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JEE Main & Advanced Physics Ray Optics Question Bank Refraction at Curved Surface

  • question_answer
    A combination of two thin lenses with focal lengths \[{{f}_{1}}\] and \[{{f}_{2}}\] respectively forms an image of distant object at distance 60 cm when lenses are in contact. The position of this image shifts by 30 cm towards the combination when two lenses are separated by 10 cm. The corresponding values of \[{{f}_{1}}\] and \[{{f}_{2}}\] are                                                  [AIIMS 1995]

    A)            \[30\] \[cm,\]\[-60\]\[cm\]   

    B)            \[20\]\[cm\], \[-30\]\[cm\]

    C)            \[15\]\[cm,\] \[-20\]\[cm\]   

    D)            \[12\] \[cm,\] \[-15\]\[cm\]

    Correct Answer: B

    Solution :

                       \[\frac{1}{60}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}\]  ?(i)                    and \[\frac{1}{30}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}-\frac{10}{{{f}_{1}}{{f}_{2}}}\]  ?(ii)                    On solving (i) and (ii)  \[{{f}_{1}}{{f}_{2}}=-\,600\] and \[{{f}_{1}}+{{f}_{2}}=-10\]            Hence \[{{f}_{1}}=20\]\[cm\] and \[{{f}_{2}}=-30\]\[cm\]


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