A dentist wants a small mirror that when placed 2cm from a tooth, will produce \[ 3 \times \] upright image. What kind of mirror must be used and what must its focal length be?
A. concave mirror, 3.0cm
B. concave mirror, 1.5cm
C. convex mirror, 3.0cm
D. convex mirror, 1.5cm

Hint: For a \[ 3 \times \] upright image to be formed, the image needs to be magnified. The mirror that is used for the purpose of magnification is always a concave mirror. An upright image is formed by a concave mirror only when the object is in the front of the focal point.

Formula used:
The magnification is given by:
 \[ m = \dfrac{v}{{ - u}} = \dfrac{f}{{f - u}}\]

Complete step by step answer:
Only plane mirrors and concave mirrors can form upright images. Since, plane mirror is not given in the options; the correct options would be those consisting of concave mirrors. But for a concave mirror to produce an upright image, the object is in the front of the focal point.
The focal length is the distance of the principal focus from the pole of the mirror. It is represented by \[ f\] and the magnification produced is given by
 \[ m = \dfrac{v}{{ - u}} = \dfrac{f}{{f - u}}\]
where \[ v\] is the distance of the image, \[ u\] is the distance of the object from the pole of the mirror and \[ f\] is the focal length of the mirror. The magnification is given as \[ 3 \times \] and the distance of the object is \[ 2cm\] . Then,
 \[ m = \dfrac{f}{{f - u}}\]
Substituting the values,
 \[ 3 = \dfrac{f}{{f - 2}}\]
Rearranging the terms in form of \[ f\]
 \[ 3\left( {f - 2} \right) = f{\text{ }} \to 3f - 6 = f \to 2f = 6{\text{ }} \to f = 3cm{\text{ }}\]
Thus, the focal length is \[ 3cm\] .

So, the correct answer is “Option A”.

Note:
The concave mirror forms a real image for \[ u \geqslant f\] . It forms a virtual image for \[ u \leqslant f\] . The real image formed by the concave mirror is diminished for \[ u > 2f\] . It is magnified when \[ u\] lies on the focal point and between \[ f\] and \[ 2f\] .
If the image formed by a concave mirror is magnified, it is always a virtual image.