Question

This section has to do with spherical mirrors and lenses. So here goes...

(a) In front of a spherical concave mirror of radius 23 cm, you position an object of height 2.8 cm somewhere along the principal axis. The resultant image has a height of 2.2 cm. How far from the mirror is the object located? Answer: ____________.__ cm

(b) In front of a spherical convex mirror of radius 23 cm, you position an object of height 2.8 cm somewhere along the principal axis. The resultant image has a height of 2.2 cm. How far from the mirror is the object located? Answer: ____________.__ cm

(c) On one side of a converging lens of focal length 12 cm, you position an object of height 0.7 cm somewhere along the principal axis. The resultant image has a height of 0.2 cm. How far from the lens is the object located? Answer: ____________.__ cm

(d) On one side of a diverging lens of focal length 12 cm, you position an object of height 0.7 cm somewhere along the principal axis. The resultant image has a height of 0.2 cm. How far from the lens is the object located? Answer: ____________.__ cm

Answer 1

(a)

radius of curvature R = 23 cm, object height ho = 2.8 cm, and image height hi = 2.2 cm

magnification hi/ho = q/p, then

q/p = 2.2/2.8

q = 0.78*p

from mirror equation

1/p + 1/q = 2/R

1/p + 1/(0.78p) = 2/23

object distacne p = 26.2 cm

(b)

from mirror equation

1/p + 1/q = 2/R

1/p - 1/(0.78p) = -2/23

p = 3.24 cm

(c)

object distance p, focal length f = 12 cm, image distance q, object height ho = 0.7 cm

image height hi = 0.2 cm, magnification hi/ho = q/p, then

q/p = 0.2/0.7

q = 0.28*p

from mirror equation

1/p + 1/q = 1/f

1/p + 1/(0.28p) = 1/12

p = 54.8 cm

(d)

magnification hi/ho = -q/p

-q/p = 0.2/0.7

q = -0.28*p

from mirror equation

1/p + 1/q = 1/f

1/p - 1/(0.28p) = -1/12

p = 30.8 cm