This is stock number 7T334B. We are here at Summit Automotive in Fond du Lac, Wisconsin. You're new and used Ram headquarters. Today we are checking out this super clean 2016 Ram 1500 Crew Cab Short Box Sport Edition. This one has the Stinger package on it. And from this HD video you will be able to tell just how clean this truck is all the way around inside and out. And get a good idea of the overall stuff that is involved with the Stinger package. So first off we got the 20 inch polished aluminum rims. These are part of the Sport package and it has the Goodyear Wrangler SRA 27560R20 tires. And they have about I'd say 50 to 60% of the tread left. A little over half the tread left on these tires. The color of course is called Stinger Yellow. Has the projector lamp headlamps, LED running lights, the blacked out headlight bezels. You can see you get the black Ram logo on the front. No dents, stings or scuffs or cracks on that front bumper. Factory fog lights, the chrome trim grille. You get the black hood decals. Which are part of the Stinger package. It has the Coal induction sport hood. Passengers side rim. Absolutely perfect. No scuffs or scrapes. These calipers have been painted yellow. I don't think they came from the factory yellow, but they have been painted on nice little accent. And as you go down this side of the truck you can see just how clean the body is. How reflective and glass like that paint is. We shoot all of our videos in 1080P so if you have HD capabilities on your computer tablet or smartphone device turn them on right now because it's like you're right here looking at the truck with me. But we take these videos so if you are far away or even if you're close by and just cannot make the truck down you can still see the truck here, the truck and have confidence in the vehicle that you're looking at before you even get here. Lower rockers are in excellent shape. Some nice aftermarket chrome step bars. Back rim no scuffs or scrapes on that. Back tires have both the same amount tried maybe just a hair less. Has the plastic wheel well guards and that keeps your frame and under body looking nice and clean. We sell these trucks all over the country so we want to make sure especially on a rare truck like this that we give you the most accurate representation of the vehicle as possible so that when you get here there's absolutely no surprises. Tailgate is in excellent shape. You get the blacked out Ram logo and the blacked out 4x4 symbol with the Stinger package dual exhaust. You get with the Hemi has the 4 pin and 7 pin wiring and the receiver hitch. This one has a drop-in bed liner as well. I'm dropping bed mat as well as a spray and liner. Very nice condition and it's very nice. It has the LED bed lighting back here as well. Didn't see any dents or dings on the tailgate. Very nice condition and as you go down this side of the truck I didn't see any dents or dings. That's why we take these videos. We want to make sure that if there are any dents or there are any imperfections you can see them in the video so that when you get here there's absolutely no surprises and you can shop from the comfort of your own home. Inside the Stinger package gives you the black and yellow leather and cloth interior. Both of these seats are heated. You get the yellow accent stitching. No rips or tears or scuffs. Power driver seat. Factory floor mats throughout. Power windows power locks. Power folding mirrors. They power fold out as well. They are heated as well and you can see you got the yellow stitching there as well. Auto headlamps power pedals and as we go inside the truck here you can see that this truck has 40,209 miles. You get the 7-inch TFT display. The heated leather wrap steering wheel that also has the yellow accent stitching on it. Cruise control on that side. Bluetooth and information center controls on that side and of course on the back of the steering wheel there's buttons for the radio. Speaking of the radio you have the 8.4 touchscreen radio. You have your heated seat controls right there. You can also do your backup camera right there as well. This one does have the factory navigation on it.  You also have heated seat and heated steering wheel buttons there. The turn dial, 8-speed automatic transmission and the push button 4-wheel drive. This one has the keyless enter and go system so you just have to have this key on you. Push this start stop button and it will start right up. It also has remote start on there as well. This one has the Alpine sound system so you get the subwoofer in back and it also has the 911 and assist buttons. Those are features of the vehicle that you never have to pay for. It has the homelink system and the power sliding rear window. Pass and receipt is absolutely perfect as well. This truck has never been smoked in. You get the ram stitched into the headrest as well as the sport on the headrest. No rips, no tears.  We'll take a quick look at the back seats and then we will check out under the hood. Back seats are just as clean as the front. I don't think anybody's even actually sat back here.  