hello everyone hope you all are doing good I am anjana from lanohub the free learning platform where you can study math science and SSD absolutely free at learnohub.com in today's class we are going to discuss icse class in physics chapter 4 refraction of light at plane surfaces we'll be discussing what refraction is and the refraction through a rectangular glass slab will also be solving problems that are related to it foreign just look at this figure this is an experiment that you can perform take a glass of water and put a pencil like this okay when you look at this from outside you will be seeing that there is a bending happening to the pencil does it really happen no right the pencil will be straight but when you look it from outside you will be seeing a bent and the reason for this is refraction okay now what is refraction we have studied in detail about reflection in class 9th when you have a smooth surface a plain mirror and when light Ray falls on this what happens to the slight Ray this light Ray touches the surface and bounce back okay then you will be having angle of incidence and angle of reflection we have studied the laws of reflection incident Ray reflected ray normal at the point of incidence or lie on the same plane and the second law was angle of incidence is equal to angle of reflection now what is this reflection when we have two mediums given light is traveling from one medium to other what happens to the light is the light path will change the direction of light will change okay a part of this will be getting partially reflected and a part will be refracted to the New Media it will be going into the new medium okay so this process of bending of light when light travels from one medium to other is called refraction the change in the direction of path of light when it passes from one transparent medium to another transparent medium is called refraction so what the basic idea of what refraction is in case of refraction this is your incident Ray foreign is happening so this is reflected ray in this chapter you won't be considering this reflected ray you can just neglect this and this is your refracted ring here you have the normal we have studied about normal when we studied about reflection normal is a straight line which is perpendicular to the surface okay which is at 90 degree the angle between the normal and the incident Ray is angle of incidence angle between the normal and the refracted Ray is angle of refraction got the basic idea so this is what a process of refraction is we'll be considering two cases that is when light travels from a rarer medium to denser medium and the second when light travels from a denser medium to rarer medium for example if we take air and water air is our error medium and water is a denser medium density is more for water so here the first cases we are considering light traveling from our error medium to denser medium we know that light will always be in a straight line path okay light has rectilinear motion light travels in a straight line motion but when light travels from one medium to another that is in case when light travels in vacuum or in air in a single medium but when light travels from one medium to another will not be considering the reflection the partial reflection we are considering only the refraction light is bending here you can see the light bending this is the actual path of light here we have the incident Ray this is the actual path of light what is happening to light here when it moves from Rider to denser medium for example air2 water what has happened the light Ray has been towards a normal now Otis is normal we know right it is a perpendicular drawn to the surface angle will be 90 degree ok now the angle between this is your incident Ray angle between incident Ray and normal is taken as I angle of incidence similarly you have angle of refraction here we have angle of refraction the angle between refracted Ray and the normal now this is the deviation that has happened from the actual path of light what has happened there is a shift okay there is a shift in the refracted Ray so this angle we will be representing as Delta angle of deviation it is how much it has been shifted from its actual path okay now when you consider this normal and the path of Light original path of light you can see that these two lines cut at this point then you will be having vertically opposite angles okay you can have I is equal to R plus Delta incident angle is equal to angle of refraction plus this angle of deviation from this you will be having angle of deviation is equal to angle of incidence minus angle of refraction clear so when light travels from air to water it bends towards the normal or when light travels from a error medium to a denser medium well array of light travels from a rare and medium to a denser medium that is air to Glass it bends towards the normal okay now the second case here you have the second case denser to iron medium with our example water to air denser to rare medium what is happening in this case you have the incident Ray normal angle of incidence okay now this is the normal and this is your angle of refraction here you have the refracted Ray from the actual path it is moving away from the normal by an angle Delta this is your angle of deviation okay now how to find angle of deviation here this is angle of incidence here we have angle of refraction and this is angle of deviation now when you consider these two lines that is actual path of light and the normal vertically opposite angles will be I and this part what is this part this part will be total angle of refraction minus the shift angle of deviation okay so you will be having I is equal to R minus Delta then what is Delta taking this Delta to the opposite side Delta is equal to R minus I okay so angle of deviation in this case that is when light travels from denser to rarer medium you can determine like this refractive angle minus angle of incidence in case when light travel from Rara to denser medium the angle of deviation will be angle of incidence minus angle of refraction okay so we can conclude that when a ray of light travels from a denser medium denser medium to our error medium glass to air it bends away from the normal first case it went towards the normal second case it bends away from the normal understood now if you are taking here we have taken the example of air and water if you are taking water and glass water and glass dryer medium is water and glass is your denser medium when light travels from water to Glass it bends the light Ray will be bending two words normal it bends towards normal okay now this the opposite that this glass to water light is traveling from glass to water in that case it bends away from the normal away from normal okay now the third case is refraction at normal incidence what does normal incidence mean here you have a surface SS Dash now A B is your normal or a B is your incident Ray here you can see this is your incident right incident Ray coincides with the normal normal will be at angle 90 degree with the surface s s Dash first medium second medium air is taken as the first medium which is a rare medium and glass or water can be taken as a denser medium second medium okay here you have the incident Ray and this is your refractory now what is the angle between the incident Ray and normal angle of incidence in this case what is angle of incidence angle of incidence is equal to 0 degrees why because Normal and the incident Ray are coinciding similarly what is the angle of refraction angle of refraction is also equal to 0 degree why because Normal and the refracted Ray are also coincide even though the light travels from one medium to another medium the light will be following a straight line path it won't be deviated when it is at normal incidence okay incident Ray coincides with the normal so that we can say the ray of light incident normally on the surface separating the two media passes and deviated okay there is no deviation angle of deviation will be equal to 0 in this case now what causes refraction we said when light travels from direct to denser medium it bends towards the normal when it travels from denser to rarer medium it bends away from the normal what is actually happening when a ray of light passes from one medium to another medium its direction or path changes because of change in speed this is important there is a change in speed of light in air we know that light travels with the speed C is equal to 3 into 10 raised to 8 meter per second okay when the light travels from this air to another medium like water or glass what is happening is there is a change in the speed of light first case rather to denser medium what is happening the light will start to slow down light which was traveling at a Speed 3 into 10 raised to 8 meter per second here this is your surface you have normal this is your incident Ray and this incident Ray in air medium is traveling at a Speed 3 into 10 raised to 8 meter per second okay when it touches the surface touches the surface here you have a denser medium let us take the case of water the speed of light is going to decrease okay speed of light will decrease in that case we know it will bend towards the normal and this is your first case light slows down rather to denser medium okay light slows down light slope and light slows down what is happening the light bends towards the normal now just the opposite case that is denser to rarer medium when light travels from a denser medium to our error medium here the air we are taking in the second part that the second medium is air air is your second medium this is your light Ray it will bend away from the normal now what is happening the speed of light is increasing okay when light Ray travels from denser to our error medium speed of light is increasing in that case it will bend away from the normal light bends away from the normal second case now even though if the speed of light Remains the Same there may be no change in Direction okay that is the case of normal incidence in normal incidence there may be sometimes there may be speed of light change but the direction will be the same this is how the light will go okay in that case that is when speed of light will remain the same in that case there is no Bend in light light does not bit clear now we'll be discussing the laws of refraction when you studied about reflection we studied two important laws the first law was the incident Ray the reflected ray and the normal at the point of incidence lie on the same plane here again we have a similar law the first law is the incident Ray the refracted Ray and the normal at the point of incidence all lie in the same plane only difference is in place of reflected ray you will have to change it to refracted ring okay so this is your first law secondly what was the second law of reflection it was related to two angles the angle of incidence and the angle of reflection what was the angle of incidence is equal to angle of reflection now here we have to take the ratio of sine of angle of incidence and the sign of angle of refraction angle of incidence angle of refraction we will be taking the ratio of the sign okay in that case when you take the ratio of their sign you will be getting a constant value and this constant will be representing by mu this is your second law these two laws are known as Snell's laws of refraction okay Snell's laws of refraction this is the first statement first is similar to the first law of reflection and the second is a law which gives a relation between angle of incidence and angle of refraction there you had angle of incidence and angle of reflection here it is angle of incidence and angle of refraction when you take the ratio of their signs you will be getting a constant value and this constant value is represented by mu okay what is Mu mu is known as refractive index mu is known as refractive index the sign of angle of incidence to the sign of angle of refraction is equal to a constant which is refractive index we'll study this in DT now we have the refractive index consider a surface this is the normal incident Ray here you have the incident Ray and we are considering two mediums medium one let us take the first Medium as air and this is the second medium medium two Let It Be glass now what happens to the ray of light it bends towards the normal