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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 38

  • question_answer
    In a YDSE, if the incident light consists of two wavelengths \[{{\lambda }_{1}}\] and \[{{\lambda }_{2}},\] the slit separation is d and the distance between the slit and the screen is D, the maxima due to each wavelength will coincide at a distance from the central maxima, is given by

    A) \[\frac{{{\lambda }_{1}}+{{\lambda }_{2}}}{2Dd}\]    

    B)         LCM of \[\frac{{{\lambda }_{1}}D}{d}\]and \[\frac{{{\lambda }_{2}}D}{d}\]

    C) \[\left( {{\lambda }_{1}}-{{\lambda }_{2}} \right)\frac{2D}{D}\]       

    D) HCF of \[\frac{{{\lambda }_{1}}D}{d}\] and \[\frac{{{\lambda }_{2}}D}{d}\]

    Correct Answer: B

    Solution :

    [b] \[\frac{{{n}_{1}}D{{\lambda }_{1}}}{d}=\frac{{{n}_{2}}D{{\lambda }_{2}}}{d};\] \[\frac{{{n}_{1}}}{{{n}_{2}}}=\left( \frac{D{{\lambda }_{2}}}{d} \right)/\left( \frac{D{{\lambda }_{1}}}{d} \right)\] LCM of \[\frac{D{{\lambda }_{1}}}{d}\]and \[\frac{D{{\lambda }_{2}}}{d}\]


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