If a lens is moved towards the object from a distance of 40cm to 30cm, the magnificayion of the image remains the same numerically. what is the focal length of the lens?

40 + 30 ÷ 2 = 35 (Answer)
  • -6
one virtual and on real image formed so
                                                                -f/u1-f =
                                                                             f/u2-f   so it comes u1+u2/2
 
  • 2
We can solve by this

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