⭐(a) At what position a candle of height 3 cm be placed in front of a convex lens so that its image of length 6 cm be obtained on a screen placed at a distance 30 cm behind the lens?
(b) What is the focal length of the lens in part (a)⭐
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height of image =6 cm
height of object=3cm
object distance (u)=30cm
magnification =height of image /heigh tof object=6/3=2 cm
m=v/u=2
30/u=2
u=30/2=15cm
1/f=1/v-1/u
1/f=1/30-(1/15)
1/30-1/15
=1+2/60=3/30
f=10cm
height of object=3cm
object distance (u)=30cm
magnification =height of image /heigh tof object=6/3=2 cm
m=v/u=2
30/u=2
u=30/2=15cm
1/f=1/v-1/u
1/f=1/30-(1/15)
1/30-1/15
=1+2/60=3/30
f=10cm
Anonymous:
that's wrong
Answer is something else ^-^
ok
now check
Still wrong :(
u is always positive ,focal length of convex lens is positive, vis positive for real and negative for virtual
ok sorry
ohh ye gdbd thi to
v given tha man eusko u man liya
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