Lateral shift and apparent depth

[00:00:01]So let's quickly see a numerical on deviation of light caused due to refraction at a glass pier. As we can see here the light ray is instant at an angle alpha at the normal. The dotted lines indicate the normals at the point of incidence and the refraction.

[00:00:22]It makes an angle alpha with the normal.

[00:00:24]Once after the refraction it makes an angle beta.

[00:00:28]Again the angle of emergence is alpha.

[00:00:31]Now we need to find the angle of deviation.

[00:00:34]Here net deviation net deviation is equals to deviation 1 plus deviation 2 since it has been deviated twice. Or we can also say that deviation is nothing but angle between the initial angle angle between the initial incident ray and the finally refracted ray that is this is the angle delta. So at the first as alpha is greater than beta. So we write it as alpha minus beta as deviation plus once again angle of emergence is greater than angle of refraction that is alpha minus beta.

[00:01:10]Hence this becomes 2 into alpha minus beta the net deviation suffered by the light ray passing through the glass slab.

[00:01:24]So let's learn about lateral shift in refractions at the plane surfaces.

[00:01:30]When light passes through rectangular parallel glass slabs where let us name this A, B, C and D. A B and C D are called as two parallel refracting surfaces. If light ever falls on the surface A at an angle, let us call it as I, it bends towards the normal and again moves away from the normal. Now here this I is called angle of incidence.

[00:02:05]This is called angle of refraction.

[00:02:09]If I call the refractive index of air being one and refractive index of glass slab, we'll call it as mu. that that doesn't matter when we derive everything. Now if you observe this angle of incidence and angle of emergence both will be equal because using the snail's law snell's law using the snell's law the incident ray and the final emergent ray incident ray and the final emergent ray are parallel are parallel.

[00:02:49]But if you observe here the instant ray because of the presence of the glass slab has been slightly displaced towards the left towards the left. So due to this there has come a deviation. So this is the deviation we'll call it as delta or simply I - r. This angle is I minus r.

[00:03:24]So now job is to find how far the incident ray has been displaced. That is this part. This part if we name the points like this point being P this being Q R. Let's take another point now.

[00:03:47]Now this distance QR that is the lateral displacement lateral displacement or shift let's call it as delta x that is in terms of qr we need to find that value qr in order to find that there is an opposite angle that is sin so that is I minus r so I can take the ratio of that so that can be written like sin Sin of I - R is equals to QR divided by PQ opposite side by hypotenuse where this comes the 90° again of PQ we don't know the value of PQ for PQ we'll take another triangle let's call this as S this is the previous one was from the triangle PQR R this is first equation.

[00:04:49]Let's pull out another equation from triangle PQS.

[00:04:57]We need the value. We know the thickness of this slab as T. Let us assume that the thickness of the glass slab is T.

[00:05:05]The distance traveled by the light ray inside the glass slab is T. It means that there are two known values. I know the angle of refraction. I know the thickness. Hence, I'll take cos cos r is equals to t by pq.

[00:05:25]So PQ is equals to that implies PQ is equals to T by cos R from 1 to 2 1 can be written as delta X by PQ Q delta X is equ= to PQ into sin I - R which would be written as T into sin I - R divided by cos R.

[00:06:16]This becomes the mathematical value of the lateral shift delta x = to t sin i - r by cos r.

[00:06:31]This is important in terms of the board.

[00:06:34]This is lateral shift of lateral shift of the light ray at the plane refracting surface.

[00:06:46]So this is for the board point of view.

[00:06:49]Let's improvise this for mains in terms of the mains.

[00:06:56]For mains we would write this with a small twist that is for small angles of incidences.

[00:07:04]We write this for the small angles of incidence. small angles of incidence we might make use of paragial approximation that is sin of I - R will be almost equal to I - R and cos of R will be equals to 1 that is paragal approximation when the values are much smaller when the angles are much smaller we make use of these approximations.

[00:07:39]In such a case, lateral shift becomes the value of lateral shift which was delta x = t into i - r / 1 / 1.

[00:07:58]This becomes x = t into i into 1 - r by i lateral shift.

[00:08:12]We can also make use of some other assumption that from snail's law.

[00:08:16]Snell's law tells us that mu1 sin i is equals to mu2 sin r.