These seats fold up for extra storage. You get a load floor which is a pretty neat deal. If you've ever used one of these you can set it down nice and flat and put both sides down if you got pets or if you got tools or muddy boots put them on there and you're not going to wreck your nice yellow and black seats. It has the drop-in has two storage trays with the drop-in bins so for easy cleaning you can take those out. You can put ice and drinks and stuff in there and then underneath the seats there is extra storage on this side. The other side has the subwoofer under it and there are little LED lights in the bottom of the seats. So when it's dark out you can see what you're doing. We'll take a quick look under the hood. Very very nice truck. Probably one of the nicest rams we have in our lot right now and the most unique. 5.7 liter V8 Hemi motor. Engine maze very clean runs very smooth. This truck has been fully safety and inspected by our service shop. It has a fresh oil and filter change. All the fluids have been checked and topped off and this truck is 100% ready to go. And to see more pictures of this truck or one of our other 400 new and used cars, trucks, SUVs, minivans, wranglers, you name it we got to go to our website www.summitauto.com. Full pictures and descriptions of every single vehicle on our lot. Videos of every single used vehicle that we have all at Summit Auto.com. Thank you so much for checking out the video.  Ask for one of our sales associates standing by to make this 2016 Ram Stinger years. Once again that number is<PHONE_NUMBER>. Thanks again for checking out the video. If you like the video want to check out more go to youtube.com slash Summit Auto. Remember to like, subscribe, and share on this video and all the videos there. In fact in a second you'll see a link to subscribe to our YouTube channel on your left. A link to more half-ton truck videos like this one on your right. If you haven't been to our website on the bottom a link to this truck on our website click those check us out and we really look forward to help you with this super clean 2016 Ram 1500 Crew Cab Short Box Sport Stinger Edition. Thanks again.


Hey guys, so today's video is going to be on my experience of taking the ACLS certification course and for those of you who don't know what ACLS it is, oh I can't talk. If you don't know what ACLS is, it is the Advanced Cardiac Life Support, ACLS. So most of you probably have your BLS, your Basic Life Support CPR certification because that's what's required for nursing school, that's what's required for most jobs, but a lot of nursing jobs require you to have your ACLS because it is a more advanced certification. So I'm going to talk about my experience and kind of what ACLS is and yeah. So ACLS is a course that really goes way beyond basic life support. You know basic life support is your CPR and so your compressions and your, your, what am I going to say, bagging, your breathing, the respirations for the patient and so that, that's your basic life support. You're doing compressions and you're helping the patient breathe, you're ventilating for them. ACLS goes above that and this is where you've learned how to shock the patients, where you learn which meds to give, where you learn which rhythms are, you know, which rhythms do you do certain things for, where you, there's different meds like as far as like if a patient is in SVT or their heart rate is super fast, you should check out my SVT video by the way that I made, if you don't know what SVT is, where, you know, you give a dentistry really quick to slow down the heart rate or where you synchronize cardiovert them in order to try and convert the heart rate back. So this is where you learn all of those things and the cool thing about ACLS is for some reason if you are in the hospital and your patient codes and let's say they are in VTAC, they don't have a pulse, you know what to do per your ACLS protocol and if there's no other doctor or anything as a nurse you can run, if your ACLS certified you can run the code, you know what to do, there's a very specific order of what you do.  So it's really kind of neat because if for some reason there's a code and there's no one else around, I'm gonna say no one else around, I always say there's people around but let's say the doctor can't get there for another two minutes because they're on the other side of the hospital and there's no other doctor available, nurses can run a code if you're ACLS certified. So that's a really cool thing about ACLS and I'm really glad I got it. The first time you're ACLS certified it's a two-day new provider course and then you get and just like your BLS CPR every two years you have to renew it but then it's only a one day course and you actually get your BLS CPR in ACLS when you renew it. So that's pretty cool. I'm really glad I got it because that means that you know I can work, I don't say I can work anywhere but you know you need it to work in the ICU, you need it to work in the ER. Again this is a nursing provider course, this is not something that you really need during nursing school but I wish I would have taken it in nursing school before I took critical care because I think it would have helped me out a lot with cardiac rhythms and knowing what meds we give and why and how they work and all that. So just a little tip bit of information if you are going into critical care you should take your ACLS. Most jobs if they require ACLS they will pay for it which is nice and my job paid for it which is even better. So anyways that's my experience taking ACLS if you have any questions leave them in the comments below give this a video a thumbs up and subscribe to my channel and I'll see you guys next time. Bye!


