Question

A lens placed $10\mathrm{cm}$in front of an object creates an upright image twice the height of the object. The lens is then moved along the optical axis until it creates an inverted image twice the height of the object. How far did the lens move?

Short answer

The lens will move $20\mathrm{cm}$.

Step-by-step solution

Given information

We have given,

object distance from the lens = $10\mathrm{cm}$,

image height = $2\times (\mathrm{object}\mathrm{height})$,

image formed = upright.

We have to find how far did the lens move to obtain the inverted image.

Simplify

Magnification of the lens

$$\begin{gathered} \mathrm{m}=\frac{\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{image}}{\mathrm{height}\mathrm{of}\mathrm{object}} \\ \mathrm{m}=2=\frac{\mathrm{v}}{\mathrm{u}} \\ \mathrm{we}\mathrm{know}\mathrm{u}=-10\mathrm{cm} \\ \mathrm{v}=-20\mathrm{cm} \end{gathered}$$

We know that for a lens to get an upright image that is 2 times magnified, the object is between the lens and the focal point.

lens equation

$$\begin{gathered} \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ \frac{1}{f}=\frac{-1}{20}-\frac{-1}{10} \\ f\mathit{=}\mathit{20}\mathit{}\mathit{}cm \end{gathered}$$

Simplify

After the lens has moved:

$$\begin{gathered} \mathrm{m}\text{'}=\frac{\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{image}}{\mathrm{height}\mathrm{of}\mathrm{object}} \\ \mathrm{m}\text{'}=-2=\frac{\mathrm{v}\text{'}}{\mathrm{u}\text{'}} \\ \mathrm{v}\text{'}=-2\mathrm{u}\text{'} \end{gathered}$$

Using lens equation:

$$\begin{gathered} \frac{1}{f}=\frac{1}{v\mathit{\text{'}}}-\frac{1}{u\mathit{\text{'}}} \\ \frac{1}{20}=\frac{1}{-2\mathrm{u}\text{'}}-\frac{1}{\mathrm{u}\text{'}} \\ \mathrm{u}\text{'}\mathit{=}\mathit{-}\mathit{30}\mathit{}cm \end{gathered}$$

then, lens has moved=$\mathrm{u}-\mathrm{u}\text{'}=-10+30=20\mathrm{cm}$