[00:08:28]For small angles, it would become sin i becomes i.

[00:08:35]For small angles, sin i would be equals to i. sin r would be equals to r.

[00:08:49]Then mu1 i is equal to mu2 r. That implies r by i gives us mu1 by mu2.

[00:09:08]And this now changes our expression for lateral shift. Now the lateral shift looks like delta x is equ= to thickness into angle of incidence into 1 minus mu1 by mu2.

[00:09:28]Now what are mu1 and mu2? mu1 is medium of incidence. Mu2 is medium of refraction from where light is incident. that is mu1 from into which light is refracted that is mu2. Now this is how we derive an expression for lateral shift in terms of ne in terms of j means and need. This is so useful for us.

[00:09:57]Let's see a numerical on this lateral shift. The question says if a glass slab a glass slab of thickness 0.1 cm.

[00:10:23]Angle of deviation of a light ray.

[00:10:36]Incident on it is 60° and angle of refraction being 60°.

[00:10:59]Find lateral shift.

[00:11:05]We quickly can jump to the formula.

[00:11:09]Lateral shift is equals to t into sin of i - r / cos r.

[00:11:17]And the given data is t is equal to 0.1 cm. I can write it as 10 - 3 m.

[00:11:26]Deviation is 60 and angle of refraction is 60.

[00:11:38]And we also know that delta is equals to i - r.

[00:11:46]We can make use of this in the formula.

[00:11:48]So this implies t is 10 - 3 into sin of i - r that is sin of 60 that is <unk>3 half divided by cos r r value is again 60 so that becomes half that equals to 1.732 into 10 to - 3 m that is 1.73 32 mm.

[00:12:21]That would be the answer. Note it down.

[00:12:28]And if ever a graph is drawn between angle of deviation and angle of incidence, deviation versus angle of incidence, there would be a condition for There would be a condition for maximum lateral shift and the graph would look like this.

[00:13:02]And this value is possible. This is the deviation on y-axis. Angle of incidence being the maximum at by2.

[00:13:11]If this is the case means since x is equ= to t into sin of i - r divided by cos r.

[00:13:25]We shall plug in the value of I equ= to<unk> by 2. Then r becomes a critical angle. A topic which we'll discuss in the next classes. t into sin of<unk> by 2 - c divided by cos c.

[00:13:46]That implies x maximum is equals to t. That is the maximum lateral shift possible. X maximum is nothing but maximum lateral shift possible.

[00:14:07]Right?

[00:14:16]As of now we have understood that as light enters from one medium to another its properties like wavelength, speed will change. So due to this the time taken by this or the path also changes. We'll study that in terms of a separate topic called as optical path.

[00:14:40]So optical path is based on characteristics of the medium. We'll study it as optical path.

[00:14:52]In simpler terms, it is nothing but the distance traveled by the light in vacuum for a given time. distance traveled by the light in vacuum for a given time or the product of refractive index.

[00:15:11]The product of refractive index of the medium and distance traveled by the light in that medium.

[00:15:41]that is called as optical path. So optical path let's call it as ddash is equals to mu into d. d is nothing but the distance traveled by that distance traveled by that in the medium.

[00:16:00]It's nothing but imagine a case where a light is traveling through a medium of refractive index mu. Distance traveled by the light is d in t seconds. The optical path can be seen as now. Now this optical path can be seen as so it it will be like let us take glass.

[00:16:26]Glass distance traveled by the light 40 seconds.

[00:16:36]Refract is mu. A is the length of the path. A is the length of the path.

[00:16:44]So time taken if it in t seconds in t seconds distance is d is d.

[00:17:01]Then assuming that D and T are numerically equal, this is in a particular medium like a glass. Glass medium.

[00:17:23]If I take air or vacuum, it would travel a distance of a dash to B dash.

[00:17:30]Now ddash is equals to the refractive index into t and the same time t seconds in tcs.

[00:17:44]So it means that optical path length is something that light would have traveled in air in the same time it had traveled in a particular medium of refractive index m.

[00:17:58]With this we understand how actually light travels in two different kind of media. So for this if I want to find optical path difference we we call it as optical path difference.

[00:18:16]Optical path difference.

[00:18:23]The difference of both the paths that is equals to delta x again optical path difference is equals to mu into t minus t that is equals to t into mu -1.

