where path AB is incidence ray and path BC is reflected ray. This makes the incident angle equal to reflection angle. Note: The laws of reflection are the fundamental principle behind the solution to this question, it states that when a ray of light reflects a smooth surface, the angle of incidence is always equal to the angle of reflection, and another law behind this rule is that at the point of incidence, the incident ray, the reflected ray and the normal ray all lie in the same plane. When light is reflected from a surface, the angle of incidence is always equal to the angle of reflection, where both angles are measured from the path of. According to Quantum electrodynamics, in case of a perfectly elastic collision the momentum is conserved and the light chooses the path which takes the least time to reach the final point which develops a path. They may appear unequal when the surface from which light reflects is unequal, yet the law holds valid when examined at extremely small surface areas. However, according to the first law of reflection, Angle of incidence equals angle of reflection at a small surface area. The angle of reflection at which light bounces off the surface should be equal to the angle of incidence at which light strikes the surface. $x_i \neq x_r$, where $x_i$ is angle of incidence and $x_r$ is angle of reflection. However, according to the first law of reflection, Angle of incidence equals angle of reflection at a small surface area. if the surface has some irregularities) then the angle of incidence is not equal to the angle of reflection. $\theta_i = \theta_r$, where $\theta_i$ is angle of incidence and $\theta_r$ is angle of reflection. As we know according to the law of reflection angle of incidence is equal to angle of reflection this is true if and only if, if the surface on which light strikes is smooth. The angle between the incoming light ray and the normal is called the angle of incidence (i) and the angle between the outgoing light ray is called the angle of.
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