An 8cm tall object is placed perpendicular to principal axes convex lens of focal length is 25cm .The distance of object from lens is 30cm.Find the position,nature and size of image formed.
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Answered by
30
focal length ( f) = 25 cm
object distance ( u ) = -30 cm
use formula ,
1/v - 1/u = 1/f
1/v - 1/(-30) = 1/( 25)
1/v = 1/(-30) +1/( 25)
= 1/150
v = 150 cm
m = v/u = ( 150)/(-30) = - 5
hence, image form real , inverted , position of image right side 150 cm from pole .
size of image is 5 times of size of object e.g size of image = 40 cm
object distance ( u ) = -30 cm
use formula ,
1/v - 1/u = 1/f
1/v - 1/(-30) = 1/( 25)
1/v = 1/(-30) +1/( 25)
= 1/150
v = 150 cm
m = v/u = ( 150)/(-30) = - 5
hence, image form real , inverted , position of image right side 150 cm from pole .
size of image is 5 times of size of object e.g size of image = 40 cm
Answered by
10
position by using lens formula
1/v- 1/u= 1/f
1/v +1/30 = 1/25
1/v = 1/25 -1/30
1/v = 1/150
v=150
so ,image distance is +150 cm
and the image is real and inverted
now magnification
m= v/u
=150/-30
=-5
1/v- 1/u= 1/f
1/v +1/30 = 1/25
1/v = 1/25 -1/30
1/v = 1/150
v=150
so ,image distance is +150 cm
and the image is real and inverted
now magnification
m= v/u
=150/-30
=-5
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