[00:18:38]This is very very important optical path difference traveled by the light in vacuum for a given time. So that that is how we calculate optical path difference. Let us see equation.

[00:18:53]If if light ray travels from If a light ray travels from water to glass, water to glass Both of both of same thickness same thickness let us call it as some 2 mm then find optical path difference.

[00:19:43]optical path difference. This is one type of question.

[00:19:53]So it will be like we should be knowing the refractive indices of glass and water.

[00:20:01]Refractive indices of glass and water.

[00:20:03]So collectively it is actually traveling through water and glass. So it should be like this. Delta X is equals to optical path difference. Delta X is equals to So it is traveling from glass and then water. So we write it like mu g / mu w -1 into t.

[00:20:27]Mu Z is equ= to 3/2.

[00:20:30]M U W is equ= to 4/3 and the distance or T is equ= to 2 mm.

[00:20:41]Substituting the values we'll get delta x is equ= to 3x2x 4x3 -1 into 2. That comes out to be 9x 8 - 1 9x 8 -1 into 2.

[00:21:06]1x8 into 2 is equals to 1/4 of a mm.

[00:21:12]That is the optical path difference or this can also be written as 0.25 m.

[00:21:22]Right? We'll see another question based on this.

[00:21:32]The question says this is optical path difference of both of them.

[00:21:44]This we see the question like the optical path of the optical path of a monochromatic light is same if it passes through.

[00:22:18]If it passes through 4 cm of glass or 4.5 cm of water, if mu g is equal to 1.53, find Mu W.

[00:22:49]Again one another example of using optical path difference.

[00:22:55]They say that optical path of light in two different media. One is glass and other is this. It means they say that optical path is delta x= to mu into 2D.

[00:23:11]It means that glass into distance traveled in the glass is equals to refractive index of water distance traveled in water.

[00:23:23]Taking down the data given 1.53 into 4 is equals to mu into 4.5 that is the distance.

[00:23:38]Solving this data, mu is equals to 1.53 into 4 / 4.5.

[00:23:48]This value turns out to be something around 1.36 with no units. The refractive index of water 1.36 approximately.

[00:24:06]As we have learned about optical path till now we also have learned that light causes some kind of illusion when it passes from one medium to another medium. The same kind happens when stars twinkle. The same thing happens when water bodies like the depth of the sea, the depth of the rivers appears to be shallow when we step in. So if we are standing near a swimming pool and if a coin is seen at the bottom of the swimming pool, it always appears to be close. But in actual scene, it actually would be much deeper than VC. It is a kind of illusion caused by light that we'll study in terms of apparent depth or if ever spear fishing people in some island countries often fish through spears they throw spears at the fishes inside the water. So if ever we are spear fishing so apparent depth of the fish should be understood. That means that this is the interface of air and water.

[00:25:26]This is air and this being water.

[00:25:30]Imagine there's a fish here.

[00:25:36]We'll make it a little deeper.

[00:25:40]Imagine a fish being present here.

[00:25:44]A fish here. Now the light rays from this the light rays from this for an observer outside might be coming the light rays would be coming like this to an observer here an observer here and if the observer is observing from here light rays from the object we'll call this as O which is the object now this is the surface the solid line is the surface and we draw a perpendicular line like This as you can see this light ray this light ray that is the refracted light ray appears to be coming in a straight line from a position above this. This is supposed to be the image and the observer sees the position of the fish to be here.

[00:26:34]Let us name these points. Let us take this as a and this as b.

[00:26:40]Now if you observe the depth original depth of the fish that is H but it appears to be at a shallower depth that is H ddash.

[00:26:55]It means that if the fish is really at at a real depth h real depth is h refractive index is mu refractive index is mu of the second medium refractive index is equal to mu of the medium a bird flying in the air for a bird flying in the air. Imagine the observer is a bird for a bird flying in the air.

[00:27:33]Now I is the position of the fish as observed by the bird. That is called as bird's view.

[00:27:42]So apparent depth while calculating the apparent depth we'll take these points now. We'll take these points. Now we can see a triangle triangle A B and O for the triangle A B and O which looks like this something like this A. So a b and o for the triangle a b and o sin i is equals to sin i is nothing but this is I angle of incidence since if you observe here if this is angle of incident this would be angle of incidence and if this is angle of refraction this would be angle of refraction.

