In: Physics
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? (10 points) 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
(a)
object distance p = ?
radius of curvature R = 23 cm
image distance q = ?
object height ho = 2.8 cm
image height hi = 2.2 cm
magnification hi/ho = q/p
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
<<<======ANSWER
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(b)
object distance p = ?
radius of curvature R = -23 cm
image distance q = ?
object height ho = 2.8 cm
image height hi = 2.2 cm
magnification hi/ho = -q/p
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
p = 3.24 cm
object distacne p = 3.24 cm
<<<======ANSWER
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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
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
object distacne p = 54.8 cm
<<<======ANSWER
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(d)
object distance p = ?
focal length f = 12 cm
image distance q (negative) = ?
object height ho = 0.7 cm
image height hi = 0.2 cm
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
object distacne p = 30.8 cm
<<<======ANSWER
==========================