Question

How to find the distance between the lens and the image if a 25 cm object is 5 cm from a convex lens that has a focal length of 15 cm.

Answer 1

If the distances from the object to the lens and from the lens to the image are $S_{1}$ and $S_{2}$ respectively, for a lens of negligible thickness, in air, the distances are related by the thin lens formula

$$\frac {1}{S _ {1}} + \frac {1}{S _ {2}} = \frac {1}{f}.$$

Note that if $S_{1} < f$, $S_{2}$ becomes negative, the image is apparently positioned on the same side of the lens as the object. Although this kind of image, known as a virtual image, cannot be projected on a screen, an observer looking through the lens will see the image in its apparent calculated position.

We have $f = 15, S_{1} = 5$. So $S_{1} < f$, $S_{2}$ becomes negative, the image is apparently positioned on the same side of the lens as the object.

$$\frac {1}{5} + \frac {1}{S _ {2}} = \frac {1}{15} \gg S _ {2} = - 7.5 \mathrm{cm}$$

Answer

The distance between the lens and the image is $-7.5\mathrm{cm}$.