[00:28:23]angle of incidence sin I is equals to AB divided by OB that is equals to A divided by OB is nothing but the real depth that is H. If the angle is much small for a small angle, smaller angles we'll take tan I equ= to sin I equ= to I the approximation we are doing this approximation. So that what happens?

[00:28:53]Sin I is equals to. So this becomes sin I equ= to X. We'll call this as A by H or simply X by H. Anything is fine for us. We'll go for the image triangle. Now image triangle. The image triangle is nothing but we'll call it as AI. IS triangle is AI. for the triangle A B I which looks something like this A B and I. Now tan r is equals to a b divided by b i that is equals to again a b by hdash.

[00:29:46]Mhm.

[00:29:47]Here slight change tan here tan i. Once again for the smaller angles I can write here tan r will be equal to sin r that implies sin r is equals to a b by hr hdash hdash is nothing but the apparent depth which is not real. So according to snail's law we know that we shall make use of snail's law here.

[00:30:16]According to snail's law, mu1 sin i is equals to mu2 sin r that is equal to keeping intact these refractive indices mu1 and mu2. Mu1 into x by h is equals to let us leave it aside like this mu2 into a b by hdash.

[00:30:45]cancel these.

[00:30:47]What do we need? We are trying to find apparent depth. Hence, apparent depth Hdash is equals to hdash is equals to mu2 by mu1 into h.

[00:31:05]What is mu21? Mu1. Mu2 is the medium of refraction. Mu1 is the incidence.

[00:31:14]So if I take this mu21 mu1 in a different way like water and air it becomes if mu2 is equals to 1 since air and mu1 as mu let it be any liquid then what happens to apparent depth then apparent depth hdash is equal to h by mu this is very important And one thing we'll understand mu is always greater than one. Mu is always greater than one. Now if you observe something else here when light travels from when light travels from denser medium to rarer medium that is from the fish light has to travel from the fish to the bird.

[00:32:18]Then real depth will be greater than apparent depth.

[00:32:31]This is to be noted.

[00:32:33]Right?

[00:32:35]The reverse can also be possible. The reverse can also be possible. So how much is the shift? Shift is nothing but you can observe that there uh the position was at O. Now with respect to the which with respect to the surface it is at I the shift is nothing but uh we'll write it like this which is not real. So we'll write it like apparent shift apparent shift due to refraction again delta x is equal to h - h ddash that is equals to h - h by mu.

[00:33:16]We know that hdash is less than h. This is h and this is hdash.

[00:33:23]H into 1 - 1 by mu this is apparent shift right when light travel from denser medium to rarer medium right now if we plot a graph for this if we plot a graph between apparent y apparent height on y-axis and the real height or the real depth on x-axis apparent height versus real height so that turns out to be a straight line passing through the origin. As we have seen earlier, tan I by tan R that is nothing but tan theta. So there should be a graph apparent height versus real height.

[00:34:20]The graph this is hdash and this is h will be a straight line passing through the origin. Angle is theta. tan theta is equals to hdash by h and tan theta is equals to we also know this 1 by mu mu is the refractive indices ratio of refractive indices means I have taken it like mu2 by mu1 otherwise we will write we'll take it like mu2 by mu1 if mu1 is mu mu2 is equal to 1 I have written it like 1 by mu.

[00:35:01]Now position of the observer does not depend on apparent depth or apparent shift.

[00:35:08]Right? So total depth total apparent depth for found by that will be equals to whatever the apparent shift we have seen.

[00:35:17]Right? Apparent shift plus the height of the bird from the surface. We'll write that. Copy that out.

[00:35:41]So total distance total apparent depth seen by the Now we have seen this. This is the surface. Here is the bird which is at a height y. Right? Now the fish was here.

[00:36:08]Fish was here. It appears to be here which is xdash some hdash or something. So that is h is equals to y + xdash. that is equals to y + x by mu. mu is the refractive index.

[00:36:25]We're trying to find the total apparent depth and how far does it change? How does it change?

[00:36:33]Right? Now, if the bird is flying towards the fish, if the bird is flying towards the fish, what happens to the height? Height reduces. Right? So, we shall differentiate that part. If the bird is flying towards the fish, if the bird flies with VB and fish with VF, fish moves with VF.

