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Carnegie Mellon University

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Problem 60 Hard Difficulty

Total Internal Reflection
In Fig. 33-58, light from ray $A$ refracts from material 1$\left(n_{1}=1.60\right)$ into a thin layer of material 2$\left(n_{2}=\right.$ $1.80 ),$ crosses that layer, and is then incident at the critical angle on the interface between materials 2 and 3 $\left(n_{3}=1.30\right) .$ (a) What is the value of incident angle $\theta_{A} ?$ (b) If $\theta_{A}$ is decreased, does part of the light refract into material 3?
Light from ray $B$ refracts from material 1 into the thin layer, crosses that layer, and is then incident at the critical angle on the interface between materials 2 and $3 .(\mathrm{c})$ What is the value of incident angle $\theta_{B} ?(\mathrm{d})$ If $\theta_{B}$ is decreased, does part of the light refract into material 3 ?

Answer

(a) $\theta_{A}=54.3^{\circ} .$
(b) Yes.
(c) $\theta_{B}=51.1^{\circ}$
(d) No.

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Video Transcript

here. Ah, we can For part A, we can use the condition in Equation 33 44. Ah, here. The condition required in the critical angle calculation here would be that 30 the angles of the status of three of the angle measure some three would be equal to 90 degrees. So this is would be the condition for equation 33 44. Thus, with ah here, we know that status of two is equaling to a critical angle. And we can simply say that according to Snell's law, for a fraction ah, and someone status of one equals and sub too. Sign of status of two and and sub three equal Sign of fate, US of three. Therefore, fate us of one with the equal to theta would be equal to arc sine of the refractive index of three divided by the refractive index of one. This is equaling 54.3 degrees. So we find Davis of one equal state of equals 54.3 degrees. This would be our final answer for part A now, for part B. If we were to reduce Seita, uh, this would lead to a reduction in status of two. And if we're reducing, um, the angle's up to, uh, it becomes they just have to becomes less than the critical angle. And if that occurs, uh, there will be some transmission of light into material three. So for party. Ah, the answer is yes. For part C, we know that here we're going to note that the compliment off the angle of refraction is the critical angle. So we can say compliment of the angle of refraction in material, too would be equal to other reconstructs. A is equal to the critical angle. But and given this, we know that sense of one sign of status of one equals answer to sign of status up to and so this would be equal to ends up to times the square root of one minus. Here it would be, and sub three divided by ends up to quantity squared. And so this would simply be equal to the square root of answer to squared minus and sub three squared, and we can then say that Seita would be equal to 51.1 degrees. This would be our answer for part C. And, uh, finally Oh, my apologies. this would simply be theta. And then this would be the critical angle. The fate of subsea. My apologies. And finally, four part D. We know that reducing Fada now ah, would lead to an increase of angle with which the light strikes the interface between materials two and three. So here it becomes greater than the than the critical angle. And, of course, if that's the case ah, there will be no transmission of light into material three. And so your answer for part through a four part D is no. No transmission of light into material three. That is the end of the solution. Thank you for watching.

Carnegie Mellon University
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Cornell University

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