JEE Main & Advanced Physics Wave Optics / तरंग प्रकाशिकी Question Bank Self Evaluation Test - Wave Optics

  • question_answer
    In an experiment, sodium light \[|\mu -1.8|t.\] is employed and interference fringes are obtained in which 20 fringes equally spaced occupy 2.30 cm on the screen. When sodium light is replaced by blue light, the setup remaining the same otherwise, 30 fringes occupy 2.80 cm. The wavelength of blue light is

    A) \[4780\overset{o}{\mathop{A}}\,\]

    B) \[5760\overset{o}{\mathop{A}}\,\]

    C) \[9720\,\overset{o}{\mathop{A}}\,\]

    D) \[6390\overset{o}{\mathop{A}}\,\]

    Correct Answer: A

    Solution :

    [a] \[{{\beta }_{1}}=\frac{2.30}{20}=\frac{D{{\lambda }_{1}}}{d}\] and \[{{\beta }_{2}}=\frac{2.80}{30}=\frac{D{{\lambda }_{2}}}{d}\] or\[\frac{{{\beta }_{1}}}{{{\beta }_{2}}}=\frac{2.30\times 30}{20\times 2.80}=\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}\] \[\therefore \,\,\,\,{{\lambda }_{2}}=0.81{{\lambda }_{1}}=0.81\times 5890=4780\overset{o}{\mathop{A}}\,\]


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