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Problem (10 points) Aircraft sometimes acquire small static charges: Suppose supersonic jet has Sxlo-7 charge ar1d flies horizontally at & specd of 660 ms over ...

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Problem (10 points) Aircraft sometimes acquire small static charges: Suppose supersonic jet has Sxlo-7 charge ar1d flies horizontally at & specd of 660 ms over Earth' north magnetic pole; where the xIO magnetic field points straight up. What magnitude of the magnetic force the plane? Problem (15 points). IEthe apex angle of prism 72" (as shown in the associaled ligure), what the Hinimuft incident ungle ifil is emerge fron the opposite side (that is, mt t0 be totally reflected), giv

Problem (10 points) Aircraft sometimes acquire small static charges: Suppose supersonic jet has Sxlo-7 charge ar1d flies horizontally at & specd of 660 ms over Earth' north magnetic pole; where the xIO magnetic field points straight up. What magnitude of the magnetic force the plane? Problem (15 points). IEthe apex angle of prism 72" (as shown in the associaled ligure), what the Hinimuft incident ungle ifil is emerge fron the opposite side (that is, mt t0 be totally reflected), given the index refraction of glass 158? Problem (15 points). 1.7-m-tall shopper standing . meters awuy fiom 4 (-[.2)-m-focal-length convex security mirtor. points) Where the shopper imagc? points) image virtual real? Explain your answer; points) How tall is the shopper '$ imnage the mirtor? For problem 2 For problem



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Additional Problems
In Fig. 33-71, two light rays pass from air through five layers of transparent plastic and then back into air. The layers have parallel interfaces and unknown thicknesses; their indexes of refraction are $n_{1}=1.7, n_{2}=1.6, n_{3}=1.5, n_{4}=1.4,$ and $n_{5}=1.6 .$ Ray $b$ is incident at angle $\theta_{b}=20^{\circ} .$ Relative to a normal at the last interface, at what angle do (a) ray $a$ and (b) ray $b$ emerge? (Hint: Solving the problem algebraically can save time.) If the air at the left and right sides in the figure were, instead, glass with index of refraction 1.5, at what angle would (c) ray $a$ and (d) ray $b$ emerge?

So here for part A. We know that the angle of incidence fate US of B one, uh, at be is a compliment of the critical angle at a and so it's we can say sign of fate. US of B one would be equal to co sign of the critical angle. Or we can say the square root of one divided by ah, the refractive index of medium three, divided by the refractive index of medium to quantity squared. And so, uh, Fada sub B two is gonna be equal to arc sine of and sub two divided by ends up three multiplied by, uh, square root of one minus the refractive index of ah, medium three, divided by ends up to quantity squared, closed parentheses. This is gonna equal arc sine of the square root of and substitute. What about ends up three quantity squared minus one close parentheses and we find out the status of B two will be equal to 35 0.1 degrees. This would be our answer for part A. For part B. We know that end sub one sign of fada would be equal to ends up to sign of fajita sub sea and this would be equal to ends up to times, ends up three divided by ends up to. And so we simply find that here theta is going to be equal to arc sine Uh, and step three divided by and serve one. And we know that this is going to be 49.9 degrees. This would be our answer for part B for part C. We know that here Ah, the angle of incidence a sub a status of a one at a is compliment of the critical angle at B s, we can say sign, Uh, Seita sub a one is equaling co sign of the critical angle fate of subsea, which would equal one, uh rather the square root of one minus and sub three divided by ends up to quantity squared, and we can then solve for the angle of refraction at a angle of incidence. Rather, at a this unequal to arc sine of things, the arc sine of ends up to over ends up three multiplied by the square root of one minus and step three over ends up to quantity squared and this will become status of a one would become arc sine of the square root of and servitude about about ends up three quantity squared minus one and this is equaling again. 35.1 degrees. This would be your answer for park. Si ah, form part D. We know that here ends of one sign of data will equal and sub to sign of status of a one. Ah, this vehicle to ends up two times the square root of one minus ends up three divided by ends up to quantity squared and this is gonna equal to the square root of and sub two squared minus and sub three squared. And then we find Seita for parties, data to be equal to arc sine of the square root of and sub two squared minus sense of three squared, divided by and someone and we find that data is going to be 26.1 degrees. This would be your answer for part D Finally, for rather four part F for party. Rather, we can say that the angle of refraction status of B two at B. This is the fate of the angle of refraction. We can say that fate us A. B two would be equal to arc sine hear of ends up two squared, divided by ants of three, divided by a rather multiplied by the square root of ends up two squared plus and sub three squared, and this is equaling 60.7 degrees. This would be our answer for Part E. And so we can then say that four part F knowing ends and someone sign of theta equals and sub, too. Sign of the Brewster's angle. This would've eagle two ends up to multiplied by ends up three, divided by answer two squared plus and sub three squared. Knowing that here, ah, the angle of incidence fate us of B one at B is the compliment of the Brewster's angle at a and so, uh, we couldn't just substitute for the Brewster's angle, and we find that here Fada would be equal to arc sine of this of and sub two times and sub three. Divided by ants of one times the square root of ends up two squared plus and sub three squared close parentheses. This is equaling 35.3 degrees. This would be our answer for part F. That is the end of the solution. Thank you for watching

