The image of a needle placed at 45 cm from a lens is formed on a screen placed 90 cm on the other side of the lens. Find the displacement of the image, if the object is moved 5 cm away from the lens. Also, find the power of the lens.
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Answered by
1
Answer:
Displacement of image = v-v' =90-75= 15 cm, towards the lens.
Answered by
1
Answer:
u = -45
v = 90
1/f = 1/v -1/u (lens formula)
therefore 1/f = 1/90 -(-1/45)
1/f = 1/90 + 1/45
1/f = 3/90 (after taking LCM)
therefore f = 90/3
focal length = 30 cms
we know,
height of image/height of object = v/u
so, height of image/5 = 90/45
therefore height of image is 10 cms (after calculation)
THANKYOU.
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