JEE Main & Advanced Physics Ray Optics Question Bank Microscope and Telescope

  • question_answer
    The focal length of objective and eye lens of a microscope are 4 cm and 8 cm respectively. If the least distance of distinct vision is 24 cm and object distance is 4.5 cm from the objective lens, then the magnifying power of the microscope will be

    A)            18  

    B)            32 

    C)            64 

    D)            20  

    Correct Answer: B

    Solution :

                       For objective lens \[\frac{1}{{{f}_{o}}}=\frac{1}{{{v}_{o}}}-\frac{1}{{{u}_{o}}}\]                    \[\Rightarrow \frac{1}{(+4)}=\frac{1}{{{v}_{o}}}-\frac{1}{(-4.5)}\]\[\Rightarrow {{v}_{o}}=36\ cm\]            \[\therefore |{{m}_{D}}\,|\ =\frac{{{v}_{o}}}{{{u}_{o}}}\left( 1+\frac{D}{{{f}_{e}}} \right)=\frac{36}{4.5}\left( 1+\frac{24}{8} \right)=32\]


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