Answer
See explanation.
Work Step by Step
The bowl surface is convex so R=-35 cm.
The focal length is half the radius of curvature.
$$f=\frac{1}{2}r=\frac{1}{2}(-35cm)=-17.5cm$$
Use the mirror equation.
$$\frac{1}{s}+\frac{1}{s’}=\frac{1}{f}$$
Solve for the image distance.
$$s’=\frac{sf}{s-f}=\frac{(60cm)(-17.5cm)}{60cm-(-17.5cm)}=-13.55 cm$$
The image is 13.55 cm behind the surface of the bowl.
Calculate the magnification.
$$m=-\frac{s’}{s}=-\frac{-13.55cm}{60cm}=+0.226$$
$$|y’|=|m||y|=(0.226)(5.0cm)=1.1cm$$
The image is 1.1cm tall, upright and virtual.
