# A dentist uses a mirror to examine a tooth that is 1.00 cm in front of the mirror. The image of...

## Question:

A dentist uses a mirror to examine a tooth that is 1.00 cm in front of the mirror. The image of the tooth is formed 10.0 cm behind the mirror.

Determine the mirror's radius of curvature and the magnification of the image.

## Concave Mirror:

Concave mirror has a positive focal point and it forms an image that is inverted and real if the object is placed at a distance greater than focal length.

Solving for the focal length, we used:

{eq}\frac{1}{f} = \frac{1}{s} + \frac{1}{s'} {/eq}

And radius of curvature of a mirror is equal to {eq}R = 2f {/eq}

Magnification is given by the formula {eq}m = \frac{-s'}{s} {/eq}

Given:

s = 1.00 cm

s' = -10.0 cm

Solution:

Magnification is given by the formula:

{eq}m = \frac{-s'}{s} \\ m = \frac{-(-10.0 \ cm)}{1.00 \ cm} \\ m = 10 {/eq}

Radius of Curvature is twice the focal length where focal length is equal to:

{eq}\frac{1}{f} = \frac{1}{s} + \frac{1}{s'} \\ \frac{1}{f} = \frac{1}{1.00 \ cm} + \frac{1}{-10.0 \ cm'} \\ f = 1.11 \ cm {/eq}

Therefore, Radius of Curvature of concave mirror is:

{eq}R = 2f \\ R = 2.22 \ cm {/eq} 