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College physics chapter 10 optical problems. How to judge whether the interference is strengthened or weakened?
First of all, I tell you that in different jamming systems, there are different explanations for whether the interference is strengthened or weakened. For example, give your first example. In this problem, it is obvious that it is a thin film interference system, which belongs to vertical incident isoclinic interference. However, both the top and bottom of the film are air. When light enters the first interface for the first time, there is no half-wave loss because the light-hydrophobic medium enters the light-dense medium, but on the lower surface of the film. There is a half-wave loss at this time. According to the interference enhancement formula, the optical path difference should be equal to an integer multiple of the wavelength, but at this time the optical path difference is 2nh+λ/2=mλ, and that λ/2 is the additional optical path difference caused by half-wave loss. But you also said in your second question that air is above the ice and glass is below, and the refractive index of glass is greater than that of ice. At this time, the optically hydrophobic medium still enters the optically dense medium on the lower surface of the ice. ! ! ! Therefore, when doing thin film interference, whether the interference is strengthened or weakened, the problem of half-wave loss should be considered. There is no half-wave loss of refracted light and reflected light when entering the optical density vertically, but there is no half-wave loss of transmitted light when entering the optical density, but there is reflected light. If the reflected light can't interfere with the light above, we should consider adding half-wave loss! !

The problems analyzed above exist not only in thin film interference, but also in any interference problem with equal thickness and inclination!

As for the fractional wavefront, such as Young's double-slit interference, because it always propagates in the air, the problem of half-wave loss is not considered, and it can be calculated directly according to the integral multiple enhancement of wavelength!