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An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens? |
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Answer» Solution :Here h = 3.0 cm, u = - 14 cm, f = - 21 cm `therefore 1/v -1/(-14) =1/(-21)` or `1/v =-1/21 - 1/14 =(-2 -3)/42 =-5/42 rArr v=-42/5 = -8.4 cm` NEGATIVE value of v means that the IMAGE is virtual and erect. If SIZE of image be h., then `h. =h(v/u) = (3.0) xx (-8.4)/(-14) = 1.8 cm` If the object is moved farther away from the lens, the virtual image moves towards the principal focus of the lens and size goes on diminishing further. |
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