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KVPY Sample Paper KVPY Stream-SX Model Paper-8

  • question_answer
    When an object is placed at a distance of 25 cm from a mirror, the magnification is \[{{m}_{1}}\]The object is moved 15 cm further away with respect to the earlier position, and the magnification becomes \[{{m}_{2}}.if\,{{m}_{1}}/{{m}_{2}}=4,\] the focal length of the mirror is:

    A) 10cm               

    B) 30cm

    C) 15cm               

    D) 20cm

    Correct Answer: D

    Solution :

    \[m=-\frac{v}{u}=-\left( \frac{f}{u-f} \right)\]
    Now \[{{m}_{1}}=-\left( \frac{f}{25-f} \right)..(i)\]
    and  \[{{m}_{2}}=-\left( \frac{f}{40-f} \right)..(ii)\]
    \[\therefore \frac{{{m}_{1}}}{{{m}_{2}}}=\frac{40-f}{25-f}\]
    Or \[4=\frac{40-f}{25-f}\]
    Or \[f=20cm.\]


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