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JEE Main & Advanced JEE Main Paper (Held On 22 April 2013)

  • question_answer
    The image of an illuminated square is obtained on screen with the help of a converging lens. The distance of the square of the square from the lens is 40 cm. The area of the image is 9 times that of the square. The focal length of the lens is :     JEE Main  Online Paper (Held On 22 April 2013)

    A)  36 cm                   

    B)  27 cm

    C)  60 cm                   

    D)  30 cm

    Correct Answer: D

    Solution :

     if side of object square \[=\ell \] and side of image square \[=\ell '\] From question,                                 \[\frac{{{\ell }^{,2}}}{\ell }=9\] or            \[\frac{\ell '}{\ell }=3\] i.e., magnification \[m=3\] \[u=-40\,cm\] \[v=3\times 40=120\,cm\] \[f=?\] From formula, \[\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\] \[\frac{1}{120}-\frac{1}{-40}=\frac{1}{f}\]                 or ,         \[\frac{1}{f}=\frac{1}{120}+\frac{1}{40}=\frac{1+3}{120}\]\[\therefore \]\[f=30\,cm\]


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