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JEE Main & Advanced Physics Ray Optics Question Bank Self Evaluation Test - Ray Optics and Optical Instruments

  • question_answer
    A luminous object and a screen are at a fixed distance D apart. A converging lens of focal length l is placed between the object and screen. A real image of the object in formed on the screen for two lens positions of they are separated by a distance d equal to

    A) \[\sqrt{D\left( D+4f \right)}\]       

    B) \[\sqrt{D\left( D-4f \right)}\]

    C) \[\sqrt{2D\left( D-4f \right)}\]

    D) \[\sqrt{{{D}^{2}}+4f}\]

    Correct Answer: B

    Solution :

    [b] Let the object distance be x. Then, the image distance is\[D-x\]. From lens equation, \[\frac{1}{x}+\frac{1}{D-x}=\frac{1}{f}\] On algebraic rearrangement, we get \[{{x}^{2}}-Dx+Df=0\] On solving for x, we get \[{{x}_{1}}=\frac{D-\sqrt{D\left( D-4f \right)}}{2}\text{     }{{x}_{2}}=\frac{D+\sqrt{D\left( D-4f \right)}}{2}\] The distance between the two object positions is \[d={{x}_{2}}-{{x}_{1}}=\sqrt{D\left( D-4f \right)}\]


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