For this problem on the topic of electromagnetic waves we are shown in the figure to light rays that are going from a through five layers of transparent plastic and then back into air. The interface interfaces between the layers are parallel and the thicknesses are unknown. However we are given the indices of refraction and we're told that rabies incident at an angle of 20° and relative to a normal at the last interface. We want to know at what angle will ray and ray be much and if we didn't want to replace the left and right sides with glass instead of a within the fact refractive index of 1.5, we want to again find the angle that Robbie and Robbie ray and ray B would emerge. Now suppose there are a total of in transparent layers and is able to five in our case well label these layers from left to right with the indices 12 all the way up until then and let the index of refraction of air B and not. We denote the initial angle of incidence of the light ray upon the earlier boundary as data I. And the angle of the emerging light ray S. D. F. We know that since all the boundaries are parallel to each other, the angle of incidence peter J at the boundary between the jade and the jet bus. One layers is the same as the angle between the transmitted light ray and the normal in the jet layer. And so for the first boundary we have in one over and not equal to sign data. I divided by sine vita one and for the second boundary we have into over in one equal to Signed the to one, oversight The two and so on. And so finally we have that and not over and N. Is equal to sign the to end for that boundary over the angle of emergency sign data. F. Multiplying these equations we get in one over in note Times into over n. one times and three over and two until and not over. An N. Is able to sign peter I over Signed the to one times sine theta one over sine the 22 all the way until sign theater N over sign with the F. Now we can see that the left hand side of the equation above can be reduced to and not over and not while the right hand side is signed Peter I. Or signed peter F. And so get that Simplifying this equation signed the two F. Is and not over and not times sine theta I. Which reduces to simply sign of the incident angle the to I. And so the incident angle the I. Is equal to the emergent angle D. To F. So for the two light rays in the problem statement, the angle of the emerging light rays are about the same as the respective incident angles. And so for a we have the two F. To be 0°. And for a B we have data F. To be The same as the incident angle, which is 20° Now for part C. In this case, all we need to do is change the value of N, not from one for a 21.5 for glass, but this doesn't change the result above. And so we still have for a that the emergent angle D to F zero degrees and the emergent angle for A B peter F. His 20 degrees.

