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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2008

  • question_answer
    In a two slit experiment with monochromatic light fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by \[5\times {{10}^{-2}}m\] towards the slits, the change in fringe width is \[3\times {{10}^{-5}}m\]. If  separation between the slits is \[{{10}^{-3}}m\], the wavelength of light used is

    A)  \[6000\,\,\overset{o}{\mathop{A}}\,\]                

    B)  \[5000\,\,\overset{o}{\mathop{A}}\,\]

    C)  \[3000\,\,\overset{o}{\mathop{A}}\,\]                

    D)  \[4500\,\,\overset{o}{\mathop{A}}\,\]

    Correct Answer: A

    Solution :

                    \[\beta =\frac{\lambda D}{d}\] \[\Rightarrow \]               \[\beta \propto D\] \[\Rightarrow \]               \[\frac{{{\beta }_{1}}}{{{\beta }_{2}}}=\frac{{{D}_{1}}}{{{D}_{2}}}\] \[\Rightarrow \]               \[\frac{{{\beta }_{1}}-{{\beta }_{2}}}{{{\beta }_{2}}}=\frac{{{D}_{1}}-{{D}_{2}}}{{{D}_{2}}}\] \[\Rightarrow \]               \[\frac{\Delta \beta }{\Delta D}=\frac{{{\beta }_{2}}}{{{D}_{2}}}=\frac{{{\lambda }_{2}}}{{{d}_{2}}}\] \[\Rightarrow \]               \[{{\lambda }_{2}}=\frac{3\times {{10}^{-5}}}{5\times {{10}^{-2}}}\times {{10}^{-3}}\]                 \[=6\times {{10}^{-7}}m=6000\,\overset{o}{\mathop{A}}\,\]


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