B Tech applied science unit one physics and I'm going to talk about intensity because this can be a little bit tricky a little bit mathematical as we shall see okay intensity so what is intensity the intensity of a wave uh it's the energy arriving every second over 1 M squar over unit area so it's the energy around ding every second so energy per second so that's power yes Jews per second is power so it's the power over every area yeah intensity capital I is the power divided by the area so intensity would be measured in watts per met squared because power is measured in watts and area is measured in me squared if we look at these pH photographs we could talk about the intensity of the light arriving from the Sun and if you're in the middle of the desert every second lots and lots of energy would shine on one meter squared so it's very very intense yeah you could make toast in the middle of the desert uh you're on your summer holiday on a Greek island and that's pretty intense but it's not as intense as in the middle of the desert so the amount of energy AR driving every second the power divided by area is the intensity and then on a dark cloudy day uh the intensity is less and then on a very very dark getting close to nighttime it's much much smaller and this is what intensity means okay it's the energy per unit area that arrives every second now here's I think this is kind of pretty straightforward before we get mathematical houses further away from the transmitter get a weaker signal so here we have a TV transmitter and we have an aerial going to your television and if the aerial is further away if you live further away from the transmitter then you get a weaker signal why do you get a weaker signal because now the energy of the Waves which is in this case we're talking radio waves the energy of the Waves emitted by the transmitter it spreads out over a bigger area so the energy received by the aerial every second uh is less if it's further away so you can imagine kind of energy the wave energy the energy of the Waves yeah coming out of the transmitter and as they move away from the transmitter they spread out over a bigger area so they get weaker and weaker so the intensity gets smaller and smaller yeah because the energy is getting more and more spread out look at this it's a another example there's a person shouting and look at the waves leaving the sound waves and the area the area if you like it's a sphere so it's the area of this sphere and the surface area of the sphere will get bigger and bigger and bigger as you get further away from this person so the same amount of energy is spread out over a bigger area and then eventually this other person can't hear because the intensity is too small yes so intensity gets less it gets smaller with distance because it has to spread out over a bigger area now consider this this is getting mathematical now imagine you're shining a torch on a piece of card uh and let's say it's 30 cm away which is about the length of a a ruler isn't it 30 cm so on the card we see a nice bright yellow the light falling on the card big yellow spot on the card now what difference will it make if we move the card further away let's move it to 60 CM now look on the card what difference has it made to the spot on the card and there are two differences one difference is it's over a bigger area and then the second difference is that it's not as bright the intensity is smaller because the same amount of energy has to spread out over a bigger area let's move the card even further away so now it's at 90 cm and we see that the the beam of light has spread out even more and we see that the intensity of the light arriving at the card is even less so what we do now is we get a light meter yeah some kind of an electronic device that measures the intensity and when we put it on the first one it gives a reading of 800 okay whatever that is it's is it in Lux is it in watts per meter squared it doesn't matter gives a reading of 800 what will it read at 60 cm if it reads 800 at 30 what will it read at 60 will it be 400 and the answer is no it will be 200 it'll be a quarter as much when we double the distance the intensity will be a quarter why because when we double the distance what's going to happen to the area of the circle it won't be double it'll be four times bigger because the area of a circle is p piun r s yes so when we double the distance the area is four times bigger so the intensity is a quarter it goes down by a factor of four what will it read at 90 cm well this time the distance is three times bigger so the intensity will be 89 why because that's 800 ided 9 yes why because if the intensity is a 9th when the distance is three times bigger if the distance is three times bigger than the intensity is if you like it's 1 3^ SAR which is a 9 if the distance between the transmitter and the receiver is doubled then the intensity of the the received wave will be a quarter yes why because the area of the sphere look what this person shouting ah and that energy is spreading out into a sphere yes a three-dimensional