A 0.15-mm-wide slit is illuminated by light of wavelength 462 nm. Consider a point P on a viewing screen on which the diffraction pattern of

Question

A 0.15-mm-wide slit is illuminated by light of wavelength 462 nm. Consider a point P on a viewing screen on which the diffraction pattern of the slit is viewed; the point is at 26.9° from the central axis of the slit. What is the phase difference between the Huygens’ wavelets arriving at P from the top and midpoint of the slit?

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Edana Edana 4 years 2021-07-13T05:35:27+00:00 1 Answers 3 views 0

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  1. thienhuong
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    2021-07-13T05:36:49+00:00
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    Answer:

    \triangle \phi=461.5rad

    Explanation:

    From the question we are told that:

    Silt width w=0.15=>0.1510^{-3}

    Wavelength \lambda=462nm=462*10^{-9}

    Angle \theta=26.9

    Generally the equation for Phase difference is mathematically given by

    \triangle \phi=\frac{2 \pi}{\lambda}(\frac{wsin\theta }{2})

    \triangle \phi=\frac{2 \pi}{462*10^{-9}}(\frac{0.1510^{-3}*sin 26.9 }{2})

    \triangle \phi=461.5rad

    \triangle \phi=146.89\pi

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