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JEE Main & Advanced Sample Paper JEE Main Sample Paper-18

  • question_answer
    A particle executes S.H.M. of amplitude 1 mm along the principal axis of a convex lens of focal length 12 cm. The mean position of oscillation is at 20 cm from the lens. The amplitude of oscillation of the image of the particle is

    A)  2.25 mm             

    B)  4.5 mm

    C)  2 mm                   

    D)  4 mm

    Correct Answer: A

    Solution :

     \[\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\cdot f=+\,12,\]\[u=-\,20\,cm.\] \[dv=dv\left( \frac{{{v}^{2}}}{{{u}^{2}}} \right)\] \[=1\times \frac{{{30}^{2}}}{{{20}^{2}}}=9\text{/}4\,mm=2.25\,mm\]


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