text{2}\sqrt{x}\text{m/s}$ , then find x. [The rays in the question are paraxial rays]

$\Rightarrow \frac{\text{1}}{\text{v}}+\frac{1}{-6}=\frac{1}{-2}$

$\Rightarrow \text{v}=-3\text{m}$

${\text{v}}_{ix}=-{\left(\frac{\text{v}}{\text{u}}\right)}^{\text{2}}{\text{v}}_{\text{o}x}\Rightarrow {\text{v}}_{ix}=2\text{m/s}$

Let height of object be${\text{y}}_0$ . Then height of image is $y_i=\frac{\text{v}}{\text{u}}{y}_0$

Y -  component of velocity of image ${\text{V}}_{iy}=\frac{\text{d}{y}_i}{\text{dt}}$

$\left(\text{or}\right){\text{V}}_{iy}=\frac{{\text{vy}}_{\text{0}}}{{\text{u}}^{\text{2}}}\frac{\text{du}}{\text{dt}}-\frac{y_0}{\text{u}}\frac{\text{dv}}{\text{dt}}-\frac{\text{v}}{\text{u}}\left(\frac{\text{d}{y}_0}{\text{dt}}\right)$

$=\frac{\left(-3\right)\left(1\right)}{{\left(-6\right)}^2}.8-\frac{1}{\left(-6\right)}\left(-2\right)-\frac{\left(-3\right)}{\left(-6\right)}\times 6$

$=-4\text{m/s}$

$\Rightarrow \text{V}=\sqrt{{\text{V}}_{ix}^2+{\text{V}}_{iy}^2}=\sqrt{20}$