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Question
A 4 cm tall object is placed on the principal axis of a convex lens. The distance of the object from the optical centre of the lens is 12 cm and its sharp image is formed at a distance of 24 cm from it on a screen on the other side of the lens. If the object is now moved a little away from the lens, in which way (towards the lens or away from the lens) will he have to move the screen to get a sharp image of the object on it again? How will the magnification of the image be affected?
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Solution
Given that
Object distance, u = −12 cm
Image distance, v = 24 cm
`1/"f"=1/"v"-1/"u"`
`1/"f"=1/24-1/((-12))`
`1/"f"=1/8`
`therefore"f"=8cm`
The focal length of the lens is 8 cm.
Now if the object is moved away from the lens, the screen has to be moved towards the lens. This is because when we move the object away from the lens, the object distance is increased. Hence, by the lens formula, the image distance decreases.
Magnification is given as
`"m"="v"/"u"`
Because the image distance (v) decreases, the value of magnification also decreases.
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