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

  • question_answer
    The periodic time of rotation of a certain star is 22 days and its radius is 7 ' 108 metres. If the wavelength of light emitted by its surface be\[4320\,\,{AA}\], the Doppler shift will be (1 day = 86400 sec)                                         [MP PET 2001]

    A) \[0.033\,\,{AA}\]

    B) \[0.33\,\,{AA}\]

    C) \[3.3\,\,{AA}\]                   

    D) \[33\,\,{AA}\]

    Correct Answer: A

    Solution :

    \[\Delta \lambda =\lambda .\frac{v}{c}\]where \[v=r\omega =r\times \left( \frac{2\pi }{T} \right)\]            \[\therefore \Delta \lambda =\frac{4320\times 7\times {{10}^{8}}\times 2\times 3.14}{3\times {{10}^{8}}\times 22\times 86400}=0.033{AA}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner