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The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3m away by means of a large convex lens. What is the maximum posible focal length of the lens required for the purpose ? |
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Answer» Solution :`(1)/(v) - (1)/(U) = (1)/(f) - u + v = 3,` here, v = 3 + u ` (1)/(3+ u) - (1)/(u) = (1)/(f)"i.e.," f = - ((u^(2) + 3u)/(3) ) ` For .f. to be MAXIMUM , `(df)/(du) ` = 0 ie., `(df)/(du) = - (1)/(3) [ 2u + 3]= 0therefore u = (-3)/(2)"" therefore v = 3 + u = 3 - (3)/(2) = (3)/(2) ` `therefore (1)/(f) = (1)/((3)/(2)) - (1)/((-3)/(2)) = (2)/(3) + (2)/(3) = (4)/(3) "" therefore f = (3)/(4) = 0.75` m |
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