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A beam of laser light with wavelength 612 nm is directed through a slab of glass having index of refraction 1.78. (a) For what minimum incident angle would a ray of light undergo total internal reflection? (b) If a layer of water is placed over the glass, what is the minimum angle of incidence on the glass–water interface that will result in total internal reflection at the water–air interface? (c) Does the thickness of the water layer or glass affect the result? (d) Does the index of refraction of the intervening layer affect the result?

a. $34.2^{\circ}$

b.$34.2^{\circ}$

c. no

d. no

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Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Hope College

McMaster University

part A of our question wants us to figure out the minimum incident angle a ray of light under under ghost, total internal reflection that's needed. So we need to find this minimum incident angle or this critical angle, considering the fact that the index refraction of the glass in sub d is 1.78 and index refraction of air is 1.0 Okay, so to, uh, To do this, we're gonna use Snell's law of refraction to find the minimum angle. So Snell's law tells us that in one times this sign of the angle one which we're gonna call data one is equal to into times this sign data too. But if things are completely internally reflected the minimum angle that that would happen, that is, when faded, too is equal to 90 degrees. If they tatoos equal to 90 degrees in the sign of 90 degrees is equal. The won their fourth data One would now become the critical angle which we could call. Say this up, see. So for part, eh, let's go ahead and solve for that critical angle, though, so data sub sea is going to be equal to the inverse sine, uh, the index of refraction to to the, uh, the ratio of the index of refraction to to index of refraction one. But here the index every fraction, too, is the air, right, cause that's the second medium and the index of refraction. One is the glass and sub G. So taking the inverse sine of these two, we find that this is equal to 34 0.2 degrees. We can go ahead and box it. And as our answer for part a part B says, if a layer of water is placed over the glass, what is the minimum angle of incident on the glass water interface that will result in total internal reflection to the water air interface? Okay, so we want total internal reflection to the water air interface. So for cart be, we want the critical angle, which here is going to be equal to the look. Let's re write that again. Using cells. Snell's law the inverse side in two times the sign data to which is index of refraction for water times the sign of the angle state and W divided by and Sigi. But since we wanted to be totally internally reflected in the air. We know that in W times, the sign of fated W has to be equal to in sub a right cause. Insulate times sign of today. Which data is 90 degrees, right? Cause we want totally internal effect reflected. That means that this sign is equal to in Sunday, Divided by and Sigi. So again, this is equal to 34.2 degrees. That could be boxing as our answer for part B. Okay, Parsi says. Does the thickness of the water layer or the glass affect the results? Well, the results are affected by Snell's law, and Snell's law on Lee has the index of refraction in the angle of incident, and there is no thickness into a dependence there, so we can say no. According to Snell's law, the thickness of the medium does not matter so we can box it in as our solution for part C. And then finally, Part D Part D says, Does the index of refraction of the inverting layer affect the results? Well, the, uh, excuse me intervening later, the intervening layer is the water. But if you look at part one or excuse me part B the index of refraction of the water times the sine of the angle of incidence of the water is just gonna be equal to the index of refraction of air. Since that's the third medium, and we want it totally internally reflected. And since that's always gonna be true, that index, every fraction of water is not going to depend on the final solution. The final solution is just gonna depend on the index of refraction of air. So we can say, uh, no, according to be part B on Lee, the index of refraction, of air and glass matters. And that committee boxed in is our solution for a party in the final solution to our question.