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JEE Main & Advanced JEE Main Paper (Held On 9 April 2016)

  • question_answer
    In Young's double slit experiment, the distance between slits and the screen is 1.0 m and monochromatic light of 600 nm is being used. A person standing near the slits is looking at the fringe pattern . When the separation between the slits is varied, the interference pattern disappears for a particular distance \[{{d}_{0}}\] between the slits. If the angular resolution of the eye is\[\frac{{{1}^{o}}}{60}\]the value of \[{{d}_{0}}\] is close to   JEE Main Online Paper (Held On 09 April 2016)

    A) 2 mm                                    

    B) 1 mm                

    C) 3 mm                                    

    D) 4 mm

    Correct Answer: A

    Solution :

                    Angular fringe width \[\theta =\frac{\beta }{D}=\frac{\lambda }{d}\] \[\frac{\lambda }{{{d}_{0}}}=\frac{{{1}^{o}}}{60}=\frac{\pi }{180\times 60}\] \[{{d}_{0}}=\lambda \left( \frac{180\times 60}{\pi } \right)\]\[=2\times {{10}^{-3}}m=2mm.\]                


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