# Light enters a substance from air at 45 degrees to the normal. It continues through the substance at 34.7 degrees to the normal. What would be the critical angle for this substance?

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### 1 Answer

**Answer: 53.6°**

**Recall Snell's Law:**

`n_1sin(\theta_1)=n_2sin(\theta_2)`

where n is the index of refraction of the material

and `\theta` is the angle the ray makes with the normal (perpendicular) of the surface.

`n_(air) = 1.0 ` by definition (since the speed of light is very close to the same as that of a vacuum)

The critical angle, `\theta_c` , is the angle of incidence of a ray such that the exiting ray will be along the surface. In other words, the ray angle of refraction is 90°. This only works when, in this case, if `n_2gtn_1`

Setting the output ray angle to 90° gives the following:

`n_1/n_2 = sin(\theta_c) `

*critical angle formula*

* Note that if `n_1 gt n_2` , the fraction will be greater than 1 - which means there is not a critical angle in that situation.*

Now that the concept is down, the solution is as follows:

let `n_1` be air, and `n_2` be our substance. `n_2 gt n_1`

Islolate n_2 in Snell's law, then plug in the result into the critical angle formula:

`n_2 = n_1(sin(\theta_1)/sin(\theta_2))`

plugging into critical angle formula:

`n_1/((n_1(sin(\theta_1)/sin(\theta_2)))) = sin(\theta_c)`

`rArr sin(\theta_2)/sin(\theta_1) = sin(\theta_c)`

`sin(\theta_c)=sin(34.7)/sin(45)`

`sin(\theta_c)~~0.805`

`:. theta_c = 53.6^o`

Hope that helps!

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