I got an eye gel for stopgap and I got a gonna get bronchoscopies ideal you give them a seven and a half or an eight two because that's what they need for bronchoscopies an eight


Welcome to the 18th lecture in the course Engineering Electromagnetics. Today we continue what we started towards the end of the last lecture, the topic listed in the course content reflection and refraction of plane waves. The topics that we have on the agenda today are reflection at the surface of a conducting medium and the reflection at a perfect conductor when the wave is incident obliquely. But before we take up any of these two topics, let us continue with a small part of the previous lecture and we consider the normal incidents of a uniform plane wave at a perfect conductor and placing the perfect conductor at x equal to 0 and considering that the incident electric field has an expression E i equal to E naught e to the power minus j beta x oriented in the y direction. The total electric field taking the reflected field also into account was found out as minus 2j E naught sin beta x so that it is 0 at x equal to 0 and at periodic distances from the interface. Similarly we found out the total magnetic field as 2 h naught cosine beta x oriented in the z direction and we said that these constitute a standing wave in front of the perfect conductor interface. Now what will be the power carried in this standing wave? That is the point of small interest and if we consider the pointing vector corresponding to these field components and we write the average pointing vector as half real of E cross h star we immediately find that since E t and h t are in phase quadrature they have no component which is in phase the result here is 0 although the medium that we have has a real intrinsic impedance or characteristic impedance. So this is another situation where the pointing vector comes out 0 when we consider the real part of E cross h star and it is a result of the fact that the total electric field vector and the total magnetic field vector have no component which is in phase they are so to say in phase quadrature. This will be the situation inside resonators and when we have this kind of perfect standing waves. Now we can go on to considering the topic that we listed for discussion today the first topic that is the reflection at the surface of a conducting medium. The difference that we have in this case from the previous case of perfect conductor is that we do not have a perfect conductor now we have in fact a general medium it is an interface between two general media which could be represented in the following manner. Let us say this is the interface between two media say medium one and medium two with constitutive parameters as mu one epsilon one and sigma one the permeability, permittivity and conductivity and similarly mu two epsilon two and sigma two and we consider that uniform plane wave is incident normally on the interface. So this is still a case of normal incidence. For this direction of propagation the electric and the magnetic field vectors must be in a plane which is normal to this direction we may represent these as follows. Let us say this is the electric field vector associated with the incident wave and let us say this is the magnetic field vector associated with the incident wave so that E cross H is in the direction of propagation. Now unlike the perfect conductor where there could be no power carried in a propagating wave or there could be no power lost because of the absorption in the medium in a general medium those restrictions are not there and therefore in general we will have part of the power of the incident wave transmitted through this interface and the part which is not transmitted should be reflected back. On this basis we can put down a transmitted wave with the corresponding electric field vector as E t and the magnetic field vector as H t and similarly for the reflected wave which is going to propagate in the opposite direction we say that let this be the electric field vector associated with the reflected wave E r and let this be the magnetic field vector associated with this way. Then corresponding to the wave propagation we will put down appropriate relations between these. These are nominal directions for the reflected field vectors and now we apply the boundary conditions at the interface boundary conditions on the tangential components of the feeds. For a general interface like this the tangential field components both electric and magnetic field components should be continuous and therefore we can write E i plus E r equal to E t and similarly H i plus H r equal to H t. However the field components within a particular wave say incident wave or reflected wave are not independent they are related through the intrinsic impedance of the medium and therefore we can write the relations E i by H i equal to E t 1 where E t 1 is the intrinsic impedance of medium 1. Similarly E t by H t should be E t 2 and as far as E r and H r are concerned they exist in the same medium that is medium 1 but this reflected wave is propagating in a direction which is opposite to that of the incident wave and that should be taken into account by writing this as minus E t 1. Now we can process these equations to find out what is the fraction of the incident electric field which is reflected back which will become the reflection coefficient. From 1 we are going to get let us say we divide by E i throughout so that we have 1 plus E r by E i equal to E t by E i. Let us say that is our equation 3 then from 2 we substitute for H in terms of E from these relations. We are going to have therefore eta 1 into E i minus E r which should be equal to it should be 1 upon eta 1 that is right and similarly here we are going to have E t by eta 2 from which by dividing through by E i and multiplying by eta 1 we can get 1 minus E r by E i which should be E t by E i into eta 1 by eta 2 which becomes our equation 4. Now both equations 3 and 4 contain E r by E i and E t by E i we can eliminate these 1 by 1 and obtain the values of these unknowns respectively. For example from 3 and 4 we can get E r by E i equal to eta 2 minus eta 1 upon eta 2 plus eta 1 considering that we are dividing 3 by 4 and then doing the standard manipulation component or dividend. The similarity with the transmission line