When a light is incident on the boundary of a medium with an angle greater than the respective critical angle, we obtain the angle of the refracted or reflected ray to be equal to the angle ofincidence following the laws of reflection.But total internal reflection occurs because of refraction so shouldn't the reflected ray follow Snell's law?
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$\begingroup$ The light that stays in the medium reflects. Refraction occurs only when passing into another medium. (For pedants: anywhere the refractive index changes.) $\endgroup$– garypCommented Jan 12, 2016 at 16:20
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$\begingroup$ Light can also refract around things like sharp edges right? $\endgroup$– Bill AlseptCommented May 5, 2018 at 6:22
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$\begingroup$ Duplicate of physics.stackexchange.com/questions/358615/… $\endgroup$– Gurbir SinghCommented Jul 17, 2018 at 3:52
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2 Answers
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we obtain the angle of the refracted or reflected ray to be equal to the angle of incidence
The angle of reflection is always the same as the angle of incidence. The angle of refraction follows Snell's law. If the incidence angle is larger than the critical angle, the angle of refraction would be greater that 90°. This is not possible so all the light is reflected, hence the total internal reflection.
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No. The law of reflection and the law of refraction (Snell's law) are two different formulas themselves. Both formulas are obtained simultaneously by electromagnetic optics (Maxwell's equations of a flat wave that changes from one medium to another). The demonstration is very common in engineering books, but I will not put it here to complicate your life.
Suppose an electromagnetic wave that travels on a medium (1) and impacts with an angle $\alpha$ over another medium (2), then there will be a reflected wave of angle $\beta$ and a refracted wave of angle $\delta$:
$$\underbrace{\alpha = \beta}_{\mbox{Reflection Law}} \quad \quad \quad \underbrace{\dfrac{\sin\left(\delta\right)}{\sin\left(\alpha\right)} = \dfrac{\gamma_1}{\gamma_2}}_{\mbox{Snell's Law}}$$
Where $\gamma_1$ and $\gamma_2$ are propagation factors, which depend on frequency ($f$) of the wave, and electrical permittivity ($\epsilon$), magnetic permeability ($\mu$) and conductivity ($\sigma$) of the medium. In particular cases where conductivities are relatively low ($\sigma_1 \cong \sigma_2 \cong 0$), Snell's law is reduced to Snell's law explained here.
Although Snell's law after the critical angle produces a reflection, conveniently called total internal reflection, this reflection is a product of refraction, and not the phenomenon of reflection itself.
In fact, the big difference is that the angle of the law of reflection does not depend on the frequency of the wave. In contrast, the refracted angle in the law of refraction does depend on the frequency of the wave.
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$\begingroup$ But how do I deduce that light obeys law of reflection during internal reflection from your answer? $\endgroup$– N.S.JOHNCommented Feb 8, 2017 at 4:30
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$\begingroup$ The law of reflection is not deduced from Snell's law. There are two totally different laws. Refraction = Refraction. Total Internal Reflection = a particular Refraction. But Reflection = only Reflection. $\endgroup$– NoirCommented Feb 8, 2017 at 4:58
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1$\begingroup$ And remember: Reflection = does not depend on the frequency of the wave or the characteristics of the medium. And Refraction = depends on the frequency of the wave and the characteristics of the medium. Therefore, a total internal reflection (particular refraction) does not necessarily have the angle of incidence equal to angle reflected, as the law of reflection says. $\endgroup$– NoirCommented Feb 8, 2017 at 5:05