s. The average person thinks of laser beams as pencil-thin rays of light, but let's calculate what actually happens to a laser beam as it prop- agates a large distance away. Let a plane wave be incident on a wall with a hole in it. To turn the plane wave into a laser beam, let's make the hole a \"Gaussian hole," which has a transmission, x] = exp[{x;"u-']3], and, to simplify the problem, just consider the x-direction (the answer is the same in the ydirection if the hole is Gaussian in that direction also}. a. Compute the shape of the beam intensity. Hx'), a large distance 2 away [by \"shape,\"l mean just the functional form of the irradiance, so you can neglect any factors that don't depend on x' and need only use the \"proportional to.\" (ac) symbol, rather than \"equals\" when calculating the electric field]. b. If we call as in the above formula the \"spot size\" of the beam at the wall, use the same definition of spot size to determine the size of the beam {call it w') a distance, 2. away as a function of z, w. and xi- {the wavelength of the beam}. In other words, set the resulting exponential of the intensity far away equal to exp[(x'fw'}3] and solve for w'. c. More precisely, the Fraunhofer approximation says that kix3+y31 {HI 2, where z is the distance away from the aperture. Explain why this allows us to neglect the xii-f term in the diffraction integral. d. As long as the Fraunhofer approximation is valid. is the beam far away larger or smaller than that nearby