Why does light not refract at 90 degrees

CBSE 10 - Physics

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Why does light not refract at 90 degrees

Asked by loversnaruto013 | 03 Sep, 2021, 08:23: PM

Why does light not refract at 90 degrees
ANSWERED BY EXPERT

CBSE 10 - Physics

Asked by rajatsinghbsr19 | 26 May, 2021, 08:11: PM

ANSWERED BY EXPERT

I don't exactly know how to ask the question but, if I understand correctly refraction occurs when the medium that light is passing through changes. So why does this not occur when the light enters the new medium at 90 degrees?

The reason that the light is not refracting is that it will take the path that will take the shortest time from through the medium. This is essentially what snell's law gives us. Since, we are looking at light at normal incidence there is no path that will take shorter time than to continue straight forward in the same medium.

You can think of it as a life guard who is trying to save a person drowning in a river. If the river is flowing and the life guard will run towards the person at an angle(on land) and he takes the current into account as he is faster on land than in the water. Thus, having different angles on land and in the water. If the water is still and the person needing help is straight out from the shore the life guard will run in a straight line to the shore and then swim straight out to rescue the person in the water.

Snell's law can be derived from Fermat's Principle which is the statement that the light will take the path that uses the least time. See the wikipedia article Snell's law for the derivation of the formula.

as mentioned by Aaron since the $\theta_1$ is zero, thus $\sin(\theta_1)=0$, $\theta_2$ needs to be zero for Snell's law to be satisfied. Below is snell's law: $$n_1\sin(\theta _1)=n_2\sin(\theta _2)$$

It is also possible to derive this directly form Maxwell's equations but that would be a boring and time consuming task.

Also note that if we are going from a medium with higher refractive index to a medium with lower refractive index we can get a situation where we will need $sin(\theta_2)$ to be greater than 1, this will result in a total internal reflection where no light will pass trough the interface.

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