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The drawing shows three layers of different materials, with air above and below the layers. The interfaces between the layers are parallel. The index of refraction ...

Question

The drawing shows three layers of different materials, with air above and below the layers. The interfaces between the layers are parallel. The index of refraction of each layer is given in the drawing. Identical rays of light are sent into the layers, and light zigzags through each layer, reflecting from the top and bottom surfaces. The index of refraction for air is nair 1.00. For each layer, the ray of light has an angle of incidence of 75.0. For the cases in which total internal refection i

The drawing shows three layers of different materials, with air above and below the layers. The interfaces between the layers are parallel. The index of refraction of each layer is given in the drawing. Identical rays of light are sent into the layers, and light zigzags through each layer, reflecting from the top and bottom surfaces. The index of refraction for air is nair 1.00. For each layer, the ray of light has an angle of incidence of 75.0. For the cases in which total internal refection is possible from either the top or bottom surface of a layer, determine the amount by which the angle of incidence exceeds the critical angle.



Answers

The drawing shows three layers of different materials, with air above and below the layers. The interfaces between the layers are parallel. The index of refraction of each layer is given in the drawing. Identical rays of light are sent into the layers, and light zigzags through each layer, reflecting from the top and bottom surfaces. The index of refraction for air is nair
1.00. For each layer, the ray of light has an angle of incidence of 75.0. For the cases in which total internal refection is possible from either the top or bottom surface of a layer, determine the amount by which the angle of incidence exceeds the critical angle.

So here we know that from Snows Law of Refraction on sub one sign of status of one equals and sub to sign of fate is up to We also know that and some two of sign of status of two equals ends up three. Sign of status of three. So essentially by transitive property and someone sign of status of one equals ends of three. Sign of status of three. Now, given this, uh, this implies simply that fate us of one equal status of three. Ah, when the refractive index of material one equals, they're effective and except three. So we're given that status of one equals 40 40 degrees. And now we're going to analyze figure 33. 50 a. And essentially, once we find, uh, we're going to look for a point and and where fate us of three equals 40 degrees. Analyzing figure 33 50 me. And we find that here the refractive index and sub one is gonna equal 1.6 because this is where ends up three is equaling 1.6. So essentially because of the transitive property making this relationship here and were given that state of sub one equals 40 degrees. Um, in figure 33. By analyzing figure 33. 50 a, we go to 33. 50 b c. The point where status of three is equaling 40 degrees. And this is where the refractive index of three equals 1.6, therefore, and someone also equals 1.6. So here, this would be our answer for part A For part B. We know that the first step in our solution to part a shows that, um here and the information concerning and sub to disappears. So here we cannot find and step two ends up to cannot be found because, um, essentially, it can't, uh, because ends up to sign of status of two is cancelled out. Bye. Algebraic manipulation nip in part. So we need more information in order to find answers to However, in part See, we can say that 1.6 sign of 70 degrees is equaling 2.4. Sign of status of three. They're forced status of three would be equal to arc sine of 1.6 sign of 70 degrees, divided by 2.4. And so status of three is equaling 39 degrees. This would be our answer for part C. That is the end of the solution. Thank you. For what

35.72. This is a pretty equal problem. So we have a Nair wedge type situation between two pieces of glass. Fix together. They're touching at one end and separated. But the other, uh, in this in this situation there held in position. And so while when there's air and between them, we see 4001 fringes and then we remove the air. But the separation between the plates of the angle between the remains the same because they're being held and we lose a dark friends. And so from this we can calculate the index of refraction of air to six significant figures. It was pretty, Uh, yeah, so we first situation we have, and 4001 and two is equal to what we're looking for. So we'll have to l the 4001 clammed up over and of air. In our second situation, we have 4000 m this case and two is one, because that's my definition. The index of refraction of vacuum. So we have two equals 4000 times divided by n to butt into his one. So let's just leave it out so we don't know or care what l and, uh, lambda are. So we can, uh, divine both of these equations by land and then set them equal to each other. So we have 4001 divided by the index of refraction of air is equal 4000. And so then just doing the rearrangement division, we find that the index of refraction of air is 1.0 zero zero two. Bye. That's it. Yeah, because this is really close to one, which is why we usually just say it's one in most problems. But if it were, if it's relevant for some reason, uh, you know, now you know that it's slightly greater than one. And, you know, like in this case, having a a vacuum versus Aaron between these piece of glass, um, you know, makes you who was a dark fringe and you might be very confused about that. He didn't remember that. Hair has, ah, the index of refraction that's greater than one

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.

So we can say for part A we're going to suppose that we have a total of end transparent layers. In this case, we have any equals five transparent layers we're going to layer the We're gonna label these layers from left to right with indices 12345 up until, uh, you know, up until n number of layers. And so the indices simply dino to the layer. And so we can say that we're goingto let the index of refraction of the air be and subzero, So this would be a WR the refractive index of the air. And we're going to denote the initial angle of incidence of the light gray upon the air layer boundary as well. Say fate, Issa bhai. So this would be the angle of Indus incidence upon the air layer. And then we can say that the angle of the emerging right Ray would be so this would be incident Ray. And then we can say status of f would be emerging right. And at this point, we can then say that Ah, since all the boundaries are parallel to one another, the the angle of incidence I subject this would be the we can say angle of incidence incident angle between the J, the jays, the J F layer and the J plus one if layer essentially. So we can say that this is going to be the same as the angle between the transmitted light ray and the normal in the J F layer again. So for the first boundary, we have n subs one divided by and not would be the again the, uh, in the er refractive index of air. So this is the sign of status of I divided by sign of data sub one for the second boundary. Uh, this would be for the first for the second we have and sub two over ends of one equaling sign of fate us of one divided by sign of fate, us of two. And then finally we have the last boundary. And so oh, divide by and some end equal sign of Fada and divided by sign of fajita sub f. So we're going to multiply these equations. Uh, we can say that here and sub one over and not times ends up to over ends of one times ends of three over ends up to and and so on will be equal to rather up until essentially and not to end to the end. This would be equal to sign of fate A sip I sign of Fada. So one times sign of fate US of one divided by sign of Fada said too up until again Sign of fade us of and divide by sign of fate a sabbeth final. Now we're going to see that the left hand side of the equation above can be reduced to an O over an Oh, so essentially, while the and the right hand side of the equation is equaling Sign of I divide by sign of F. So essentially we can say sign of data some f would be equal to and not over and knocked. This would be equal to sign of status of I equaling sign of fate us of I given that these cancel out so essentially the incidents Ray would be equal to the final angle. So for the two light raising the problem statement, we can say that the angle of the emerging light rays are both the same as their respective incident angles. Um, this would be our answer essentially, for part a four part be given that data f equals status of ethical. Zero for Ray A We can say for part B, that's it for part B. Status of F equals 20 degrees for Ray Be Oh, sorry. This would be my apologies. This would be our answer for party. This would be an answer for part C and then four part for part B for part C. However, we all we need to do here is change the value of and not equaling 1.0, too, and not equal in 1.5 for glass. So, uh, but again, because in the initial, uh, part A they both cancel out. So it doesn't really matter what the medium is. Um, So essentially, for part C rather fada sub f equals zero for Bray A once again and four part D. We can say status of F equals 20 degrees for Ray. Be so that would be our final answer. Um, we know that here it is the independent of the number of layers and the thickness of the layers and the refractive index of each layer. It's on Lee, dependent on the respective incident angles. That is the end of the solution. Thank you for watching


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