is unmistakable. If we consider the reflection coefficient based on the magnetic field components one can see easily that this is going to be minus H r by H i and similarly E t by E i is going to be twice eta 2 by eta 2 plus eta 1 whereas H t by H i is going to be twice eta 1 upon eta 2 plus eta 1 which respectively become the reflection coefficients or the transmission coefficients depending on which field we are interested in and these are coming out in terms of the intrinsic impedances of the 2 media involved. For the case of normal incidence the relations are really quite rate forward. To understand these better let us take a particular example yes or no. We have nominally put these directions their values will come out appropriately from the reflection coefficients etc and that they constitute a wave propagating in the opposite direction that we have taken care of here. In fact this is something quite similar to what we did for the transmission line right. We take an example and we can do that on the projector. We consider to a particular media. Let us say medium 1 is air alright and let us say that medium 2 is copper. So that here we have mu 1 equal to mu naught epsilon 1 equal to epsilon naught and sigma 1 equal to 0. If we have written free space that would have been more accurate. Here we have once again mu 1 and epsilon 1 that of free space as for good conductors mu 2 is mu naught and epsilon 2 is epsilon naught. However sigma 2 is very large it is 5.8 into 10 to the power 7 mose per meter. The corresponding impedances can be calculated we will have eta 1 equal to mu naught by epsilon naught which value is 120 pi or 377 ohms. For eta 2 some calculations need to be done. What will be the expression for the intrinsic impedance? The general expression is j omega mu upon sigma plus j omega epsilon for conducting media in general which can be simplified as j omega mu upon sigma for good conductors. And therefore the intrinsic impedance for this medium for copper requires specification of a particular frequency before we can numerically calculate the intrinsic impedance. So let us say that omega corresponds to 1 megahertz okay. So 2 pi into 10 to the power 6 radians per seconds and for this value of omega eta 2 comes out to be 3.69 into 10 to the power minus 4 with the associated angle of 45 degrees. A very small value okay it is only very slightly different from a short circuit alright. So let us see what kind of results we get for the various parameters that is reflection coefficient, transmission coefficient etc. So for the reflection coefficient involving the electric fields E r by E i which we have written as eta 2 minus eta 1 upon eta 2 plus eta 1 substituting the values of eta 2 and eta 1. We get a value which is minus 0 point and then 5 nights 1, 2, 3, 4, 5, 8, 6 at a very small angle minus 0.4079 or simply 8 so many degrees. The reflection coefficient is practically minus 1 which is what we expect from a perfect short circuit or something which is very close to a short circuit. The reflection coefficient based on the magnetic field components is going to be similar simply the negative of this. We can also calculate what is transmitted through such a virtual short circuit and we get E t by E i which is twice eta 2 upon eta 2 plus eta 1 the calculations can be done and we get result which is 0.5012345196 at an angle of 45 degrees okay. So there is very little transmission quite as expected. h t by h i can also be calculated and that comes out to be 1.5986 at an angle of minus 0.40 and then very close to 4 degrees okay and therefore one can say that the copper shield is virtually a short circuit. Of course it is conceivable that if we increase the frequency sufficiently this behavior may change which is what happens at very high frequencies alright. With this example we have considered the reflection and transmission at the surface of a conducting medium and now we go on to consider reflection at a perfect conductor okay. We have already considered this reflection at a perfect conductor for normal incidents and now we consider the more general case of oblique incidents. We consider let us say this is the interface between one medium and the other medium and let us say the lower medium is the perfect conductor okay. This is the normal drawn at a point where a ray representing an incident uniform plane wave strikes this interface let us say at an angle theta. Now in this case we need to distinguish between two possibilities. What are these two possibilities? This is the direction in which the uniform plane wave is incident. The electric and the magnetic field vectors have to be in a plane which is normal to this direction of propagation utilizing the properties of uniform plane waves. One possibility is one of the many possibilities is that the incident electric field is in this direction normal to this direction okay. Let us say this is the it is normal to the plane of the board. Let us say if this is y and this is z then the incident electric field is oriented along the x direction alright. And the corresponding incident magnetic field intensity vector is in this direction consistent with the fact that the vector e cross h should have a direction which is the direction of propagation. This is one of the many possibilities this possibility can be identified or specified as the case of perpendicular polarization because the polarization of the incident wave is normal to the plane of incidence. What is the plane of incidence? Plane of incidence contains the incident ray and the normal drawn from the point at which the incident ray strikes the interface okay. So in this case we have perpendicular polarization. Other possibilities are also easy to see. For example the magnetic field vector could be oriented like this that is in the direction x and the electric field vector may be like this. This would be the case of parallel polarization because the polarization of the incident wave would be in the plane of incidence or would be parallel to the plane of incidence as we shall see there are differences in these two cases that arise okay. Now a question can come up that well we have a uniform plane wave incident in this direction why do we have to consider only these two specific cases there could be a whole lot of cases in between right. Now the simple answer to