[00:37:07]Well, h is equals to d by dt of h is equals to d by dt of y that is the surface from the surface bird is at a distance of y + 1 by mu into d by dt of x. Let us call it as x. X is nothing but the actual depth of the fish. That is equals to this becomes velocity of bird with respect to fish is equals to velocity of bird plus velocity of fish divided by mu.

[00:37:51]We'll see a question on this.

[00:38:00]Hope you have copied.

[00:38:03]Let's see the question now. Given velocity of fish is 3 cm/s.

[00:38:18]Velocity of bird 6 cm per second towards the surface of water.

[00:38:36]Surface of water.

[00:38:42]And this is being observed by the fish.

[00:39:08]She has 9 cm per second.

[00:39:11]find refractive index of medium that is water. So it means that if the bird was moving with 6 cm/s for the fish which was moving with 3 cm/s it was seen like the bird was coming towards it at a speed of 9 cm/s.

[00:39:29]We'll go for the formula we have velocity of the bird with respect to fish is equals to velocity of bird plus velocity of fish divided by refractive index. Plugging in all the data we have 9 is equals to velocity of the bird is 6 + 3 by mu.

[00:39:49]Solving these terms we'll get a mu equals to 1. That is the refractive index of water. Refractive index of that water is to be 1.

[00:40:06]All right, let's see the other question.

[00:40:20]What other question says that?

[00:40:33]A bird flying in the air.

[00:40:43]with at a height h at a height h.

[00:40:55]A fish is inside for a fish inside for a fish inside water.

[00:41:10]Apparent shift of the bird.

[00:41:13]Apparent shift of the bird. Apparent shift of the bird.

[00:41:20]This is the convas case of the previous one that is this application talks about light travels from travels from When light travels from rarer medium to denser medium, rarer medium to denser medium, that is when light is traveling from the bird. If the bird is here, if the bird is here for a fish somewhere here. So if I if I see that it is actually refracting like this.

[00:42:20]When light is traveling from rarer medium to denser medium that is air inside water. Now if we take a reference line like this for an observer here it appear it appears to be coming from a point. This is to be a straight line. This is I. Now if you observe this this is the surface of water.

[00:42:50]Now if you observe this is the real height. This is the apparent height.

[00:42:54]When light travels from rare medium to denser medium apparent height is greater than rare apparent height is greater than real height.

[00:43:10]It changes due to the optical media.

[00:43:12]Optical characteristics of the media. Hence hdash should be greater than h. So we define this as hdash is equals to mu into h. So apparent shift apparent shift of the bird bird which was supposed to be seen at O will be seen at I. So the the fish will be under illusion. So the shift is equals to it is in terms of delta y height change. So that is equals to hdash minus h that is equals to mu into h minus h that is the shift.

[00:43:54]So delta y is nothing but apparent shift which was not real. Mu -1 into h. This is when light travels from rarer medium to denser medium. The previous one was for denser medium to rarer medium. Even here the graph between apparent height and the real height is always a straight line passing through the origin.

[00:44:22]Right?

[00:44:24]So the graph again apparent height versus real height.

[00:44:42]So it would be a straight line passing through the horizont. So tan theta this is hdash this is h tan theta is equals to hdash by h that is equals to mu t and theta is equals to mu right uh in this case what will be the total apparent depth seen by the seen by the fish let us suppose if The fish was at a depth of we'll write this here.

[00:45:21]If fish were at a depth of at a depth of x.

[00:45:32]Then total apparent depth is equals to total apparent depth depth is equals to we we'll write it like a mu into y + x.

[00:45:50]The actual height of the bird was y. The apparent height was mu into y plus the depth of the fish was x. What do we need? The bird was coming towards the fish. It will be seen by the bird as let us call this as h d h by dt is equals to mu into dy by dt plus dx by dt.

[00:46:16]x is we'll write the terms here. x is depth of fish from the surface.

[00:46:29]Y is height of the bird from the surface. This becomes velocity of the bird with respect to fish is equals to mu into velocity of bird plus velocity of fish.

[00:46:52]Right.

[00:47:02]So one another application of this an application.

