**Concept:** Refraction of Light

Hey guys, in this video we're going to talk about something called the refraction of light which is the change in the lights angle as it passes from one boundary to another. Alright let's get to it, remember guys that at a boundary light can do two things it can reflects off of that boundary or it can transmit, it can transmit through the boundary and make it into the new medium and remember that light in general does a combination of the two. When light crosses a boundary into a new medium it turns out that it will change its direction, remember that at the boundary we always measure angles from the normal to that boundary so if this is theta 1, then the light is going to come out at some new angle theta 2 and theta 1 is not necessarily going to equal theta 2 in general they will not. Remember that in each medium light still travels in a straight line those lines just aren't parallel because the ray has changed direction. Now the reason that the light changes its direction is due to one thing and one thing alone light travels at different speeds in different media. In a vacuum light travels the fastest it will ever travel the speed of light in a vacuum is a fundamental constant of the universe but in a medium light will always travel more slowly than in a vacuum, remember that a fundamental fact about waves in general all waves is that their speeds are determined by properties of the medium and the type of wave in this case since all light is electromagnetic waves the speed is determined simply by the properties of the medium the property of the medium that tells us how fast light is going to travel in it is known as the index of refraction, this is a very important word sometimes it's called the refractive index just so you don't have to call it the index of refraction and the index of refraction of any medium is always defined the same way, it's defined as the speed of light in a vacuum divided by the speed of light in that medium. Now remember because light always travels slower in a medium than it does in a vacuum the index of refraction is always going to be bigger than 1 right the speed of light in a vacuum is always going to be larger than the speed of light in any medium so the index of refraction is always going to bigger than 1. The two indices of refraction that you need to remember are the index of refraction for a vacuum which is obviously 1 right how fast does light travel in a vacuum C, C divided by C is 1 but for air the index of refraction is 1.00029 it's so so close to 1 that we always just say it's simply 1. So the index of refraction of air is the same as the refractive index of vacuum which is 1 now I'm going to talk about an analogy to explain why refraction occurs.

I'm going to minimise myself so we can all see this figure. You can think about refraction like a barrel rolling on a smooth surface to a rough surface, so imagine this barrel right here is rolling on smooth fresh pavement and it's rolling very very quickly and then it encounters some grass which is rough and it slows the barrel down, when the barrel is rolling entirely on the smooth surface all of the points along the barrel are moving at the same speed so the barrel rolls in a straight line but once the edge of the barrel hits the grass that edge is now moving more slowly than the opposite edge. Now that this is rotating more slowly than this edge is it hook the barrel right this barrel hooks inward and it causes it to change direction. Once the barrel is entirely on the rough surface once again all points on that barrel are moving at the same speed so it moves in the straight in a single direction again it moves straight again but since it hooked when it transitioned from the pavement to the grass it is now moving in a new direction this is the idea with refraction it's due solely to the fact that as you transition from one boundary to another the speed that you're travelling with changes and the speed in the case of the barrel change once the barrel hit the rougher grass and that caused it to hook inwards and move along the grass at a different angle. Let's do an example to further illustrate this, we want to explain refraction using Huygens principle before I do refraction though I want to talk about what Huygens principle says about a single ray of light moving through a vacuum so imagine that that boundary wasn't there. Let me minimise myself for this because I'm going to be drawing over where I am right now, alright so this ray is coming in and there's a boundary right here that separates the media on the left side, on the right side I'm still indicating where the boundary is but I'm not sorry let me redraw that really quickly I want to end at where the imaginary boundary is. I've indicated where the boundary is but I don't actually have a boundary here in reality we're still looking at this light ray without changing the media that it's in. So remember how to do these problems we're going to draw wavefronts as they approach the boundary so there's a wave front at one time after a certain amount of time there it is right, after another amount of time there it is and there it is and then it hits the imaginary boundary and then it hits it again and then hits it again and remember that this was the first time that it hit that boundary. Actually let me write that over here this is the first time that it hit that boundary right here this is the second time that it hit that boundary and right here this is the third time that it hit that boundary.

