DUMET Medical DUMET Medical Solved Paper-2004

  • question_answer
    In Youngs double slit experiment, the fringe width is found to be 0.4 mm. If the whole apparatus is immersed in water of refractive index 4/3, without disturbing the geometrical arrangement, the new fringe width will be :

    A)  0.30mm       

    B)  0.40mm

    C)  0.53mm      

    D)  450 urn

    Correct Answer: A

    Solution :

     Key Idea: When light passes from one medium to another, its wavelength changes. In Youngs double slit experiment if distance between coherent sources is d, and distance between screen and source is D, then fringe width W is given by \[W=\frac{D\,\,\lambda }{d}\] ?. (i) where \[\lambda \]is wavelength of monochromatic light source. Also, from Huygenss principle \[\mu =\frac{\lambda }{{{\lambda }_{w}}}=\] refractive index In water \[W=\frac{D\,{{\lambda }_{w}}}{d}\] .... (ii) Dividing Eq. (i) by Eq. (ii), we get \[\frac{W}{W}=\frac{\lambda }{{{\lambda }_{w}}}\] \[\Rightarrow \] \[W=\frac{{{\lambda }_{w}}}{\lambda }W\] \[=\frac{W}{\mu }\] Given, \[W=0.4\,mm,\,\,\mu =4/3\], we have \[W=\frac{0.4}{4/3}=0.3\,\,mm\]


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