[00:47:09]So due to this application what would happen? We often see on the office counters and the desks glass slabs with currency notes under that the alphabets the letters under that appears to be bulged means their actual position is much farther than their real position.

[00:47:27]Actual position is much farther than the apparent position. So they appear to be bulged. So the question is like we'll call it as application one.

[00:47:37]If an object If an object is placed in front of a glass slab at a distance x, then if you consider the refractive index of the glass slab being mu, Apparent shift of the position when seen from the other side. Apparent shift of the object. Position of the object.

[00:48:25]means that we'll take a small object a point object a glass slab whose thickness is t object here we'll take the object here this is O light travels from Okay, at this point it refracts again it refracts. So two times it gets deviated due to refraction. Now if you observe this, if we extrapolate the light rays, the light rays appear to be coming from a point here. That is the image will be seen by an observer on this side by an observer on this side.

[00:49:29]But for an observer on this side the other side of this there will be a shift in the position of the object. So it means that this oi let us call this as x from the edge of this refractive index of this is mu. So the shift the apparent shift we'll call it as delta x. This is due to the refractive index of this material. So the shift of this is equal to shift the direct application delta x is equals to t into 1 - 1 by mu because the object is traveling from rarer medium to denser medium then again the rarer medium object is in the rarer medium hence it is 1 minus 1 by mu this is a very important application copy down we'll go for the other one we'll see a different arrangement there composite slabs. We call it as the composite slabs. If n slabs are arranged, hope you have done with this. We'll go for the second application.

[00:50:44]The second application goes like this. A number of glass slabs are arranged like this. a stack number of class lines. Let us call them as mu1 refractive index, mu2, mu3, mu4 and so on. And an observer is here from the top. He's observing from the top.

[00:51:10]The thickness of each one is x1, x2, x3, x4, so on. We are taking infinite number of glass slabs.

[00:51:19]Two things they could be asking us. One is effect to refractive index. Other is apparent depth. Effect to apparent depth. Effect to apparent depth. So this turns out to be an object being placed at the bottom here. This is the object O. Now first thing we need to find apparent depth. For finding apparent depth firstly what would be the real depth?

[00:51:48]Real depth is equals to x1 + x2 + x3 + x4 so on.

[00:52:02]Right? So apparent depth becomes if this is X.

[00:52:08]This is X.

[00:52:14]Apparent death xdash is equals to x1 by mu1 + x2 by mu2 + x3 by mu3 + x4 by mu4 so on.

[00:52:33]What would be the apparent shift?

[00:52:36]Apparent shift. Shift is nothing but apparent shift.

[00:52:43]The difference of real and apparent depths.

[00:52:48]Delta x is equ= to x1 into 1 - 1x mu1 + x2 into 1 - 1 by mu2 + x3 into 1 - 1 by mu3 + x4 and so on.

[00:53:10]for effect to refracto index that is the second part in this let's call this as a and this is b effect to refractive index now as light sees the whole setup as one single arrangement we need to find the effect to refract index that is nothing but real depth divided by apparent depth.

[00:53:39]We can always check this ratio because when light travels from denser medium to rarer medium, real depth is greater than apparent depth. Hence, we can say that mu should be greater than 1. So, mu is equals to x1 + x2 plus sum of the thicknesses of the slabs divided by the apparent depths.

[00:54:11]This is the effective refract.

[00:54:21]Now equation can also be built based on this parall slabs.

[00:54:32]Try this question as homework.

[00:54:37]If you take three glass slabs, three glass slabs separated by 20 cm between each slab and refractive indices as mu1 is equals to 3x2, mu2 is equ= to 4x3 and mu3 is equ= to 2 respectively. and thicknesses. Let the thicknesses be. X1 is equals to 10 cm. X2 is equals to 15 cm. X3 is equals to 15 cm. Yes.

[00:55:29]Apparent shift would be how much? If an object is placed behind that apparent shift of object you'll see that question like three glass slabs X1 this is 10 cm this is of refraction to index 3x2 with a distance of 20 cm between them. This being 4x3 and this is 15 cm and there is another glass slab. This is again 15 cm and the refractive index is 2 and there is an observer here whose distance is 10 cm and the object is placed here at a distance of 10 cm.

[00:56:37]object apparent shift.

[00:56:46]What would be the apparent shift? Try this question as homework.