Total Internal Reflection Suppose the prism of Fig. 33-53...

Total Internal Reflection Suppose the prism of Fig. 33-53 has apex angle $\phi=60.0^{\circ}$ and index of refraction $n=1.60 .$ (a) What is the smallest angle of incidence $\theta$ for which a ray can enter the left face of the prism and exit the right face? (b) What angle of incidence $\theta$ is required for the ray to exit the prism with an identical angle $\theta$ for its refraction, as it does in Fig. 33-53?

Total Internal Reflection
Suppose the prism of Fig. 33-53 has apex angle $\phi=60.0^{\circ}$ and index of refraction $n=1.60 .$ (a) What is the smallest angle of incidence $\theta$ for which a ray can enter the left face of the prism and exit the right face? (b) What angle of incidence $\theta$ is required for the ray to exit the prism with an identical angle $\theta$ for its refraction, as it does in Fig. 33-53?

Key Concepts

Prism Geometry

The study of the geometric properties of prisms, including the apex angle, which influences the angles at which light enters and exits the prism. This understanding is key to predicting how light behaves inside the prism and to solving problems involving refraction and total internal reflection.

Critical Angle

The minimum angle of incidence in the denser medium at which total internal reflection occurs. This angle is determined using Snell's law and is crucial for determining the conditions under which light is completely internally reflected.

Snell's Law

A fundamental principle in optics that relates the angles of incidence and refraction to the indices of refraction of two media. It explains how light bends when transitioning between different materials and is central to analyzing the behavior of light in prisms.

Total Internal Reflection

A phenomenon that occurs when light traveling within a denser medium strikes the interface with a less dense medium at an angle greater than the critical angle. Under these conditions, all the light is reflected back into the denser medium instead of being refracted.

Transcript

00:01 We can say that the apex angle of the prism is going to be phi, and this is equaling 60 degrees.

00:07 And we have the index of, the refractive index of the prism, n equally 1 .6.

00:14 So we can say that for part a, if the ray has to minimally exit the second surface, then at the second surface incident angle, that should be the critical angle.

00:28 So we can say that here, we can first say that we are going to let the incident angle at the first face be theta sub i.

00:40 And this would be refracted at the first surface.

00:44 This would be theta sub r.

00:48 We can say that theta sub c would then be the critical angle, critically incident angle at the second surface.

00:57 So now we can use the snell's law of refuge.

01:00 We can say sign of the critical angle would be equal to 1 over n and so the critical angle subc would be equal to arc sign of 1 over 1 .6 the refractive index of the prism itself and we find that the critical angle is 38 .68 degrees at this point we know that theta sub c prime would simply be equal to 90 degrees minus theta sub c.

01:30 This is going to be equal to essentially 51 .32 degrees.

01:40 We can say similarly we can say that theta sub r prime would be equal to 90 degrees minus theta sub r.

01:51 And we can say that phi plus theta subr prime plus theta subc prime.

02:00 This should be equal to the total angle.

02:02 And a triangle 180 degrees.

02:06 We can then say that theta sub r prime would be equal to 180 degrees minus 60 degrees minus 501 .32 degrees.

02:20 This is equaling theta sub r prime 68 .68 degrees.

02:31 And so we can say that the refracted angle theta sub r would simply be equal to 90 degrees minus 68 .68.

02:40 This is equaling 21 .32 degrees.

02:46 And then we're going to apply the angle of refraction for the first interface of the prism.

02:53 So we can say that sine of theta sub r would be equal to n sine of theta sub i.

03:05 And so 3.

03:06 Theta sub i would be equal to arc sign of 1 .60 times sign of 21 .32 degrees.

03:18 This is equaling 35 .6 degrees.

03:23 So the smallest angle of incidence at which the ray can enter the left end and reach the right end of the prism is going to be theta sub i.

03:32 Again, 35 .6 degrees...

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