that is that any other general case can be considered as a superposition or as a combination of these two orthogonal polarizations. So one could consider the components of the incident wave with these two linear polarizations okay. And once we have understood the behavior for these two cases we will be able to analyze or understand any other general case alright. In fact that becomes the basis of some very interesting phenomena which we shall talk about later. So to start with we consider the case of perpendicular polarization okay. Now since we have once again come back to the perfect conductor the arguments that we invoke for normal incidence would again apply and therefore there is not going to be any power which is transmitted through the interface either as a propagating wave or as something which supplies power to power losses. And therefore whatever power is incident or is associated with this incident wave must remain above the interface. In fact one could go a little more further one could consider this incident wave also as composed of two parts okay. Some part of the power is falling normally on this interface and some part of the power is flowing parallel to the interface and physically what we expect is that the power flow parallel to the y direction parallel to the interface should not be affected. There is really no constraint on that and the power flow which is normal to the interface should be completely reflected back and as our experience is it should form a standing wave. Physically that is what we expect let us see how we can obtain those results proceeding more systematically. The first question that needs to be answered is that what is the direction of the reflected wave okay. Intuitively we feel that the angle of reflection should be the angle of incidence but there is a way out we can handle this a little more rigorous. For that purpose we make the following construction maybe we can make some space here. We consider the wave front associated with the incident wave and similarly the wave front associated with the reflected wave. And for that purpose let us say this is the interface and let this be one of the rays representing the incident wave and let this be the associated second rate both incident in this manner and let us say this angle is theta 2. The angle of incidence is theta 2 and as a result there is a reflected wave which is generated. Let us say which makes an angle theta 1 with the normal at that point. There will be a reflected ray corresponding to the second ray also which we may draw like this okay. Now let us draw the perpendicular I think we will have to redraw this there is a slight problem in this we have to consider more inclined waves so that we can show this better. This is angle theta 2 these two rays are incident in this manner and let us say this is angle theta 1 okay. We begin with the assumption that the angle of incidence and the angle of reflection are different. And now we consider the wave fronts associated with the incident wave and the reflected wave and let us say that these are drawn like this okay. So let us say this is point A and this is point D. A D represents the wave front associated with the incident wave on this wave front there is a uniform phase. Similarly there will be a wave front associated with the reflected wave let us call that BC okay where this angle is going to be theta 2 and this angle is going to be theta 1 alright. Now let us consider the distance AC and the distance DB or BD. This is the reflected wave wave front this is the incident wave wave front during the time the ray striking at point A travels to point C okay. In that time the other ray should travel from point D to point B that is how the wave fronts would be constituted like this okay. And therefore AC must be equal to BD because these are travels in the same media okay. So the travel is with the same velocity and therefore in the same time the distance travel must be the same and one can see that this is going to be AC is AB sin theta 1 and BD is AB sin theta 1. And therefore we get the simple relation that the angle of reflection is equal to the angle of incidence okay. So using this kind of arguments one can make out what will be the angle of reflection given the angle of incidence in such a situation or later on what would be the angle of refraction when we consider a general medium other than perfect conductor okay. And therefore we can now safely say that the angle of reflection here is the same as the angle of incidence. And now let us put down an expression for the incident electric field which is going to be EI equal to let us say some magnitude EI and then we will have e to the power minus j beta y sin theta okay. And minus z cos theta considering the direction in which the wave is propagating the direction cosines of this direction are sin theta and minus cos theta. And therefore multiplying the phase shift constant with that we get this expression for the incident electric field which is going to be in the x direction. Now we also want to put down the expression for the reflected wave electric field. And we are aware that at the interface say at y and z equal to 0 the total tangential electric field must be 0 this being a perfect conductor. And therefore EI and ER at the interface must completely cancel each other. And therefore the direction for ER that one may put down is this. Now this is slightly different from what we did last time. Now we are taking the direction into account in putting the electric field vector for the reflected wave. And once we have identified the direction of the reflected wave electric field to be consistent with the direction of propagation we must have the magnetic field vector in the reflected wave which is perpendicular to both of these. And E cross H should be in the direction of propagation and therefore we have the reflected wave magnetic field coming out like this. The magnitude of ER is to be the same as that of EI as we have just found out. And therefore HR magnitude must also be the same as that of HI using the power conservation arguments that we mentioned a little while ago. And therefore now we have ER equal to minus EI e to the power minus j beta the y direction remains unchanged consistent with our intuitive arguments it is y sin theta. But the z direction is now reversed