because we are traveling the light is traveling from rarer medium to a denser medium this is your angle of incidence angle between century and normal this is your angle of refraction okay now when you take the sine of angle of incidence and the sine of angle of refraction ratio you will be getting a constant value this will be a constant and this is equal to Mu as we said before mu is the refractive index refractive index of the second medium with respect to First medium this is how you represent it okay either you can write mu 2 1 like this or mu 2 both are correct okay refractive index of the second medium with respect to First medium clear so sign of angle of incidence to the sign of angle of refraction ratio is a constant which is a refractive index the refractive index of second medium with respect to the first medium is defined as the ratio of sine of angle of incidence in the first medium to the sine of angle of refraction in the second medium this is how you can Define refractive index next is the effect on speed frequency and wavelength due to refraction of light when refraction of light takes place what is happening to speed frequency and wavelength of light the first point is related to the speed or velocity of light when a ray of light gets refracted from a rarer to denser medium we have studied this rarer medium to denser medium minus ray of light travels its velocity will decrease okay speed of light will decrease when it travels from a denser to a error medium its velocity City will increase okay here the refraction is increasing and in this case the refraction is decreasing velocity increases refraction decreases now the second is related to frequency the frequency of light depends on the source of light so it does not change on refraction it is only dependent on the source of light for a given set of medium the frequency will be say frequency of light is a constant it does such change for a given source of light okay frequencies are constant the third is the speed of light V in a medium is related to its wavelength Lambda and the relation is V is equal to F Lambda if the first medium is air then we'll be having cuc equal to F Lambda from which we can write Lambda is equal to V by F okay frequency we know is a constant which means Lambda is directly proportional to Velocity when velocity increases velocity increases mean denser medium to error medium velocity increases Lambda also increases okay when velocity decreases Lambda also decreases when you have with June which color has the highest wavelength highest wavelength is for red light okay maximum Lambda and violet has minimum Lambda okay maximum wavelength maximum wavelength mean speed of light is maximum then in that case refraction will be minimum foreign minimum wavelength means speed of light will be less which means refraction is maximum okay so when wavelength increases refraction will decrease clear so this is a what is absolute refractive index relationship between refractive index and speed of light we know that speed of light is different in different medium and the maximum speed of light is when light travels in vacuum or in air the speed of light is maximum in vac Cube and that is equal to 3 into 10 raised to 8 meter per second all other velocity of light in different mediums will be less than 3 into 10 raised to 8 meter per second okay now we can Define refractive index as the ratio of speed of light in vacuum or air to speed of light in that Medium this is what absolute refractive indexes when you do not consider air as the first medium any metal mediums you are considering we just call it refractive index but it is absolute refractive index when light is traveling from Air to any medium either it can be air to glass or air to water in all these cases we'll be having absolute refractive index because the first medium first medium is air light is traveling from Air to any medium okay so when it is from Air to Glass we represent as Mu G okay which means the refractive index of glass with respect to air or to find the value how can you take will be having the speed of light in vacuum or air that is C and we have to find the speed of light in glass when it is traveling from Air to glass so let it be VG VG is the velocity of light in grass so mu G will be equal to C by VG if it is traveling from a or to water air to water in that case let us represent it as Mu W let us refractive index of water with respect to air and that will be equal to speed of light in air or vacuum divided by the speed of light in water ok now we know that maximum speed of light is in vacuum yes that is 3 into 10 raised to 8 meter per second so when you will be having a value Less Than 3 into 10 raised to 8 meter per second in the denominator anyway VG will be less than C also v w will be less than C why is it so because we know that maximum velocity is when light travels in air or vacuum therefore the velocity of light in glass should be less than the velocity of light in air velocity of light in water should be less than the velocity of light in air in that case that is when denominator is less than the numerator what will be the ratio is it greater than 1 or less than one the ratio will surely be greater than 1 right the value will be greater than 1 which means the refractive index of glass or the refractive index of water will have a value which is greater than one just remember this the ratio will be greater than one for example that is we can say that absolute refractive index or refractive index of any medium will always be greater than 1 why because the velocity of light in any medium will be less than the speed of light in air or vacuum okay refractive index value will allow always be greater than 1 let us take some examples to find refractive index the speed of light in air is 3 into 10 raised to 8 meter per second C is given C is equal to 3 into 10 raised to 8 meter per second and in glass it is 2 into 10 raised to 8 meter per second velocity of light in glass is equal to 2 into 10 raised to 8 meter per second what is the refractive index of glass we have to find the refractive index of class refractive index of glass mu G is equal to C by VG which is 3 