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NEET Sample Paper NEET Sample Test Paper-67

  • question_answer
    In Young's double slit experiment, the 8th maximum with wavelength \[{{\lambda }_{1}}\] is at a distance \[{{d}_{1}}\] from the central maximum and the 6th maximum with a wavelength \[{{\lambda }_{2}}\] is at a distance \[{{d}_{2}}\]. Then \[({{d}_{1}}/{{d}_{2}})\] is equal to

    A) \[\frac{4}{3}\,\left( \frac{{{\lambda }_{2}}}{{{\lambda }_{1}}} \right)\]  

    B) \[\frac{4}{3}\,\left( \frac{{{\lambda }_{1}}}{{{\lambda }_{2}}} \right)\]

    C) \[\frac{3}{4}\,\left( \frac{{{\lambda }_{2}}}{{{\lambda }_{1}}} \right)\]              

    D) \[\frac{3}{4}\,\left( \frac{{{\lambda }_{1}}}{{{\lambda }_{2}}} \right)\]

    Correct Answer: B

    Solution :

    Position of m maxima from central maxima is given by \[{{x}_{n}}=\,\,\frac{n\lambda D}{d}\] \[\Rightarrow \,\,\,{{x}_{n}}\propto n\lambda \,\,\,\,\Rightarrow \,\,\,\frac{{{d}_{1}}}{{{d}_{2}}}\,=\,\frac{{{n}_{1}}{{\lambda }_{1}}}{{{n}_{2}}{{\lambda }_{2}}}\,=\,\,\frac{8{{\lambda }_{1}}}{6{{\lambda }_{2}}}\,=\,\frac{4}{3}\,\left( \frac{{{\lambda }_{1}}}{{{\lambda }_{2}}} \right)\]


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