(Solved): 1. Fraunhofer diffraction [40 pt] Consider a sinusoidal amplitude grating wit ...

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1. Fraunhofer diffraction [40 pt] Consider a sinusoidal amplitude grating with a very small thickness. It can be defined by the amplitude transmittance function below: $$t_A\left(x_0,y_0\right)=\left[\frac{1}{2}+\frac{m}{2}\cos \left(2?f_o{x}_0\right)\right]\mathrm{rect}\left[\frac{x_0}{2w}\right]\mathrm{rect}\left[\frac{y_0}{2w}\right]$$ The grating structure is bounded by a square aperture of width $$2w$$ in both $$x_0$$ and $$y_0$$ direction, as indicated by the rect function. The parameter $$m$$ represents the peak-to-peak change of the amplitude transmittance across the grating, and $$f_o$$ is the spatial frequency of the grating in $$x_0$$ direction. (1) The grating is located at plane $$z=0$$. A plane wave transmits through the grating in the surface normal direction. Using Fraunhofer diffraction integral, calculate the expression of the electrical field $$E(x,y,z)$$ at plane $$z$$. (2) If $$f_o?1/w$$, for a fixed $$y_1$$ and $$z_1$$, roughly sketch the light intensity $$I\left(x,y_1,z_1\right)$$ versus $$x$$. Please make necessary labels on the $$x$$ axis in the plot. piane $$U(z=U)$$