into 10 raised to 8 meter per second divided by 2 into 10 raised to 8 meter per second 10 raised to it and raised to it can be cancel 3 by 2. that's 1.5 okay now the second part the speed of light in water is given VW is equal to 2.25 into 10 raised to 8 meter per second we know what c is speed of light in air is 3 into 10 raised to 8 meter per second we need to find the refractive index of water refractive index of water mu W is equal to C divided by VW that is 3 into 10 raised to 8 meter per second divided by 2.25 into 10 raised to 8 meter per second when you calculate this you will be getting a value 1.33 now check all these values 1.5 1.33 these values are greater than 1. it is getting greater than one since refractive index is a ratio there is no unit for refractive index there is no unit it is a ratio here here you can see that meter per second will get canceled numerator and denominator when you have the same unit say gets canceled therefore refractive index has no unit next is relate the refractive index we understood what absolute refractive indexes in case of absolute refractive index the first medium will be here relax relative refractive index is when there are two mediums taken it is not necessary that the medium should be air one medium should be a you are considering the light traveling from first medium to second medium okay we know that refractive indexes ratio of the speed of light in first medium to the speed of light in second medium speed of light in medium 1 to the speed of lighting medium to or it can be represented by mu 2 1 can also be written as Mu 2 1. the refractive index of second medium with respect to the first medium speed of light in first medium by speed of light in second medium we know that absolute refractive index the absolute refractive index of first medium will be C divided by V1 and absolute refractive index of second medium will be equal to C divided by V2 where V1 and V2 are the speed of light in the two mediums okay from this we can write V1 is equal to C by mu 1. from this we can write V2 is equal to C by mu 2 just crossing okay say by mu 1 and C by mu 2. now we have mu 2 1 is equal to V 1 by V2 will be equal to V1 V1 is C divided by mu 1 divided by V 2 V 2 is C by mu 2. taking the reciprocal C by mu 1 into mu 2 divided by C just taking the reciprocal of this denominator part then C C gets canceled you will be getting mu 2 by mu 1. therefore the refractive index of second medium to with respect to the first medium is refractive index of the second medium divided by the refractive index of first medium let's take an example light is traveling from water to glass okay speed of light in water VW is equal to 2.25 into 10 raised to 8 meter per second and speed of light in glass is equal to 2 into 10 raised to 8 meter per second okay 2 into 10 raise to 8 meter per second is the speed of light in glass and speed of light in water is 2.25 to 10 raised to it meter per second now if we need to find the refractive index of glass with respect to water why because water is a first medium and glass is your second medium okay will be equal to we can write this as velocity of light in water divided by velocity of lighting glass okay that is equal to 2.25 into 10 raised to 8 divided by 2 into 10 raised to it when you take the ratio you will be getting 1.125 okay 1.125 is the refractive index of glass with respect to water now just the opposite when light is traveling from glass to water glass to water we know the values will be finding the refractive index of water with respect to glass or mu w g this will be equal to velocity of light in glass divided by velocity of light in water what is the velocity of lighting glass that is equal to 2 into 10 raised to 8 divided by inverter it is 2.25 into 10 raised to 8 2 divided by 2.25 you'll be getting a value 0.89 okay this is a refractive index of water with respect to Glass they both are different okay refractive index of glass with respect to water and the refractive index of water with respect to Glass you will be getting two different values you remember this point now we'll be studying the two condition for array to pass undeviated on refraction and if it means the normal instance case so here you have the surface this is your incident Ray this is your refractory so when the angle of incident at the boundary of two medium is 0 we know that the normal incidence case here you are having the incident Ray this is the same as your normal therefore angle of incidence will be equal to zero if angle of incidence between the two medium is equal to between the boundary of two medium is equal to 0 the ray of light will be undeviated it will be going along the same straight line path first one second when the refractive index of medium 2 is same as that of medium one we know that refractive index related refractive index is represented by mu 2 1 what is Mu 2 1 mu 2 1 is Mu 2 divided by mu 1. okay when the refractive indexes is same you will be getting the ratio to be equal to 1. we know that according to Snell's law sine I by sine r is equal to Mu 2 1 refractive index of the second medium with respect to First medium I use the angle of incidence and R is the angle of refraction sine I by sine R will be equal to Mu 2 1 okay now when this mu 2 1 when mu 1 and mu 2 are equal in that case this value will be 1 which means the ratio of sine I by sine R is equal to 1 what does it mean sine I will be equal to sine r so sine I is equal to sine R then angle of incidence should be equal to angle of refraction so two cases when incident angle is equal to 0 and reflected angle equal to 0 in that case you will be getting the rate to pass and deviated and the second is when the angle of incidence is equal to angle of refraction that is Mu 2 and the both the mediums are the same mediums okay medium one and medium two are same with medium one and medium two are same the refractive indexes will be same then angle of incidence will be equal to angle of refraction in that case also the ray will go and deviate it relationship between wavelengths in the two different mediums OK here we have two mediums let's say air and water this is your first video Let's represent it by mu 1 and this is Mu 2 a refractive index mu 1 and mu 2 first medium of refractive index mu one second medium of refractive index mu 2. now we are considering array of light of wavelength Lambda and frequency F okay so here you have to remember one point frequency is never going to change when light Ray travels from one medium to other wavelength can change but frequency will always remain C now what about velocity of light velocity of light will also be different okay in different medium velocity of light is different in air we'll be having velocity of light to be equal to C okay and in water let's say the velocity is taken as V okay now we can write a relation that is we have velocity is equal to wavelength into frequency so first in the case of air first medium is air in case of air we'll be having C is equal to Lambda F from which the frequency f is C by Lambda okay now the light Ray travels from this medium to the second medium light travels from air to water rarer to denser medium what happens is the velocity is changing with this the wavelength is also changing let's say now the wavelength is Lambda Dash here velocity V velocity of light in water V is equal to Lambda Dash frequency is the same we can represent it by F okay which means here we'll be having frequency f is equal to V divided by Lambda Dash Okay so this is your first equation and this is your second equation when you check these two equations you can find the left hand side is frequency left hand side is f okay therefore right hand side should also be equal c bar Lambda will be equal to V by Lambda Dash cross multiplying we get C Lambda Dash is equal to V Lambda and then Lambda Dash is equal to V Lambda divided by c what does this mean you can find the wavelength of light in the second medium by knowing the wavelength in the first medium velocity of light in the second medium and the velocity of light in first medium here the first medium is air okay we have a relation refractive index mu is equal to velocity of light in air divided by velocity of Latina given medium okay here velocity of light air divide by velocity of light in water OK for example if mu 2 1 represents the refractive index of second medium with respect to First medium we can write this is equal to Mu 2 divided by mu 1 okay now check this here we have Lambda Dash is equal to V by C into Lambda okay we have C by V is equal to refractive index mu which means we can write take the reciprocal of this 1 by mu will be equal to V by C taking reciprocal of the left hand side and right hand side okay now substituting here we get Lambda Dash to be equal to Lambda divided by refractive index of the medium okay now consider this case if the light is moving from a iron medium to a denser medium which means the refractive index of first medium is less than the refractive index of second median okay second medium is denser in that case you will be getting the refractive index to be greater than one yes refractive index will be greater than one why because numerator is greater than denominator okay the refractive index is greater than one what does it mean here this value mu value will be greater than 1 so when you divide wavelength with a value which is greater than 1 we will be getting Lambda dash at that Lambda Dash is less than Lambda because we are dividing with the number which is greater than 1 okay now if the refractive index of the first medium is greater than the refractive index of second medium which means the denominator is greater okay if denominator is greater then you will be getting refractive index to be a value Which is less than 1 Okay so so if you are dividing wavelength with the value Which is less than 1 then Lambda Dash will be greater than Lambda understood if the refractive index mu is greater than 1 then Lambda Dash is less than Lambda and if refractive index mu is less than 1 then Lambda Dash is greater than Lambda what does it mean when refractive index mu increases wavelength decreases okay here error two denser medium light is traveling from rarer to denser medium which means the refractive index is increasing what happens to wavelength wavelength is decreasing for a given frequency of light okay next the factors on which the refractive index depends the first one is nature of medium that is velocity of light we know that when light travels from one medium to another the velocity of light is changing right for example the velocity of light in glass is equal to 2 into 10 raised to 8 meter per second velocity of light in water is equal to 2.25 into 10 raised to 8 meter per second what about refractive index we have determined that refractive index in case of glass is equal to what is the value that we are obtaining will be getting 1.5 1.5 is the refractive index of glass and the refractive index of water is equal to 1.3 so when you check this you will understand velocity of light in glass is less than velocity of light in water what about refractive index refractive index of glass is greater than refractive index of water which means the velocity and refractive index are inversely proportional when velocity increases refractive index decreases and when refractive index increases velocity decreases clear second is physical condition such as temperature what happens when temperature increases in case of air we have studied that hot air is less denser okay hot air is rarer compared to normal air so when temperature changes what is happening we set hot air hot when temperature increases what is happening to density density is decreasing density is decreasing when temperature increases okay density decreases means what is what about refractive index refractive index is also decreasing therefore temperature is inversely proportional to refractive index second condition the third condition is the color or wavelength of light how does wavelength depends on the refractive index when you take vibjured maximum wavelength is for red light okay minimum is for Violet minimum wavelength is for violet light okay refractive index we know refractive index mu 2 1 is Mu 2 divided by mu 1 which is equal to V1 by V2 is equal to Lambda 1 by Lambda 2 frequency is a constant okay therefore what will be the dependence of wavelength and refractive index the medium that is for violet light there will be having maximum new value why because refractive index is inversely proportional to wavelength wavelength and velocity are directly proportional therefore refractive indexes inversely proportional to Velocity and refractive index is inversely proportional to wavelength and red light will be having minimum refractive index so here when you take these colors the wavelength is increasing wavelength of light is increasing from with Violet to Red what about refractive index refractive index will be decreasing and the minimum refractive index is obtained for red light principle of reversibility of the path of light now consider here you have the surface this is your incident Ray normal and this is your refracted light is traveling from a rarer medium to a denser medium okay this is the path of light now according to this principle of reversibility it states that the light will follow exactly the same path if its direction of travel is Twist what is direction of travel reversing mean here if this light is traveling from the opposite mediums that is light is starting from here and traveling to this New Media okay rarer two dancer medium is the first case then this is your path of light and denser to rare medium if the light is traveling the light is starting from here if this is your incident Ray then the refracted Ray will be the same okay for a given set of medium this is your error medium and this is your denser medium okay so this is how it is you will be having in case of air and glass the refractive index of glass with respect to air because light is traveling from Air to Glass the first medium is air and second medium is glass okay mu GA that will be equal to Mu G divided by mu a yes and this is sine I divided by sine R okay angle of incidence I angle of refraction r mu G A will be sine I divided by sine n now take the case if the light is starting from here the initial cases when light is starting from here this is your incident Ray now light is starting from here here you have the instant Ray and this is your refracted Ray in that case this angle will become the angle of incidence and this angle will be the angle of refraction for the convenience we'll be taking the same okay in that case the first medium is glass glass to air when the light travels you will be having the refractive index of air with respect to glass is equal to Mu a divided by mu G and that is equal to sine r divided by sine I refractive index of first medium here the first medium is this one and the second medium is this one okay so sine of angle of incidence in this case angle of incidence is the refracted angle okay here we have the first medium in first medium the angle of incidence will be formed and this is your second medium in the second media you will be having the angle of refraction now we will be taking it as I okay now when you multiply these two this is your first equation and this is your second equation when you multiply these two what do we get you will be getting mu G A into mu a g for our convenience we are taking the mediums as air and glass if you are taking mediums as two and one refractive index of the second medium with respect to First medium into refractive index of first medium with respect to second medium will be equal to sine I divided by sine R into sine r divided by sine I on canceling you will be getting that equal to 1 which means mu 2 1 will be 1 by mu 1 2. the refractive index of the first medium with respect to the second medium will be equal to 1 by the refractive index of the second medium with respect to First medium this is a principle of reversibility clear next is the refraction of light through a rectangular glass block okay here you have a rectangular glass block p q r s the first medium is air then the light travels through glass and the third medium is again air so first here you have the incident Ray this incident Ray when it strikes the surface p q what happens when you draw a normal you will be getting angle of incidence so here what is happening the light is traveling from Air to glass Air 2 glass which means a rarer medium to a denser medium when light travels from a rare medium to denser medium we know that it bends towards the normal right the speed of light is decreasing in this case speed of light is decreasing and it bends towards the normal you will be getting the angle of refraction with respect to the normal okay the angle between normal and the refracted Ray now what happens is this refracted Ray again falls on the opposite side okay opposite surface it falls that is s r and then comes out to air this is a second refraction two refraction is happening first is from Air to Glass and the second is from glass to air so when it travels from glass to air what is happening in this case the light is traveling from a denser medium to our error medium denser to error medium the speed of light is increasing when it comes from glass to air the speed of light is increasing and it moves away from the normal okay now we have angle of incidence and angle of refraction in this case clear here this Ray is called emergency this Ray is called emergentry when you draw a line a straight line extending the incident Ray this is your straight line when you extend the incident Ray this straight line will be parallel to the emergency the ray which is coming out from the glass when it comes to the air is the emergent Ray emergent Ray and the incident Ray extended will be parallel to each other the angle between the emergency and the normal is called angle of emergence so 2 times refraction is happening in this case when you take the case of a rectangular glass slab or a rectangular glass block okay this is called when there is a shift happening right so here bending towards the normal is happening and here bending away from the normal is happening so we'll be getting a distance a shift between the emergent Ray and the instant Ray extended so this shift is a perpendicular distance and this is called lateral displacement lateral displacement okay so first is from Air to Glass and the second is from glass to air we have studied the principle of reversibility okay here we have two parallel surfaces then according to the principle of reversibility air to Glass glass to air is how the light is traveling okay first from Air to Glass and second from glass to air so principle of reversibility when used you will be getting angle of incident equal to the angle of emergence okay what was principle of reversibility you had mu 2 1 is equal to 1 by mu 1 2 okay here first medium is here second is glass in this refraction glass is the first medium and the second is air okay when you take this with the sign when you consider the snail slow you will be having sine of angle of incidence and sine of angle of refraction when you take the ratio and you will be getting angle of incidence is equal to angle of emergence here this angle and this angle are the same these are the angle of refraction represented by R here this is the angle of incidence incidence for this ray of light and it is represented as R Dash these value will be seen why because normals will be parallel to each other and here you have a transversal okay so these two this whole angle will be same then we are having this 90 degree and this part 90 degree in that case r and r Dash will be equal clear angle of incidence and angle of emergence will be same so we have discussed what is lateral displacement here you have the incident Ray incident Ray when extended you will be getting parallel to the emergency okay angle of incidence and angle of emergence are equal this parallels when you take the perpendicular distance between the emergent Ray and the incident Ray you will be getting the lateral displacement so we will be saying that the emergent Ray is laterally shifted from the original path of light lateral displacement can be defined as the perpendicular distance between incident Ray and emergency okay now what are the factors on which the lateral displacement depends on the first factor is the thickness of block so when you are using a glass block when the thickness of this glass block increases thickness increases you will be having the lateral displacement to be increased lateral displacement will also increase thickness and lateral displacement are directly proportional I mean one point that is when thickness increases the lateral displacement will also increase this is your first point now the second one the angle of incident consider this case here you have angle of incidence this is your angle of emergence here you have the lateral displacement now take the case when you are increasing the angle of incidence here we are increasing the angle of incidence this is your new incident Ray and the angle of incidence has increased okay then it will undergo refraction somewhere here you will be getting let's say so this is your refracted Ray okay so refraction angle will decrease in that case and here you draw a normal this is how you get the emerge entry this is how you will be getting the emergency okay here you have the angle of emergence new angle of emergence now when this is extended that is the incident Ray is extended here you have a straight line this is the straight line we'll be having the emergency and the instant Ray parallel to each other and the perpendicular distance between them will be giving you the lateral displacement here this is your lateral displacement now when you check this you will find in the first case the lateral displacement is just this much here the lateral displacement has increased so what happens when the angle of incidence is increased lateral displacement will also increase angle of incidence increases lateral displacement will also increase okay they are also directly proportional now the third is refractive index of the block how does the lateral displacement depend on the refractive index of block we know that the refractive index and wavelength are inversely proportional now when refractive index increases the lateral displacement will also increase will also increase what is happening when refractive index increases the density is changing right so here when you have glass so this is a new medium with a greater refractive index the density has increased bending towards the normal will increase yes the refractive angle will decrease it will bend towards the normal refractive angle will be decreasing in that case you will be getting somewhere here you will be having the emergence emergent Ray will be somewhere here so the distance of also increasing from the normal this distance this distance has increased right this is what is happening here when density increases or refractive index increases the lateral displacement will also increase when we relate it to wavelength when wavelength decreases the lateral displacement lateral displacement foreign decreases the lateral displacement will increase so when you consider which your violet light is having the least wavelength okay violet light has least wavelength the full lateral displacement will be maximum when you are using violet light okay now let us do problems here you have the first problem refractive index of water is four by three UW is equal to 4 divided by 3 we have to calculate the speed of light in water VW speed of light in vacuum is given which is equal to 3 into 10 raised to 8 meter per second okay refractive index in water will be equal to speed of light in vacuum divided by speed of light in water from this we can write the speed of light in water is equal to speed of light in vacuum divided by refractive index of water put the values 3 into 10 raised to 8 meter per second divided by mu W mu W is 4 by 3. 4 by 3 taking the reciprocal 3 into 10 raised to eight into three divided by 4. 3 into 3 is 9 9 divided by 4 9 by 4 into 10 raised to eight the unit is meter per second nine by four you will be getting a value 2.25 into 10 raised to 8 meter per second therefore the velocity of light in water is equal to 2.25 into 10 raised to 8 meter per second okay this is how you will be finding here you have the absolute refractive index because the first medium is air or vacuum now the second problem orange light or wavelength 6600 angstrom traveling in air gets refracted in water at the speed of light in air is 3 into 10 raised to 8 meter per second and refractive index of water is four by three we have to find the first one the frequency of light in air second the speed of light in water and the third the wavelength of light is water okay given orange light of wavelength 6600 angst and traveling in air Lambda a is equal to 6600 answer which is equal to we know one angstring is tender is to minus 10 meters therefore 6600 into 10 raised to minus 10 meter meter is the standard unit what else is given if the speed of light in air is given C is equal to 3 into 10 raised to 8 meter per second and refractive index of water is given refractive index of water with respect to air is equal to 4 divided by 3. first question is to find the frequency of light in air the frequency of light in air and the frequency of light in water will be the same because frequency is not changing frequency is a constant for a given source of light okay we just need to find the frequency of light in air we have the relation speed is equal to frequency in two wavelength from this we can write frequency is equal to Velocity by wavelength here we have speed of light in air divided by wavelength of light in air put the values 3 into 10 raised to 8 divided by Lambda a is equal to 6600 into 10 raised to minus 10. okay when you calculate this you will be getting a value 4.54 into 10 raised to 40 and the unit of frequency is Hertz okay Hertz is the unit of frequency now the second part we need to find the speed of light in water we have a relation refractive index of water with respect to air is equal to refractive index of Water by refractive index of air that is equal to velocity of light in air divided by velocity of light in water which is equal to wavelength of light in air divided by wavelength of light in water so this is a relation if you remember this relation you can find any value here we need to find the speed of light in water we have to find the speed of light in water so take the relation Lambda W A A is not necessary because here it is the absolute refractive index is equal to VA is C divided by v w which means VW is equal to C divided by Mueller mu W mu W is a refractive index of water what is the value of c 3 into 10 raised to 8 divided by 4 by 3 okay which is equal to 3 into 10 raised to a taking the reciprocal and multiplying that is 3 divided by 4 3 into 3 you will be getting 9 into 10 raised to 8 divided by 4 9 into 10 to 8 by 4 will be equal to 2.25 into 10 raised to 8 and this is a velocity of light in water therefore the unit is meter per second meter per second is the unit next part we have to find the wavelength of light in water wavelength of light in water how to find again we'll be using the same relation mu W is equal to Lambda a divided by Lambda W from which we will be having Lambda W is equal to Lambda a divided by mu W put the value Lambda a wavelength of light in air is 6600 into 10 raised to minus 10 divided by mu W mu W will be equal to 4 by 3. 4 by 3 that is 6 6 double 0 into 10 raised to minus 10 into 3 divided by 4. okay so on dividing you should find the value will be equal to 198 double 0 into 10 raised to minus 10 divided by 4 on dividing this you will be getting 4000 950 into 10 raised to minus 10 meter so 10 raised to minus 10 we know it is and strength for 4950 answering is your final answer clear problem 3 the speed of light in diamond is 1 lakh 25 000 kilometer per second what is its refractive index we have to find the refractive index okay here we know the speed of light in vacuum which is 3 into 10 raised to 8 meter per second and the speed of light in diamond is given which is 1 lakh 25 000 kilometer per second so this kilometer per second should be first converted into meter per second we know one kilometer is thousand meters therefore this becomes one two five triple zero into thousand meter per second or this can be written as 1.25 into 10 raised to 8 meter per second okay refractive index we have to find refractive index of diamond as refractive index of diamond with respect to air will be equal to velocity of light in air by velocity of lighten diameter okay V A by v d V A will be equal to C C by weed for the values C is equal to 3 into 10 raised to H and v d is equal to 1.25 into 10 raised to 8. when you calculate this 10 raised to a 10 raised to 8 can be canceled 3 divided by 1 1.25 will be getting 2.4 as the answer refractive index is a ratio remember there is no unit 2.4 is your final answer next problem number four array of light of wavelength 5400 angstrom suffers refraction from Air to Glass taking the refractive index of glass with respect to air is equal to 3 by 2 find the wavelength of light glass we have to find the wavelength of light in glass what is given the wavelength of light in air is given which is equal to 5400 and strength okay what else is given the refractive index of glass with respect to air is 3 by 2. refractive index of glass with respect to errors 3 by 2. okay you will be having refractive index of glass with respect to error is equal to the wavelength of light in air divided by wavelength of light in glass okay what we need to find is wavelength of lighting glass which is Lambda G is equal to Lambda a divided by mu GA on Crossing you will be getting this one what is Lambda a Lambda is equal to 5400 divided by refractive index is given which is 3 by 2. taking the reciprocal and multiplying 5400 into 2 divided by three five thousand four hundred into two divided by three so on canceling you will be getting 1800 here 1800 into 2 is 3600 and the unit is angstrom here you are not changing the unit to meters because here this quantity that we have is one refractive index refractive index has no unit so it is not going to make any changes okay when it is velocity when velocity meter this meter is in meter per second in that case you will have to change this to meters that's all for today in today's class we have discussed basic concepts about refraction and the refraction of a light rate through a rectangular glass slab you have also solved a few problems hope you all enjoyed the session I'll be back in the next session until then stay tuned to learn thank you