sphere and the area of that sphere will be four times bigger the the the surface area of a sphere is actually 4 pi r s Yeah so basically if you double the radius the area is four times bigger if the distance between the transmitter and the receiver is trebled if it's three times as much then the intensity of the wave received will be uh hopefully you can tell me it will be a 9th why because the area of the sphere will be nine times bigger yes if the radius of the sphere is three times bigger then the area will be nine times bigger so we say that intensity follows an inverse Square law uh in physics I mean if you were doing a level physics there's loads of things follow an inverse Square law uh this is the only one that we need to worry about in applied science and basically what it's saying is if the distance is double the intensity is a quarter because it's 1 / 2^ sared if the distance is time 3 then the intensity is a 9th because it's 1 over 3^ SAR now if you're lucky you'll get a very simple question but I think you're more likely if a question comes up on this it's likely to be a bit more tricky and you're going to have to use this equation now it's i = k / r² squ where K is a constant okay so the value of K doesn't change K is a constant I is the intensity and R is the distance and looking at it the intensity is proportional to the distance squared actually one over the distance squared which is why it's called an inverse Square law it's the inverse of the distance squared uh doesn't matter about the units it's the relationship that we're interested in and the point is that if we know the value of K if you're given a value of K or you're given enough information to work it out then you can work out the intensity at any distance or you can work out the the distance at which the intensity has a certain value and to work out the value of K it would be that just rearranging the equation so k = i * R2 and I think on most questions if this comes up it will involve working out the value of K and then using that value of K to either work out the intensity or to work out the distance right let's have a go at this question here a student measures the light intensity from a bulb at different distances using a light meter uh at 30 cm from the bulb the intensity is 6.6 what will it be at 60 cm what will it be at 72 cm do this yourself don't be lazy pen paper calculator have a go I'll show you the answer in 5 4 3 2 1 and here we go there so the first one is easy because uh we're going from 30 to 60 yeah so the distance is doubling so the intensity will be a quarter so a quar of 6.6 is 1.65 so that's nice and easy okay uh the second one at 72 cm from the bulb well this is a little bit trickier now we need to work out the value of K so K is I * R 2 so 6.6 so what I'm going to do is I'll use use this 30 cm here 0.3 now you can leave it in cenm actually uh you'll get a different value for K but at least it will still be the same value of K it'll be constant but in this one I'm just changing it to meters you don't have to you can leave it in centimeters actually uh so K is I * R 2 so uh at 30 cm 0.3 it's 6.6 so 6.6 * 0.3 s so that's my my value for K there is 0.59 now we can work out the intensity at any distance so it's going to be 0.59 / 0.72 again I've changed it to meters uh 0.72 squar 1.15 watts per met squared and that is the answer now here's an exam question from June 17 uh and this is a stinker uh have a Go With It Yourself uh I'll talk through the answer in a few seconds but I'd like you to have a go yourself first of all uh and the answer is in 5 4 3 2 1 there you go okay a painting is displayed in the dark room uh an electrician fits a single light source to illuminate the painting So at 1 meter from the light source uh it's 100 watts per met squared now the intensity of the light falling on the painting must not be greater than 30 so we want to know if if at 1 meter the intensity is 100 at this distance X we want the intensity to be 30 and it's asking us to work out the value of x okay it says calculate using the inverse Square law the minimum distance so it's asking us to work out the distance where the intensity is 30 so what we're going to do is we're going to work out the value of K yes so i = k over R2 so K is I * R 2 which is 100 * 1 squar okay uh my intensity is 100 uh I've just left that as one I've left that in meters okay so 100 * 1^ 2 is 100 so my value of K is 100 now uh what value of R will the intensity that's intensity there be 30 so I want to know the value of R so I've got to rearrange the equation for r R so hopefully your maths is up to this rearrange that equation for R and it's R is the square root of K over I I know a lot of students will will have struggled with this question it makes it a hard question to be honest you'll get a mark for working out the value of K but to get all of the marks Ral k/ I which is 100 yes uh over 30 yeah which is 1.83 because my value of K is 100 now my intensity is 30 so the sare < TK of 100 over 30 1.83 and that was a stink of that question usually they're not that hard