Question

CE CHP 22 QS 13

A 4.7 cm tall object is placed 32 cm in front of a spherical mirror. It is desired to produce a virtual image that is upright and 3.3 cm tall.

Part B

Where is the image located?

Part C

What is the focal length of the mirror?

Part D

What is the radius of curvature of the mirror?

Answer 1

Height of object (h) = 4.7 cm

Height of image (h') = 3.3 cm

Object distance (u) = - 32 cm

Magnification (m) = h'/h = 3.3/4.7 = 0.702

m = - v/u

v = um = 32 x 0.702 = 22.464 cm

Image distance = 22.464 cm

m = f/(f-u)

0.702 = f/(f-(-32))

0.702f + 22.464 = f

f = 75.30 cm​​​​​​

Focal length = 75.30 cm

Radius of curvature (R) = 2f = 2x75.30 = 150.6 cm

Radius of curvature (R) = 150.6 cm