JEE Main & Advanced JEE Main Paper (Held on 7 May 2012)

  • question_answer
    In a Young's double slit experiment with light of wavelength\[\lambda ,\] fringe pattern on the screen has fringe width \[\beta .\] When two thin transparent glass(refractive index \[\mu \])plates of thickness \[{{t}_{1}}\] and \[{{t}_{2}}\]\[({{t}_{1}}>{{t}_{2}})\]are placed in the path of the two beams respectively, the fringe pattern will shift by a distance   JEE Main Online Paper (Held On 07 May 2012)

    A) \[\frac{\beta (\mu -1)}{\lambda }\left( \frac{{{t}_{1}}}{{{t}_{2}}} \right)\]             

    B)                        \[\frac{\mu \beta }{\lambda }\frac{{{t}_{1}}}{{{t}_{2}}}\]

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

    D)                        \[\left( \mu -1 \right)\frac{\lambda }{\beta }\left( {{t}_{1}}+{{t}_{2}} \right)\]

    Correct Answer: C

    Solution :

                    Shift\[=\frac{\beta \left( \mu -1 \right)}{\lambda }{{t}_{1}}-\frac{\beta \left( \mu -1 \right)}{\lambda }{{t}_{2}}\] \[=\frac{\beta \left( \mu -1 \right)}{\lambda }\left( {{t}_{1}}-{{t}_{2}} \right)\]


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