A dentist uses a small concave mirror with a focal length of 20\text{ mm} to inspect a pat… — JAMB UTME

A dentist uses a small concave mirror with a focal length of $20\text{ mm}$ to inspect a patient's tooth. If the mirror produces an upright image that is magnified $4$ times, how far from the tooth is the mirror placed?

$15\text{ mm}$ Answer
B
$25\text{ mm}$
C
$80\text{ mm}$
D
$5\text{ mm}$

Answer

The mirror must be placed

15\text{ mm}

from the tooth.

For a concave mirror, an upright image is virtual. Linear magnification

m = 4

implies

v = -4u

. Substituting

f = 20\text{ mm}

into the mirror formula

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

gives

\frac{1}{20} = \frac{1}{u} - \frac{1}{4u} = \frac{3}{4u}

, solving to

u = 15\text{ mm}

Step-by-Step Solution

Identify given values and apply sign conventions.

Focal length of concave mirror

f = +20\text{ mm}

. Magnification

m = +4

because the image is upright (virtual).

Upright images formed by single optical mirrors are always virtual, requiring a negative image distance.

Express image distance

v

in terms of object distance

u

Linear magnification

m = -\frac{v}{u} \implies +4 = -\frac{v}{u} \implies v = -4u

The linear magnification formula relates orientation, object distance, and image distance.

Substitute

f

and

v

into the mirror formula.

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \implies \frac{1}{20} = \frac{1}{u} - \frac{1}{4u} = \frac{3}{4u}.

The mirror equation determines object and image locations relative to the focal length.

Solve for the object distance

u

4u = 60 \implies u = 15\text{ mm}.

Cross-multiplying gives the required distance between the mirror and the tooth.

Key Concept

Reflection and image formation by concave spherical mirrors (virtual magnified image)

Question

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