Find the position and nature of the image formed by an object placed at a distance of 30 cm from a convex lens of focal length 15 cm.
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1
U = 30 cm, f = 15cm, V = ?
1/V - 1/U = 1/F
1/V - 1/30 = 1/15
1/V = 1/15 + 1/30
1/V = 2 + 1/30
1/V = 3/30
V = 30/3
V = 10cm. distance of image.
nature of image is real and inverted.
1/V - 1/U = 1/F
1/V - 1/30 = 1/15
1/V = 1/15 + 1/30
1/V = 2 + 1/30
1/V = 3/30
V = 30/3
V = 10cm. distance of image.
nature of image is real and inverted.
Answered by
0
Explanation:
u=-30cm f= -15cm
1/f=1/v-1/u
1/v=1/f+1/u
on substituting v=-10cm
m=v/u
m=-10/-30
m=1/3=.33
so the magnification is positive so it is a virtual erect image .
m isless than 1 so the image is diminished .
and v is negative so the image is formed on the same side of the lens.
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