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2015-05-18T16:21:16+05:30

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We want a real image on the wall.  We keep the lens in between the two walls, horizontally at level with the bulb.

   let u is -ve  as per sign convention,  v is positive.  and f is positive.  Let u = - U.
   =>   v + U = 3 meters
     =>  v and U are  between 0 and 3 meters.

  1/f  = 1/v - 1/u  = 1/v + 1/U  = (v+U)/ (vU)  = 3 / ( v U )

   f = U v / 3  = U (3 - U) / 3  = U - U²/3
   f is maximum or minimum when  d f / dU  =  0,
             =>  df/dU = 1 - 2/3 U  = 0  =>    U = 3/2 meter = 1.5 meters.

  Since df/dU > 0  for  U < 3/2  and    df/dU < 0  for  U > 3/2 m,  focal length f is maximum for  u = - 1.5 meters.  That is when the focal length is half the distance between the screen and the object.


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