A dentist uses a concave mirror to examine a patient's tooth. When the mirror is placed 12… — JAMB UTME

A dentist uses a concave mirror to examine a patient's tooth. When the mirror is placed $12\text{ cm}$ in front of the tooth, it forms an erect image that is magnified $3$ times. What is the focal length of the mirror?

$18\text{ cm}$ Answer
B
$9\text{ cm}$
C
$24\text{ cm}$
D
$36\text{ cm}$

Answer

The focal length of the mirror is

18\text{ cm}

For a concave mirror, an erect image is always virtual, located behind the mirror. The magnification formula

m = -v/u = +3

gives an image distance of

v = -36\text{ cm}

when the object distance

u = 12\text{ cm}

. Applying the mirror equation

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{12} - \frac{1}{36} = \frac{2}{36} = \frac{1}{18}

yields a focal length of

18\text{ cm}

Step-by-Step Solution

Identify the nature of the image and apply the magnification relationship

Since the concave mirror forms an erect image, the image must be virtual. Therefore, linear magnification

m = -\frac{v}{u} = +3 \implies v = -3u

An erect image formed by a spherical mirror is always virtual, which corresponds to a negative image distance under standard sign conventions.

Calculate the image distance

v

v = -3 \times 12\text{ cm} = -36\text{ cm}

Given object distance

u = +12\text{ cm}

, multiplying by

-3

yields the position of the virtual image behind the mirror.

Substitute

u

and

v

into the mirror formula to find

f

\frac{1}{f} = \frac{1}{12} + \frac{1}{-36} = \frac{3 - 1}{36} = \frac{2}{36} = \frac{1}{18} \implies f = +18\text{ cm}.

The mirror equation

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

relates object distance, image distance, and focal length.

Key Concept

Focal length calculation for spherical mirrors producing virtual images

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic