00:10
So let's start with the problem the problem given is from optics and here we have an optical fiber okay, so optical fiber has one core okay and there will be claddy always the core it will be denser okay and claddy is rarer okay so we have been given that n1 i'll just label down whatever is within the problem so just one second sorry for the disturbance okay i'll continue okay this is the imaginary line passing through the center okay so there is an incident ray okay at the core okay so it comes like this it is incident and it is making an angle of theta i okay it enters the core so this is it is coming from here so here is rarer and this is denser so the bending will be towards the normal okay the refraction will be towards the normal and this was our normal okay so it will be towards the normal so it goes like this this is the path okay and then it hits the cladding and core interface and from there according to the figure it is reflected internally okay and we have been given the core is n1 and the refractive in the cell of cladding is n2 and we have been given n1 is equal to 1 .465 and n2 is equal to 1 .450 it is given that this is of silica and germanium and this is of silica we need not go into that so what we have to do is we have to find the critical angle so we have to find theta critical and then we have to show that sine theta i is equal to root n1 square minus n2 square okay and then we have to show we have to find the value of theta i okay these are the things that we have to do so let's start okay so i'll tell you the light entered and the core and air interface okay so core is denser so it was refracted towards the normal okay so bending is towards the normal it then goes to the cladding and core interface and then it is internally reflected because this angle will be greater than c critical angle theta critical okay so i'll make the normal at this point also okay so like this i'll make okay so i'll take the angle over here as theta r and this is theta c okay so okay that is for us for a time being you consider this as theta c okay this will be something else, okay, i'll write theta, light c, light c, it's not theta c, light c, theta c is your critical angle, okay? okay, so when the angle over here, is it, it is greater than theta c, okay? when it is greater than theta c, then only total internal reflection will take place, okay? so, okay, so let's start.
03:38
So we have to find the theta c, okay? theta c is theta critical is what the critical angle so at critical angle what happens is there will be uh what refraction along the normal okay sorry along the surface so the angle will be 90 okay so this is theta c okay so this is taking place at this okay imagine this over here okay so what we will do using snail's law we will write law of refraction we can write this is n1 this is n2 okay this was core and this is cladding this is taking place at this point okay so i light n1 sine theta c equal to uh into sine of 90 okay the angle the angle refractor angle over here is actually 90 degree okay so sign 90 is 1 so this becomes n1 sine theta c is equal to n2 okay sign theta c is i have written critical as theta c okay so sine theta c is n2 by n1 so if you know the definition of formula you can straight away use it okay this is actually the derivation for it so okay so i'll write sine theta c is equal to n2 by n1 okay so over here we have n2 as how much n2 is 1 .45 okay n2 is 1 .45 and n1 is 1 .46 so i'll substitute the value 1 .46 okay so this comes around 0 .98 okay so i have sine theta c as 0 .98 okay so if i need theta c it will be sine inverse of 0 .98.
05:30
So if i need theta c it will be sine inverse of 0 .9.
05:35
98 okay sign inverse of 0 .98 and this comes around 78 .5 degree okay so we have found theta c as 78 .5 degree so let's move on to the b part okay where we have to derive the what you have to derive sign this numerical aperture this is actually numerical aperture i light it over here this is actually numerical aperture we have to derive it okay so i'll copy this figure onto that place okay copy and let's start deriving it okay so it hasn't come okay let's start from again okay so we solved that problem so i'll draw a condition okay this is the normal okay the ray is entering we have this is uh n2 and this was n sorry the core was n1 okay am i right yes the core is n1 and the cladding is n2 okay so it will come it will be incident over here so it will bend towards the normal and then from here it underwent t ir total internal reflection okay so this is we are assuming at c and c is always greater than theta c okay if then only it will get reflected internally otherwise uh what at theta c what happens is it is refracted along the surface okay so let's start proving it and this is theta i okay okay so let's start proving it okay so what i'll do is i'll assume n -not as the i'll write everything okay as the refractive index of index of air okay so at the air core interface okay air core interface is this point okay so here i'll write okay i forgot to write the angle over here let us assume it as theta r, refracted, okay, it is refracted from the normal, okay? so let's write the snell's law, okay? using snell's law, okay? i'll write the equation for this.
07:50
So what i'll write is n0, n0 theta i, n not, n.
07:56
N not, sine theta i, n not is, this is for air medium, n0, sine theta i equals the refractive index for core is n1.
08:07
So n1, sine theta r, okay, sine theta r.
08:13
This is what i can write, okay? so i'll write sine and we know n -0, a refractive index of air is 1.
08:21
Okay, so this will become sine theta i equals n1, sine theta r.
08:28
Okay, so what i can see is this triangle, if you can look carefully, okay, this is this triangle, i'll take it over here.
08:35
So this angle is theta r and this is c and this is 90 degree okay so can i write 90 plus theta r plus c as 180 okay so i can write i can take this c and 180 over here so theta r becomes 90 minus c okay i have no this concept okay i have no have uh what calculated theta r as 90 minus c okay so i'll substitute over here 90 is 90 degree okay so sine theta i equal n1, sine theta, instead of theta r, i'll put 90 minus c.
09:10
Okay, if you know, 90 is nothing but our pi by 2.
09:13
Okay, so we know sine pi by 2 minus theta is what? sign pi by 2 minus theta is cost theta.
09:19
Okay, so we also know sine 90 minus theta is cost theta.
09:23
So i'll write it, sine theta i is equal to n1, sign, sorry, this will change to cost, cos c, okay? this is one relation that we have now what we'll do is we will apply this nelson at this point okay okay so i'll write at this point i'll draw it okay so it is like this is the normal and this is okay so this is c okay so i'll write this was n1 this was n2 this is our core this is our cladding okay so at core cladding interface phase i'll write okay this was c and this is assume it at and going at 90 degree okay at theta c okay so i write a sign sorry sorry sorry i'll multiply the refractive index n1 sine c okay n1 sine c equal to n2 sine 90 okay this will be n2 sine 90 okay and what i can do is sign c i can write it this will be 1 okay and i'll take n1 there so this will become n2 by n1 so i have got this relation okay and also i know okay and what we got from the previous step is this thing okay so i'll write it over here i'll write sign theta i equal to n1 cos c okay cos c so we know what is sign c we don't know what is cos c so we have learned is sorry we have learned is sine square theta plus cost square theta is equal to 1 so cost square theta can we write it like 1 minus sine square theta okay and cost theta is nothing but under root 1 minus sine square theta okay this is the formula that we are going to use it over here so instead of see we will be putting this okay so this becomes i can write it as n1 okay under root 1 minus sine square c okay this is sine theta i itself okay and instead of sine c i'll substitute these values okay so this becomes sine theta i equal to n1 and i'll what right this substitute this value into by n1 square okay now i'll expand it okay this becomes n1 under root okay denominator it will become n1 square minus n2 square upon n1 square okay now what i'll do is i'll take this n1 out okay so this becomes sine theta i uh sorry i need not write sign tita i i'll light n1 upon n1 under root n1 square minus n2 square okay since i took this n1 square it will come out in denominator as n1 okay now i'll cancel this n1 and n1 i use different color okay i'll cancel them so i'll get what i'll get sine theta i is equal to under root n1 square minus n2 square this is what we needed to prove in part b i hope you got it okay so i'll take a quick recap okay we will write the uh what using snail's law we will write the equation at the core and air interface and then we will write it at the core cladding interface so when we write at air interface we get this relation and when we write at core cladding interface we get this relation okay and this relation okay and this thing we will substitute in equation one okay i'll name it this is equivalent equation 1 and this was equation 2 okay so i'll substitute 2 in equation 1 okay so we substituted this and we got this relation okay this was what we had to find in the second part okay now this is the second part so part 1 we have done theta critical we have found it was 78 .5 we have done sine theta we have proved it now we have to find theta i okay so it is simple we will just use this formula okay, so moving on...