so we put plus z cos theta here and this remains x cap. And now we can consider the behavior of the total electric field which will be a combination of the fields electric fields in the incident wave and the reflected wave. And we are going to have EI into e to the power minus j beta y sin theta. And then we have e to the power j beta z cos theta minus e to the power minus j beta z cos theta x cap which gives us 2 j EI e to the power minus j beta z cos theta y sin theta. And then sin of beta z cos theta overall direction remaining x cap from which one can make out the behavior of the overall electric field. It will be somewhat easier if we consider the corresponding time varying total electric field as against the phasor notation in which we have put down the total electric field. We have the time varying total electric field it will be twice EI sin of beta z cos theta. And what else do we have sin of omega t minus beta y sin theta. Which indicates that in the y direction we have a propagating wave with an argument which is t minus y by v that velocity we will identify shortly. And in the z direction it is a standing wave with the field value total field value which is 0 at z equal to 0. And as we have progressed up it is going to show periodic nulls and maxima nodes and anti nodes which are the characteristic of standing waves. We can introduce new symbols for briefer notation it is going to be twice EI sin of beta z z and sin of beta z cos theta. Omega t minus beta y y x cap where beta z is beta cos theta whereas beta y is beta sin theta. One can put down the corresponding wave lengths and the phase velocities lambda z would be 2 pi by beta z. So, that it is 2 pi by beta cos theta or in other words it is going to be lambda by cos theta where lambda is the wave length associated with free space. The corresponding phase velocity let us call it v z is going to be omega by beta z and one can see that that is going to be the velocity of the wave in free space divided by cos theta. It is going to be greater than the velocity of light in free space, but this is just the phase velocity. Energy or information is not going to be transmitted at this velocity. Correspondingly we have lambda y as lambda by sin theta. So, this is going to be the and v y as v by sin theta. The reason for these z directed and y directed velocities being greater than v can be understood by considering an example like this. And the distance swept along the y direction or along the z direction is greater than the distance that the wave travels in its direction of propagation and the time involved is the same. Therefore, these phase velocities come out greater than the velocity of light in free space. This kind of expression in the z direction indicates a standing wave with the periodic nodes and anti nodes and the distance between adjacent nodes is going to be lambda by 2 cos theta or lambda z by 2. Quite consistent with what we have done earlier in similar situations. And as we had argued out intuitively, the incident wave can be considered to be composed of two parts. One part incident normally which should be completely reflected back this being a perfect conductor forming a standing wave and other part travelling parallel to the interface and going unaffected by the interface. So, our final result is quite consistent with those intuitive arguments. Now, we are in a position to consider the second case corresponding to this, the case of parallel polarization. We make the following construction. These are the y and the z axis. The interface is coincident with the z equal to 0 plane like last time and this is the wave incident at an angle theta and this time the incident wave electric field is parallel to the plane of incidence and the corresponding magnetic field is like this. As far as the direction of the reflected wave is concerned, the angle of reflection that should from similar arguments come out the same and therefore, we can mark the reflected wave in this manner and similar intuitive arguments are going to apply to this case also. Now, of course, we expect the reflected wave fields to be in a plane normal to the direction of the propagation of the reflected wave. And from this point of view, one can consider some simple cases. To be able to put down more clearly the field orientations, let us consider that the incident electric field is composed of two parts. One is the y component, the other is the z component which we label as E z i E y i. At the corresponding point here, now E y i is a field component which is tangential to the interface. And when we apply the boundary condition that the total tangential electric field at the perfectly conducting interface must be 0. E y i must come out E y r must come out as negative of E y i. So, incorporating this direction in the reflected wave y component of the electric field, the field component is going to be like this. This becomes our E y r. Then as we have been arguing out the part of the wave, the component of the wave, notional component of the wave which is travelling in the y direction should remain unaffected by this interface, because this is not imposing any constraints on that. And therefore, the z component of the electric field in the reflected wave must remain unchanged. And this is how E z r must look like, so that we get the reflected wave total electric field which is oriented in this manner, normal to the direction of propagation with these components. Once the electric field has been obtained, then since h must be normal to the electric field and to the direction of propagation and E cross h must be in the direction of propagation, we can say that the reflected wave magnetic field should be like this. And now, one can combine the various field components and consider their behavior along similar lines which part we shall postpone till the next lecture. In today's lecture, we have considered the case of normal incidence at an interface between two general media. And we have considered the reflection coefficient and the transmission coefficient. And next we considered the case of oblique incidence at a perfect conductor. We saw that we need to consider two different cases. We have considered the case of perpendicular polarization today. And we shall consider the case of parallel polarization in the next lecture. Thank you.


