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Light enters a substance from air at 45 degrees to the normal.  It continues through...

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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?

Posted by lkehoe on April 24, 2013 at 11:35 PM via web and tagged with air, critical angle, optics, physics, science, substance

High School Teacher

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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!

Posted by quirozd on April 25, 2013 at 1:27 AM (Answer #1)

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