An object is placed at a distance of $30\text{ cm}$ in front of a convex lens of focal length $15\text{ cm}$. Use the lens formula to determine the position of the image. What will be the size of the image in this case?

Using the sign convention of Section 9.3.6 (focal length of a convex lens is positive, object placed to the left is negative):

Object distance, $u = -30\text{ cm}$

Focal length, $f = +15\text{ cm}$ (convex lens)

By the lens formula (Eq. 9.8): $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Since the object is placed at $u = -30\text{ cm} = -2f$, i.e. at $2F_1$, Table 9.4 of the chapter tells us the image forms at $2F_2$, is same size, real and inverted — the calculation below confirms this.

Magnification: $m=\dfrac{v}{u}=\dfrac{30}{-30}=-1$. The negative sign confirms the image is real and inverted, and $|m|=1$ confirms it is the same size as the object.