Prime Minister Lee and Mrs. Lee honored guests who are here tonight. Nancy and I welcome you to the White House. It was a great pleasure for me today to renew a valued friendship with Prime Minister Lee. I first met the Prime Minister on a trip that I took on the behest of President Nixon. And when we stopped in Singapore I was amazed at the dynamic society that I found there. How could a country with such a small area and few resources be making such strides? And then I met Prime Minister Lee and my questions were answered. He is a man of principle and vision. His leadership has provided the vigorous and creative people of Singapore the means to move ahead, to achieve, and to build. Singapore's experience has been in stark contrast to developing countries where political power has been derived from terror and brute force. Instead, Prime Minister Lee's authority has rested on his capacity to mold the opinion of his countrymen and build consensus. He has used his position to free the talents and energy of his people so they could be channeled into constructive society-building activity. Fortunately, Prime Minister Lee's sound judgment does not stop at the water's edge. American leaders, including this one, have frequently benefited from his wise counsel. Our meetings today were no less beneficial. Our exchange was cordial, reflecting a mutuality of interests and a harmony of views. Mr. Prime Minister, I want to express my personal admiration for your recognition of the contributions America makes to world peace. As the world's most powerful democracy, our people carry a heavy military and diplomatic burden, an often thankless task. But you have demonstrated an appreciation and understanding that makes it all worthwhile. This spirit of mutual respect was evidenced in our meetings today. None of this should be reason for surprise. Our two peoples may at first glance seem worlds apart, both in geographic location and culture. But a closer look reveals that Singapore and the United States are nations made up of hardworking immigrants and their descendants who came to a new homeland to improve their lot and build a decent life for their families. We're both democratic nations committed to peace and to the preservation of human liberty. And these bonds are being bolstered by continued cultural and educational exchanges. And of course, the many commercial ties between our peoples. Mr. Prime Minister, we're aware that your people are now faced with severe challenges brought on by international economic conditions. The United States faced economic adversity not long ago. Tough decisions had to be made. It's heartening to see that you're moving forward, Mr. Prime Minister, with an eye toward the long run well being of your people. I understand full well this is not always easy to do, but I want you and your citizen to know that the people of the United States want you to succeed and prosper. Our meetings today confirmed again the people of Singapore, as we say here, are our kind of people. So would you all join me in toasting the people of Singapore and the distinguished leader, Prime Minister Lee and Mrs. Lee? Mr. President, Mrs. Reagan, ladies and gentlemen, my wife and I are much honored and delighted to be here with you. Enjoy your warm hospitality. We would like to express our special thanks to Mrs. Reagan for your learned of her personal interests in the preparations for this splendid occasion. It is a rare and gracious first lady who would personally settle and approve the menu, the wines, the floral arrangements, and the entertainment. I read of Mrs. Reagan in her campaign against drug abuse. Her personal efforts and her attention to the details of the cause that she as champion has won her wide acclaim. I watched her on television standing amidst the rubble of Mexico City a few days after the earthquake, bringing succour and comfort to the victims. After nearly five years in the White House, I noticed that the latest opinion, Paul, puts her approval rating at 71%, ahead of the precedents, and by 9%. Mr. President, your staff has to shape up, or you may have to borrow the first lady's staff. Mr. President, I've been a regular visitor of the United States for about two decades. It was an ordeal to watch America arrive in the self-inflicted agony at home. During the years, she tried to fight a war in Vietnam. And even after the war, she did not bounce back from her depression. And morale dropped to a new low when American hostages were held in Tehran. When you were taking your oath of office as president, the hostages were released. It was proof to me that the Iranian mullahs were not as crazy as the media had made them out to be.  