let
focal length = f
object distance = do
image distance =di
we know that
1/f= 1/do+1/di
given focal length = f= 37.4 cm
the radius of curvature is R = 2 f = (2)( 37.4 cm)=74.8 cm
given that
distance between the object and its image is = 52.8 cm
do+di = 52.8 cm
di = do-52.8 cm
1/f= 1/do+1/di
becomes
1/37.4= 1/do+1/ do-52.8
1/37.4 =( (do-52.8)+do) / (do)( do-52.8)
do2-52.8do= 74.8do-1974.82
do2-127.6do+1974.82 =0
by solving we get
do=109.578 cm or
do=18.0221 cm
case 1)
when the object lies beyond the center of curvature then
object at do =109.578 cm
image at di =do-52.8cm = 56.778 cm
a) do =109.578 cm
approximately
do=109.58 cm
b) di = 56.778 cm
approximately
di= 56.78 cm
case 2)
when the object is lies between the focal point and the mirror
then
c)
object is at do = 18.0221 cm
d)
image at di = do-52.8 cm = 18.0221 cm--52.8 cm = -34.7779 cm = -37.78 cm
approximately
di = -37.78 cm
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