JEE Main & Advanced JEE Main Paper Phase-I (Held on 09-1-2020 Morning)

  • question_answer
    The aperture diameter of a telescope is 5 m. The separation between the moon and the earth is\[4\times {{10}^{5}}\text{ }km\]. With light of wavelength of 5500\[\overset{\text{o}}{\mathop{\text{A}}}\,\], the minimum separation between objects on the surface of moon, so that they are just resolved, is close to:      [JEE MAIN Held on 09-01-2020 Morning]

    A) 20 m    

    B) 200 m

    C) 600 m  

    D) 60 m

    Correct Answer: D

    Solution :

    [d] \[\theta =1.22\frac{\lambda }{a}\] Distance \[{{O}_{1}}{{O}_{2}}=\left( \theta  \right)d\] \[=\left( 1.22\frac{\lambda }{a} \right)d\] \[=\frac{\left( 1.22 \right)\left( 5500\times {{10}^{-10}} \right)\times 4\times {{10}^{5}}\times {{10}^{3}}}{5}\] \[=5368\times {{10}^{-2}}m\] \[=53.68m\]


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