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

  • question_answer
    A lens of refractive index \[n\] is put in a liquid of refractive index \[n'\] of focal length of lens in air is \[f\], its focal length in liquid will be                                                             [MP PET 1999]

    A)                    \[-\frac{fn'(n-1)}{n'-n}\] 

    B)                    \[-\frac{f(n'-n)}{n'(n-1)}\]

    C)                    \[-\frac{n'(n-1)}{f(n'-n)}\]                                          

    D)                    \[\frac{fn'\,\,n}{n-n'}\]

    Correct Answer: A

    Solution :

                       \[\frac{1}{f}=\left( \frac{n-1}{1} \right)\ \left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] and \[\frac{1}{f'}=\left( \frac{n-n'}{n'} \right)\ \left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\]                                                               \[\therefore \] \[\frac{f'}{f}=\frac{n-1}{1}\times \frac{n'}{n-n'}\]\[\Rightarrow \] \[f'=-\frac{fn'(n-1)}{n'-n}\]


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