A 4 cm tall object is placed perpendicular axis of a convex lens of focal length 20 cm.The distance of the object from the lens is 15 cm. Find the nature,position and size of the image(please don't provide link of similar question, i want ans of this question only)
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nature is virtual and erect.
position is in the same direction of object i.e. in front of convex lens.
size of image is 16 cm.
position is in the same direction of object i.e. in front of convex lens.
size of image is 16 cm.
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given
h=4cm
f=20cm
u=-15cm
use lens formula,
1/v-1/u=1/f
1/v-1/(-15)=1/20
1/v=1/20-1/15=-1/60
v=-60cm
hence image from object side which is virtual ,erect and enlarged.
m=v/u= height of image/height of
object
(-60)/(-15)=height of image/4
hence, height of image=16cm
h=4cm
f=20cm
u=-15cm
use lens formula,
1/v-1/u=1/f
1/v-1/(-15)=1/20
1/v=1/20-1/15=-1/60
v=-60cm
hence image from object side which is virtual ,erect and enlarged.
m=v/u= height of image/height of
object
(-60)/(-15)=height of image/4
hence, height of image=16cm
abhi178:
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