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An equibiconvex lens has a focal length of 20 cm. A ball pin of length5 cm is placed on one side of the lens, such that the mid point of the pin is at a distance of 30 cm from the centre of the lens. Calculate the size of the image of the pin and its magnificationf = 20 cm l = 5 cm. |
Answer» Solution : `(1)/(F)=(1)/(-u)+(1)/(V)` Given : `-u_(A)=27.5 cm` `(1)/(f)=(1)/(-u_(A))+(1)/(v_(A))` `(1)/(20)=(1)/(27.5)+(1)/(v_(A))` `(1)/(20)-(1)/(27.5)=(1)/(v_(A))` `v_(A)= 73.33 cm` `(1)/(f) = (1)/(-u_(B))+(1)/(v_(B))` `-u_(B)=32.5 cm` `(1)/(20)-(1)/(32.5)=(1)/(v_(B))` `(1)/(v_(B))=(12.5)/(650)` `v_(B)=52 cm` `v_(A)-v_(B)=21.33cm` magnification`m=(v_(1))/(v_(0))=(21.33)/(5)=4.266` |
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