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An object is kept at rest on the principal axis of a lens. Initially the object is at a distance three times the focal length 'f' of the lens. The lens runs towards the object at a constant speed u, until the distance between the object and its real image becomes 4 f. If the iamge always forms on a moving screen then express the velocity of the screen as a function of time. |
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Answer» `u("x-co-ordinate of object")=-[3 F-u t]` by lens FORMULA `rArr(1)/(v)-(1)/(u)=(1)/(f)` `rArr (1)/(v)=(1)/(f)+(1)/(u)=(1)/(f)+(1)/(-(3f-u t))` `=(-3f+u t+f)/(f(3f-u t)), v=(f(3f-u t))/(u t- 2f)`  Now by differentiating the lens formula once we get `v_(iL)=(v^(2))/(u^(2))v_(oL) rArrv_(i)-v_(L)=(f^(2)(3f-u t)^(2))/((ut-2f)^(2))xx(1)/((3f-ut)^(2)).(v_(o)-v_(L))` `=u+(f^(2))/((ut-2f)^(2))(0-u) =u[1-(f^(2))/((ut-2f)^(2))]` Ans `THEREFORE v_(i)=u[1-{(f)/(ut-2f)}^(2)]` |
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