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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2002

  • question_answer
    When one of the slits of Youngs experiment is   covered   with   a transparent sheet of thickness 4.8 mm, the central shifts to a position originally occupies by the 30th bright fringe. What should be the thickness of the sheet if the central fringe has to shift to the position occupied by 20th bright fringe?

    A)  1.6 mm

    B)  3.8 mm

    C)  3.2 mm

    D)  7.6 mm

    Correct Answer: C

    Solution :

     t = thickness of sheet introduced between one of the plates of Youngs slits. \[\beta =\] width of one fringe. \[{{X}_{n}}=\] position of central fringe shifted. \[\therefore \] \[{{X}_{n}}=\frac{D}{2d}(\mu -1)t\] Where D = distance between slit and screen. \[\mu =\] refractive index. 2d= slit width. \[\therefore \] \[\] \[{{X}_{30}}=30\beta \] \[{{X}_{20}}=20\beta \] \[\therefore \] \[\frac{{{X}_{30}}}{{{X}_{20}}}=\frac{{{t}_{1}}}{{{t}_{2}}}\] \[\frac{30\beta }{20\beta }=\frac{4.8}{{{t}_{2}}}\] \[{{t}_{2}}=3.2mm.\]


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