Physics - Geometrical Optics Question with Solution | TestHub

$\mathrm{f}=8\mathrm{cm}$, when the particle is at mean position, $\mathrm{u}=-14\mathrm{cm}$ $$\begin{array}{rl}& \frac{1}{\mathrm{V}}=\frac{1}{\mathrm{f}}+\frac{1}{\mathrm{u}}=\frac{1}{8}-\frac{1}{14}=\frac{3}{56}\\ & \therefore \mathrm{V}=\frac{56}{3}\approx 19\mathrm{cm}\end{array}$$ When the particle is at one of the extreme positions its distance from the lens is $14+1=15\mathrm{cm}$ $$\therefore \mathrm{u}=-15\mathrm{cm}$$ Again, $\frac{1}{\mathrm{V}}=\frac{1}{\mathrm{f}}+\frac{1}{\mathrm{u}}=\frac{1}{8}-\frac{1}{15}=\frac{7}{120}$ $$\therefore \mathrm{v}=\frac{120}{7}\approx 17\mathrm{cm}$$ Amplitude of the image $=19-17=2\mathrm{cm}$