You have never allowed any problem, however daunting, to weigh you down. Now, as imports surge into America because of an overstrong dollar, Congress in its pessimism moves towards protectionism. You have not yielded to such despair. You will astound your critics yet again when you return the spell of apparent adversity to advantage by opening up foreign markets and creating new jobs for Americans. The Reagan years will surely be a noticeable landmark in American history, for you have restored American leadership in the maintenance of a just and equitable world order. Mr. President, I have been privileged to share some of your thoughts today. The friendly relations between the United States and Singapore are at their best, indeed, as they should be. I have been a privileged guest in these surroundings under different precedents, as I explained to the young lady, so judiciously selected to be my companion tonight. I have never felt more relaxed and more at home, and I think I owe that to the other attractive young lady on my left. I have been here and have been impressed. Tonight I have been here and have enjoyed myself. Mr. President, ladies and gentlemen, I now ask you to join me in a toast to the President and the First Lady to wish them good health and happiness. Not a day goes by, not a single...  Get much better, much longer, but it only gets better and stronger and deeper and nearer and simpler and freer and richer and but I just go on thinking and sweating and cursing and crying and turning and reaching and waking and dying. And now something in French that I know will be familiar to many of you. Say a thank you to Karen Akers, DeMarc, for a little magic that we've had here this evening. It was a few years ago that a musical called Nine Broadway, and someone named Karen Akers dazzled the audiences of Broadway, and then there were performances at Carnegie Hall. It was a movie called The Purple Rose, Cairo, and then here tonight. Now Karen has entertained here for us before, but tonight is something special and I'll tell you why. Incidentally in that musical Nine, you were a 10. The thing that's special about tonight is, and I think you should know this, that someone else was scheduled to be here and entertaining and just about 24 hours ago was rushed to the hospital. And for someone to do as Karen has done and on such short notice to step in and entertain as she has here tonight, I think that's service above and beyond the call. So again, there's just no words to thank you being here. And I'm thinking that in the future, several times I'm going to pick out someone of ill health and invite them to come. So we'll be sure to get you again. All right. Thank you very much.


address the media post game. Let's hear what John Carlos Stanton has to say. Hey John. Carlos. How did your quad hold up tonight? I feel good. It's definitely tested a lot of swings. Check swings. Didn't run too much but everything checked out great. How difficult is it when you miss 13 games to come back, step in that batter's box and see live pitching for the first time? It really doesn't matter how difficult it is. I have to, from being the lineup, I got to get done. So I had two opportunities to help us win this game and not have the outcome of tonight and wouldn't even have been in a walk-off situation there. So it doesn't matter how long I've been out. That has to get done. What did you think was going on in the batter's box tonight? I was picking it up okay. I'd say just my, in between pitches, adjustments, I wasn't as fast as I usually am from being down a couple of weeks. So I said just the mental part of getting back in, making a quick little adjustment and making sure I get the ball in play. Thank you. Marley Rivera, you have the next question. Giancarlo, when you first started feeling up, you know, your latest injury, one of the things that Boone kept telling us what they wanted to, you wanted to get ahead of it. Can you tell us a little bit about that process when you started feeling it, when you start saying let me, let me get ahead of this before it becomes something more serious? I felt it in my second class at bat in Tampa and of course in the last at bat and then I just figured usually this is something that I'll run myself out there and try to push through it and then I'm adding another month to this simple two weeks. So luckily it's worked out so far and I don't want to miss any time but two weeks is much better than longer than that. And one of the things that you just said is that if I'm in the lineup, I have to contribute, but then how do you also balance the fact that you do have to be patient with yourself after being out 13 games, you know, and trying to stay healthy? Oh yeah, I understand that. At the same time, it's nobody's excuse to have. The picture doesn't care if I've been out. The team cares but it's still, I need to be in their producing so it's nothing of, I don't take that excuse at all. Brendan Cutty, you have the next question. John Carlo, obviously you're off the aisle, but is this situation something you still have to manage coming back off the aisle or is it behind you? Do you feel like it's completely healed? I'll have to pace out the game a little while, but it shouldn't be too long.