The direction is defined by the plane of the mirror, with the virtual source being the reflection of the real source over the plane of the mirror. The waves produced by this virtual source continue in the same direction they would be going if there was actually a source on the other side of the mirror, which is the angle of reflection. FIGURE 171 A light ray reflecting from a mirror shows that the angle of incidence equals the angle of reflection. This is why when you look at in a mirror, it looks like you are in the mirror, because it is as if the image of you is a new source that is reflected over the mirror surface (a "virtual" image). The waves do constructively interfere in the air so that its like the source of the wave is reflected over the metal surface (the source looks like it is "inside" the metal). In the case of a metal, these waves destructively interfere inside the metal and so very little of the wave penetrates into the metal. The angle of incidence O is always greater than the angle of reflection O must equal the angle of reflection O is always less than the angle of reflection O may be greater than, less than or equal to the angle of reflection This problem has been solved Youll get a detailed solution from a subject matter expert that helps you learn core concepts. Whenever a light wave is incident on a metal or an insulator like water or glass, each point on the surface radiates a wave that is either 180 degrees out of phase (for a metal or from a low refractive to a high refractive index) or in phase (from a high refractive index to a low index, like from inside glass to air). Answer: A) must equal angle of reflection Explanation: 'The angle of incidence is equal to the angle of reflection'. Ie the quickest path for the photon to travel between these two points is with the same incident and reflected angle, and according to Fermat's principle, that's what they will always do. So if calculate the derivative you will find that the answer looks exactly like incident path: (length of opposite side)/ hypotenuse = reflected path (opposite/hypotenuse) If you can find where t is minimal by when the derivative is zero. If surface has a length of 1 and the location where the photon gets reflected is X, A is at height h1, B at height h2, then the time it takes it to go from A to B is, If you start at point A, sometime during the journey reflect of a surface to bounce back to reach point B, then the angle of incident, reflection are completely unknown because you don't know at what point it hits the surface and gets reflected. I remember having to derive this in waves and optics, it's pretty easy to reach the conclusion that the angles mist be the same just with Fermat's principle, which states "the path taken by a ray between two given points is the path that can be traversed in the least time."
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