I assume that by blocking an arm you mean that you prevent light from going through that arm. The Michelson Interferometer causes interference between two parts of the incident wave, that has been divided by the beam splitter. If you prevent light to go through one arm of the Interferometer, for instance by blocking it, you can't get the interference anymore, so what you get at the revealing component is the intensity of the light that goes through the allowed arm, no matter how long it is. So, if you represent the transmittance as a function of the arm's length, you will see that it is constant. In the second question, I think that the spectrum you are talking about is the transmittance, as function of the frequency, of the interferometer. In this case, for a fixed difference in the length of the two arms, you have to evaluate the difference in the phase of the two waves (the one through the first arm e the one through the second one) for a fixed difference of optical path. If this difference of path is an integer multiple of the wavelength, you have constructively interference and a maximum in transmittance; if it is an odd multiple of half the wavelength, waves destructively interfere and you get a minimum. For every other difference (that is, for every other wavelength), you have an intermediate value of transmittance. Therefore, you can see that the transmittance is periodic, with maximum every multiple of the wavelength.