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 =$10\mathrm{cm}$

image height is twice the height of the object.

We have to find the distance of the image at which we get the inverted and twice size oaf the image.

Simplify

We know that the magnification of the lens is write as ,

$$\begin{gathered} \mathrm{m}=-\frac{\mathrm{v}}{\mathrm{u}}=\frac{\mathrm{h}\text{'}}{\mathrm{h}} \\ -\frac{\mathrm{v}}{10\mathrm{cm}}=2 \\ \mathrm{v}=-20\mathrm{cm} \end{gathered}$$

Since the image is formed upright then the image will be at same side of the lens as the object.

Then lens equations given by,

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

Then the image distance for the image to form inverted and twice in size is found out as, remember the inverted image will form opposite to the lens.

$$\begin{gathered} \mathrm{m}=\frac{\mathrm{h}\text{'}}{\mathrm{h}}=-\frac{\mathrm{v}}{-\mathrm{u}}=2 \\ \mathrm{v}=2\mathrm{u} \end{gathered}$$

putting this value in the lens equation then,

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

Then the distance of the lens is

$$\begin{gathered} \mathrm{d}=30\mathrm{cm}-10\mathrm{cm} \\ \mathrm{d}=20\mathrm{cm} \end{gathered}$$