JEE Main & Advanced Sample Paper JEE Main Sample Paper-4

  • question_answer
    In YDSE screen is kept 0.8 m from the slits. The coherent sources are 0.016 cm apart & fringes are observed on the screen. It is found that with a certain monochromatic source of light, the fourth bright fringe is situated at a distance of 1.06 cm from the central fringe. The wavelength of light used is

    A)  480nm                                

    B)  530 nm  

    C)  650 nm                               

    D)  580 nm

    Correct Answer: B

    Solution :

    For maxima, \[\Delta x=n\lambda \]\[\Rightarrow \]\[\frac{dy}{D}=n\lambda \] For fourth maxima, n = 4 \[\frac{dy}{D}=4\lambda \Rightarrow \lambda =\frac{dy}{4D}=\frac{0.016\times {{10}^{-2}}\times 1.06\times {{10}^{-2}}}{4\times 0.8}\] \[\Rightarrow \lambda =530nm\]

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