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An object is placed at a distance of f over 2from a convex lens. The image will be

Physics-General

  1. At one of the foci, virtual and double its size    
  2. At 3f / 2, real and inverted    
  3. At 2f, virtual and erect    
  4. None of these    

    Answer:The correct answer is: At one of the foci, virtual and double its sizefraction numerator 1 over denominator f end fraction equals fraction numerator 1 over denominator v end fraction minus fraction numerator 1 over denominator u end fraction (Given u equals fraction numerator negative f over denominator 2 end fraction)
    rightwards double arrow fraction numerator 1 over denominator f end fraction equals fraction numerator 1 over denominator v end fraction plus open parentheses fraction numerator 1 over denominator f divided by 2 end fraction close parentheses rightwards double arrow fraction numerator 1 over denominator v end fraction equals fraction numerator 1 over denominator f end fraction minus fraction numerator 2 over denominator f end fraction
    rightwards double arrow fraction numerator 1 over denominator v end fraction equals fraction numerator negative 1 over denominator f end fraction and m equals fraction numerator v over denominator u end fraction equals fraction numerator f over denominator f divided by 2 end fraction equals 2
    So virtual at the focus and of double size.

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