JEE Main & Advanced Physics Ray Optics Sample Paper Topic Test - Refraction of Light Through Curved Surfaces

  • 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

    A) \[30\text{ }cm,\text{ }-60\text{ }cm\]

    B) \[20\text{ }cm,\text{ }-30\text{ }cm\]

    C) \[15\text{ }cm,\text{ }-20\text{ }cm\]

    D) \[12\text{ }cm,\text{ }-15\text{ }cm\]

    Correct Answer: B

    Solution :

    Initially, \[F=60\text{ }cm\] (Focal length of combination)
    Hence, by using \[\frac{1}{F}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}\]
    \[\Rightarrow \]   \[\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}=\frac{1}{60}\Rightarrow \,\,\frac{{{f}_{1}}{{f}_{2}}}{{{f}_{1}}+{{f}_{2}}}\] ...(i)
    Finally by using \[\frac{1}{F'}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}-\frac{d}{{{f}_{1}}{{f}_{2}}}\]
    where \[F'=30\,cm\] and \[d=10\text{ }cm\]
    \[\Rightarrow \]   \[\frac{1}{30}=\frac{1}{{{f}_{1}}}+\frac{1}{{{f}_{2}}}-\frac{10}{{{f}_{1}}{{f}_{2}}}\] ...(ii)  
    From (i) and (ii),  \[{{f}_{1}}{{f}_{2}}=-600\]
    From equation (i) \[{{f}_{1}}+{{f}_{2}}=-10\] ...(iii)
    Also, difference of focal lengths can written as
    \[\Rightarrow \]   \[{{f}_{1}}-{{f}_{2}}=50\] ...(iv)
    From (iii) \[\times \] (iv), \[{{f}_{1}}=20\] and \[{{f}_{2}}=-30\]

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