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2. (20 pts.) Fresnel Diffraction by a Thin Wire: Fresnel diffraction is observed on a screen 3 m behind a wire of thickness 0.37 mm, which is placed 2 m from the light source of 543.5 nm. For parts a and b, treat the regions above and below the wire as if they are two apertures. a. Determine the value of the Fresnel Integrals for point P from the light coming from the region above the wire. b. Repeat part a for the region below the wire. c. When waves from above and below the wire reach point P, the total irradiance at P goes as the following: 1 = lo [(C(V21) - C(V11) + C(V2u) - C(V₁u))² + (S(V21) - S(V11) + S(V2u) - S(V₁u))²] where the subscript u refers to the contributions from the upper region (above the wire), and the subscript I refers to the contributions from the lower region (below the wire). Using this equation and the results of parts a and b, what is the total irradiance at point P? Compare this irradiance to the unobstructed wave irradiance, lu. Screen 0.37 mm Wire Source S 3 m 2 m