JEE Main & Advanced Physics Wave Optics / तरंग प्रकाशिकी Question Bank Critical Thinking

  • question_answer
    A point source of electromagnetic radiation has an average power output of 800 W. The maximum value of electric field at a distance 4.0 m from the source is

    A)            64.7 V/m                                

    B)            57.8 V/m

    C)            56.72 V/m                             

    D)            54.77 V/m

    Correct Answer: D

    Solution :

               Intensity of EM wave is given by \[I=\frac{P}{4\pi {{R}^{2}}}={{v}_{av}}.c=\frac{1}{2}{{\varepsilon }_{0}}E_{0}^{2}\times c\] Þ \[{{E}_{0}}=\sqrt{\frac{P}{2\pi {{R}^{2}}{{\varepsilon }_{0}}c}}\]          \[=\sqrt{\frac{800}{2\times 3.14\times {{(4)}^{2}}\times 8.85\times {{10}^{-12}}\times 3\times {{10}^{8}}}}\]           = 54.77 \[\frac{V}{m}\]


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