JEE Main & Advanced Sample Paper JEE Main Sample Paper-43

  • question_answer
    A photosensitive material is at 9 m to the left of the origin and the source of light is at 7 m to the right of the origin along \[x\]-axis. The photosensitive material and the source of light start from rest and moves with 8i m/s and 4i m/s respectively. The ratio of intensity at \[t=0\] to\[t=3\]s as received by the photosensitive material is

    A)  16: 1                          

    B)  1 : 16           

    C)  2 : 7                           

    D)  7 : 2

    Correct Answer: B

    Solution :

     The   separation   between   source   and photosensitive material at \[{{\phi }_{1}}=\] is 16m, therefore, intensity received by photosensitive material at \[=BA\] is \[{{\phi }_{2}}=\], where P is the power of source of light. At \[||\]s, the source is at (15, 0) and detector is at (19, 0), so the separation between them is 4 m. \[=0\]             So,       \[\therefore \,\,|\xi |\,=\,\left| \frac{-\Delta \phi }{\Delta t} \right|=\,\left| -\left( \frac{0-BA}{\frac{T}{4}-0} \right) \right|=\frac{4BA}{T}\]

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