Question

# The image of the needle placed 45 cm from a lens is formed on a screen placed 90 cm on the other side of the lens. The displacement of the image, if the needle is moved by 5.0 cm away from the lens is

A
10 cm, toward the lens
B
15 cm, away from the lens
C
15 cm, towards the lens
D
10 cm, away from the lens

Solution

## The correct option is C 15 cm, towards the lensHere, u = -45 cm,  v = 90 cm $$\therefore$$ $$\dfrac{1}{f} = \dfrac{1}{v}-\dfrac{1}{u} = \dfrac{1}{90} + \dfrac{1}{45} =\dfrac{1}{30} \Rightarrow f = 30 cm$$when the needle is move 5 cm away from the lens,u=-(45 + 5) = -50 cm$$\therefore$$ $$\frac{1}{v'} = \dfrac{1}{f} +\dfrac {1}{u'} = \dfrac{1}{30} + \dfrac{1}{(-50)} = \dfrac{2}{150} = \Rightarrow v' = 75 cm$$$$\therefore$$ Displacement of image = v-v' =90-75= 15 cm, towards the lensPhysics

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