A 4cm 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 12cm and a 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 will we have to move the screen to get a sharp image of the object again? How will magnification be affected?

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Answers

2016-02-25T14:42:18+05:30

h = 4cm , f = 20 cm and u = -15cm

Using the lens formula:

(1/v) - (1/u) = (1/f)

(1/v) - (1/-15) = (1/20)

(1/v) + (1/15) =  (1/20)

(1/v) =  (3- 4)/60

(1/v) =   (- 1/60)

v  =  - 60cm

Negative sign shows that image is virtual

v/u = h'/h   h' = (v × h) /u = (-60 × 4)/ - 15 = 16

m = h'/h = 16/4 = 4 ⇒ Positive sign shows that image is erect and magnified.

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