Example 13.4: Diverging lenses

Question: How far must an object be placed in front of a diverging lens of focal length $45$cm in order to ensure that the size of the image is fifteen times less than the size of the object? How far in front of the lens is the image located?
 
Answer: The focal length $f$ of a diverging lens is negative by convention, so $f=-45$cm, in this case. If the image is fifteen times smaller than the object then the magnification is $M=0.0667$. We can be sure that $M=+0.0667$, as opposed to $-0.0667$, because we know that images formed in diverging lenses are always virtual and upright. According to Eq. (364), the image distance $q$ is given by

$$q = - M\,p,$$

where $p$ is the object distance. This can be combined with Eq. (367) to give

$$p = f\left(1-\frac{1}{M}\right) = -(45)\,(1-15) = 630\,{\rm cm}.$$

Thus, the object must be placed $630$cm in front of the lens. The image distance is given by

$$q = -M\,p = -(0.0667)\,(630) = -42\,{\rm cm}.$$

Thus, the image is located $42$cm in front of the lens.