The object of 2cm tall stands vertically on the principal axis of a converging lens of focal length 4cm.the object is placed 9cm from the lens.
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Answered by
1
Here,ho=2 cm
F=-4 (concave )
U=-9(u is always in neg)
Now,
1/v=1/u+1/f
1/v=1/-9+1/-4
=1/-9-1/4
=-4-9/36
=-13/36=-2
Therefore v =-2
Hii wt hv to find
F=-4 (concave )
U=-9(u is always in neg)
Now,
1/v=1/u+1/f
1/v=1/-9+1/-4
=1/-9-1/4
=-4-9/36
=-13/36=-2
Therefore v =-2
Hii wt hv to find
Answered by
2
Height of object = 2cm
Object distance = 15cm
Focal length = 10cm
1/f = 1/v +1/u
-1/10 = -1/15 +1/v
1/v = -3 +2 /30
v = -30 cm
Height of image/Height of object = -v /u
Height of image / 2 = 30/-15
Height of image = - 4cm
Nature - Real and inverted
Position - 30 cm from concave mirror
Size = 4 cm
Object distance = 15cm
Focal length = 10cm
1/f = 1/v +1/u
-1/10 = -1/15 +1/v
1/v = -3 +2 /30
v = -30 cm
Height of image/Height of object = -v /u
Height of image / 2 = 30/-15
Height of image = - 4cm
Nature - Real and inverted
Position - 30 cm from concave mirror
Size = 4 cm
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