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Your are given three lenses L_(1), L_(2) and L_(3) each of focal length 20cm. An object is kept at 40cm in front of L_(1) as shown. The final real image is formed at the focus I of L_(3). Find the separation between L_(1) and L_(2) and L_(3). |
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Answer» Solution :For lens `L_(1)` `(1)/(f_(1)) = ( 1)/( v_(1)) - ( 1)/( u_(1))` `(1)/( 20) = ( 1)/( v_(1)) - (1)/( -40) rArr v_(1) = 40CM` For lens `L_(3)``(1)/( f_(3)) = ( 1)/( v_(3)) - ( 1)/( u _(3))` `u_(3) = ? , f_(3) = + 20 cm, v_(3) = 20cm` `(1)/( 20) = ( 1)/( 20) + ( 1)/( u_(3))` `u_(3) = oo` It shows that `L_(2)` must render the rays parallel to the common axis. It means that the image `(I_(1))`, FORMED by `L_(1)`, must be at a distance of 20cm from `L_(2)` ( at the focus of `L_(2)` ) Therefore, distance between `L_(1)` and `L_(2) = ( = 40 + 20) = 60 cm` and distance between `L_(2)` and `L_(3)` can have any value . |
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