# 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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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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