Question

An oral surgeon wishes to inspect a patient's tooth with a magnifying mirror. He places the mirror 1.00 cm behind the tooth. This results in an upright, virtual image of the tooth that is 6.00 cm behind the mirror. (a) What is the mirror's radius of curvature (in cm)? cm (b) What magnification describes the image described in this passage?

Answer 1

The radius of curvature of the mirror is 12/7 cm and the magnification of the image is -6.

Step-by-step explanation:

To find the radius of curvature of the mirror, we can use the mirror equation:

1/f = 1/do + 1/di

Where f is the focal length, do is the object distance, and di is the image distance.

Given that the object distance (do) is 1.00 cm and the image distance (di) is 6.00 cm, we can substitute these values into the equation:

1/f = 1/1 + 1/6 = 7/6

Simplifying this equation, we find that the focal length (f) is 6/7 cm. Since the radius of curvature (R) is twice the focal length, the radius of curvature is:

R = 2 * f = 2 * 6/7 = 12/7 cm.

Therefore, the mirror's radius of curvature is 12/7 cm.

To find the magnification, we can use the magnification equation:

magnification = -di/do

Given that the image distance (di) is 6.00 cm and the object distance (do) is 1.00 cm, we can substitute these values into the equation:

magnification = -6/1 = -6.

Therefore, the magnification of the image is -6.