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AIIMS AIIMS Solved Paper-2002

  • question_answer
    Light of wavelength 6000\[E=2V,\,r=0.1\Omega ,R=3.9\Omega \] is reflected at nearly normal incidence from a soap films of refractive index 1.4. The least thickness of the fringe then will appear black is:

    A)  infinity    

    B)                                                        200\[V=2-\left( \frac{2}{3.9+0.1} \right)\times 0.1\]                       

    C)        2000 \[V=1.95V\]         

    D)                        1000\[{{m}_{1}}{{u}_{1}}={{m}_{2}}{{v}_{2}}\]

    Correct Answer: C

    Solution :

                    Key Idea: Black fringe means destructive interference. For destructive interference, we have \[=\frac{(2n+1)\lambda }{2}\] where \[E=\sigma {{T}^{4}}\] is refractive index, \[\sigma \] is wavelength and is thickness. \[\sigma \]   \[\frac{E}{E}=-\frac{2Q}{Q}\] For minima, we have \[\Rightarrow \] \[E=-\frac{E}{2}\]    \[3.9\Omega \] \[\therefore \]


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