For this problem On the topic of electromagnetic waves in the figure we are shown to light rays passing from air through five layers of transparent transparent plastic and then back into air. The layers have interfaces that are parallel and unknown thicknesses. But we are given the indices of reflection Rabies incident at an angle to to be of 20° relative relative to normal. At the last interface we want to know at what angle will ray and Ray B emerge If the at the left and right side of the figure were replaced with a glass which has a reflective index of 1.5. We want to now know the angle at which Ray and Ray B would emerge. Now we have to support their total of capital and transparent layers in our case and is equal to five. And we label these layers from we have to write with indices 12 all the way up until then the index of refraction of the air will be enought and we denote the initial angle of incidence of the light ray upon the lower boundary. As data I the angle of the emerging light ray will be peter F. And note that since all the boundaries of parallel to each other, the angle of incidence data. J. At the boundary between the J and the J planners one players is the same as the angle between the transmitted light ray and the normal in the J player. That's for the first boundary we have in one over and not equal to sign theater I over Sign Theatre one for the second boundary we have into over in one equal to signed 2-1 over. Sign the 22 and so on. And so for the last boundary we have and note over and N. To be sign peter N divided by sine of the angle of emergency to F. If we multiply these equations, we get In one over and not times Into over in one times and three over in two and so on all the way up until and not over and in is actually equal to sign theater I over sign 3 to 1 times Sign the to one over sine theater two and so on. Up until sign peter in over sign the to F. And so we can see that the left hand side of the equation can be reduced to an auto and not while the right hand side is equal to sine theater I over side theatre F. And so we are left with sine of the angle of emergency to F. Is equal to and not over and not times sign the to I, which is simply the sign of the to I. Which gives us the angle of emergence theater F. To be equal to the angle of incidence data I For the two light treason the problem statement. Now the angle of the emerging light trees are about the same as the respective incident angles and therefore for a we have peter F equal 20 and for a B we have the angle of emergence theater F. To be the same as the angle of incidence, which is 20° next. For part C. If you replace the air by Glass, all we do is change the value of N, not from one, which is for air to 1.5 which is for glass, but this doesn't change the result above. And so again we have for A D to F to be zero, and for a B, peter F is equal to 20 degrees, and we can see that the result of this problem is general and independent of the number of layers and the thickness, an index of refraction of each layer.

This question. We have a Raider system ah, attempting to detector our stuff aircraft and first part. They want to know the intense stuff being when the being reaches their aircraft. Location the aircraft for effective reader waves, I thought he has across sexual era off. They're pointing to mirrors. So the intensity of the being you were related to the power invited by the area off the We're being the spread. It's always a few and, well, this is a upper atmosphere of this fear. So we can substitute violence here and we get 100 Katie time. Stand to the three watts, divided by two by my I understand the three isn't a distance of their craft square. Everything's in municipal. This gives the street Quint filed by instead to the monastics boats by Middle Square. They want to know for part B. Ah, what is it? Four off the aircraft for affection. And we have to assume that the being is reflective informally over atmosphere backed the Raider side, so the power a fraction is the intensity times the area off. The reflections of this is 3.5 bank standard and minus six times area that you saw it was Europe into two million square. So this gives a 7.8 understand? To the minor. Seven. What for? Potsie. They want to know. Ah, back at the reader side. Where is the intensity? In the matter of value? Off the electric fictive actor in the area mass value off the magnetic field. Let's do this. The intensity is again the intensity Ah, given by this power? No, because we're back so intensities. Okay, so this supposed to be a way? All right, so p r divided by two by r squared that this is the same. So the 7.8 times stand to the minus stallion If I did by two pi 19 times 10 to the three square. There's this 1.5 times at the 90 17. What premiere considered intensity drops Damascus Because we are The intensity is off the reflective part. Only now they want to know their letter magnetic field so we can find this from the intensity. So the letter magnetic amplitude politic, um, amplitude is square root ruins off to see mew zero the intensity. So this is square root off two times to enter. Understand? to the eight for by 10 to the minus seven times 1.5 10 to the minus 17. This is the power. This is this there square old here. And this gives us 1.1 times, then to the my seven boats premiere and the air mass value for the magnetic few dishes. So for party B. R s is related to our best value off the athletic field divided by sea. And there are mass value for the amplitude just divided by the square root. So am I bothered by a square root of two Valorous e. And there's this 1.1 time stand to the minus seven by the by square. Root off three. I'm spent eight and we gets 2.5. I'm stand to the miners. 16 Tesler.


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