A candle is placed 15.50 cm in front of a convex mirror. When the convex mirror is replaced with a plane mirror, the image moves 7.0 cm farther away from the mirror. What is the focal length of the convex mirror?

What the image shift tells you

A plane mirror always forms a virtual image the same distance behind the mirror as the object is in front. So the plane-mirror image position is fixed by the given object distance.

The problem says that when you switch from the convex mirror to the plane mirror, the image moves 7.0 cm farther from the mirror. That means the convex-mirror image was 7.0 cm closer to the mirror than the plane-mirror image.

Find the image distance for the convex mirror

Object distance: $d_o = +15.50\,\text{cm}$.

Plane mirror image distance: $$d_{i,\, plane}=-d_o=-15.50\,\text{cm}$$

“Farther away by $7.0\,\text{cm}$” after switching to the plane mirror means: $$|d_{i,\, plane}| = |d_{i,\, convex}| + 7.0$$ So, $$|d_{i,\, convex}| = 15.50-7.0=8.50\,\text{cm}$$ Because a convex mirror makes a virtual image behind the mirror, $$d_i=-8.50\,\text{cm}$$

Use the mirror equation to solve for $f$

Mirror equation: $$\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$$ Substitute $d_o=15.50$ cm and $d_i=-8.50$ cm: $$\frac{1}{f}=\frac{1}{15.50}+\frac{1}{-8.50}$$ $$\frac{1}{f}=0.064516-0.117647=-0.053131$$ $$f=\frac{1}{-0.053131}\approx -18.8\,\text{cm}$$

Quick sign check

A convex mirror must have a negative focal length in the standard sign convention, and the result $f\approx -18.8\,\text{cm}$ matches that.