Question

jb

2) This section has to do with spherical mirrors and lenses. a) In front a spherical concave mirror with a radius of 41 cm, you position an object of a height of 1.6 cm somewhere along the principal axis. The resultant image has a height of 1.2 cm. How far from the mirror is the object located? b) In front of a spherical convex mirror with a radius of 41 cm , you position an object of a height of 1.6 cm somewhere along the principal axis. The resultant image has a height of 1.2 cm. How far from the mirror is the object located?     c) On one side of a converging lens of a focal length of 9 cm , you position an object of a height of 0.9 cm somewhere along the principal axis. The resultant image has a height of 0.6 cm. How far from the mirror is the object located? d) On one side of a diverging lens of a focal length of 9 cm , you position an object of a height of 0.9cm somewhere along the principal axis. The resultant image has a height of 0.6 cm. How far from the mirror is the object located?

Answer 1

Given

   spherical concave mirror ofradius R = 41cm, object height y = 1.6 cm,

   image heighty' = 1.2cm

from lateral magnification m = -s'/s = y'/y

           ==> s = -s'*y/y'
           s= - s'*1.6/1.2

           s = 1.33*s'
form the relation

       1/s+1/s' = 2/R

       1/1.333s'+1/s' = 2/41

       s' = 35.87 cm

so the object should be placed on the principle axis about 1.33 times the image distance

that is s = 1.33*35.88 cm = 47.72 cm

b) for convex mirror radius of curvature is negative so

spherical concave mirror ofradius R = - 41cm, object height y = 1.6 cm,

   image heighty' = 1.2cm

from lateral magnification m = -s'/s = y'/y

           ==> s = -s'*y/y'
           s= - s'*1.6/1.2

           s = 1.33*s'
form the relation

       1/s+1/s' = -2/R

       1/1.333s'+1/s' = 2/41

   s' = - 35.87 cm image distance
so the object should be placed on the principle axis about 1.33 times the image distance

that is s = 1.33(-35.88) cm = - 47.72 cm

c)
  
   focal length of the converging lens is f = 9 cm,

   object height y = 0.9cm, image height y' = 0.6 cm

from lateral magnification

   1/s +1/s' = 1/f
   from lateral magnification y'/y = -s'/s ==> 0.6/0.9 = -s'/s = 0.66 s = s'

   1/s+1/0.66s = 1/9

   s = 22.63 cm
d) for diverging lens the focal length is -ve f = -9 cm

   object height y = 0.9cm, image height y' = 0.6 cm

from lateral magnification

   1/s +1/s' = 1/f
   from lateral magnification y'/y = -s'/s ==> 0.6/0.9 = -s'/s = 0.66 s = s'

   1/s+1/0.66s = -1/9

   s = -22.63 cm