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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?

Sagot :

okpalawalter8

Answer:

[tex]\triangle \phi=461.5rad[/tex]

Explanation:

From the question we are told that:

Silt width [tex]w=0.15=>0.1510^{-3}[/tex]

Wavelength [tex]\lambda=462nm=462*10^{-9}[/tex]

Angle [tex]\theta=26.9[/tex]

Generally the equation for Phase difference is mathematically given by

[tex]\triangle \phi=\frac{2 \pi}{\lambda}(\frac{wsin\theta }{2})[/tex]

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

[tex]\triangle \phi=461.5rad[/tex]

[tex]\triangle \phi=146.89\pi[/tex]

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