1. A dentist uses a curved mirror to view the back side of teeth in the upper ja

Question

1. A dentist uses a curved mirror to view the back side of teeth in the upper jaw. Suppose she wants an upright image with a magnification of 1.5 when the mirror is 1.6 cm from a tooth.

Part B

What focal length should it have?

2. A 2.0-cm -tall object is placed in front of a mirror. A 1.0-cm -tall upright image is formed behind the mirror, 120 cm from the object.

Part B

What is the object distance?

Part C

What is the image distance?

Part D

What is the focal length of the mirror?

3. A concave mirror has a 47 cm radius of curvature. The focal length is half the radius.

Part A

How far from the mirror must an object be placed to create an upright image three times the height of the object?

Explanation / Answer

1. part B:

magnification M = -v/u

1.5u = v

v = 1.5*1.6 = 2.4 cm

u = 1.6 cm

so 1/f = 1/u + 1/v

1/f = 1/2.4 - 1/1.6

f = 4.8 cm

-------------------------------

2.

u +v = 120 cm

magnification m = v/u = 1/2 = 0.5

v = 0.5 u

so

u + 0.5 u = 120

u = 120/1.5 = 80 cm

v = 0.5* 80 = 40 cm

1/f = 1/80 + 1/40

f = 26.67 cm

--------------------------

3.

f = R/2 = 47/2 = 23.5 cm

m = v/u = 3

v = 3 u

so

1/f = 1/u + /3u

1/23.5 = 1/u + 1/(3u)

u = 31.33 cm

v = 3* 31.33 = 94 cm

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