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?

Answer 1

`\triangle \phi=461.5rad`

Step-by-step 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=(2 \pi)/(\lambda)((wsin\theta )/(2))`

`\triangle \phi=(2 \pi)/(462*10^(-9))((0.1510^(-3)*sin 26.9 )/(2))`

`\triangle \phi=461.5rad`

`\triangle \phi=146.89\pi`