Question

In: Physics

This section has to do with spherical mirrors and lenses. So here goes... (a) In front...

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

Solutions

Expert Solution


(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

===================


(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


===========================


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


==========================


(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


==========================


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