So that third point was the most recent time it hit so it produces the smallest ray sorry the smallest wavelet. The green dot is the second time it hit so it produces the second smallest or the second largest wavelet because there's only three of them. The blue is the most recent sorry the earliest time that it hit so it has the longest time to propagate the wavelet, so it has the largest wavelet and what ends up happening is that this wave front is going to be parallel to all of those other wavefronts and this ray is going to go through undisturbed right that's exactly what we expect because as it passes through no boundary right this is no boundary right here I drew it as an imaginary boundary but in fact it's travelling in the same medium because there's no boundary it shouldn't change direction it should absolutely move in the same direction. Now let's see what happens when we encounter a boundary where the speed of the light does change, so once again here's a wavefront then a little time later another wavefront a little time after that another wave front then away from hits the boundary then another time after that another wave front hits and then it hits again. So it hits at those three places the first place is here, the second place is here and the third place it hit is here. So the most recent encounter is at the third point the earliest encounter is at the first point this is no different than every time we've applied Huygens principle Now what's the difference between the situation on the left with a boundary and the situation right here with no boundary. On the left side in the new medium light travels more slowly then when you don't change media, because light travels more slowly these wavelets don't travel as far so that third wavelet the red one is going to be smaller over here. Than it was right here, the green wavelet is also going to be smaller over on the left than it was here and finally the blue wavelength too is going to be smaller on the left than it is on the right, and that's simply due to the fact that light travels more slowly in this new medium. So what happens the new wavefront is at a different angle than the old wavefront because it's at a different angle now when I draw my ray, my ray is moving at a new angle. Remember that this process which is called refraction is due solely solely to the fact that the speed of light is different between these two boundaries. That wraps up our discussion on refraction. Thanks for watching guys.

**Concept:** Snell's Law

Hey guys, in this video we're going to talk about something called Snell's Law, Snell's law is the mathematical law that governs refraction across a boundary, let's get to it. Remember that when light transmits through a boundary into a different medium the angle of its path must change, what I have here is an image that shows one medium indicated by an index of refraction N1 and another medium indicated by the index of refraction N2 so the light travels at different speeds in these two media so long as in N1 and N1 don't equal. If they do equal then they travel at the same speed and no refraction occurs. Remember that refraction depends only upon a change in speed so what we have is some incident angle theta 1 and some refracted angle theta 2. The question is how did those angles relate to one another? Well Snell's law gives us this angle of refraction and it says that the number of the index of refraction times sine of the angle is conserved across the boundary. So N times sine of theta on the left side is also equal to N times sine of theta on the right side. So that quantity the index of refraction times sine of the angle is conserved across a boundary. Now this is where our definition of angles becomes very very very important, your angle which appears here and here has to be measured from the normal. Theta has to be measured from the normal, it has to be measured from the normal notice these two angles are both measured with respect to this normal line that I drew this is the normal. If you measure that angle let's say with respect to the surface of the boundary like so that is not going to give you the right answer. Your answer will absolutely unequivocally be wrong every single time you measure the angle with respect to the boundary, do not do this always measure it with respect to the normal.

Let's do a quick example, a light ray is incident on an air-water boundary at 33 degrees to the normal, with the refractive index of water being 1.33. If the light ray passes from air to water, what is the refracted angle? What if the ray were to pass from water to air? So we have two different scenarios, we have air to water and then we have water to air. So let's start with air to water, we know what the incident angle 33 degrees to the normal which is always how we have to measure these angles and we know the index of refraction of water 1.33. We're going from air to water so theta 1 the angle the incident angle is 33 degrees. N1 is our index of refraction of our initial medium, N2 is the index of refraction of the medium we're passing into, which in this case is water right air to water 1.33 and then theta 2 is our unknown. The only thing here that we're missing is that index of fraction of our original medium which is air, always remember the index of refraction of air is 1 So now we have enough to use Snell's law to solve this problem Snell's law says N1 sine theta 1 equals N2 sine theta 2. So I need to solve for sine of theta 2. So rewriting this I get sine of theta 2 is N1 over N2 sine of theta 1 which is N1 the incident medium is just 1, N2 the medium that we're passing into which is water is 1.33 and the incident angle was 33 degrees and this is equal to 0.41. Which means using the arc sine function on our calculators that this angle is simply 24 degrees, relative to the normal in the second medium which in this case is water all right that's the first part.

What about water to air, now the media are flipped right we're going from water to air but the incident angle is still the same we're still incident at 33 degrees from the normal so theta 1 is still 33 degrees but now N1 the refractive index of our original medium is now that of water 1.33 and N2 the index of refraction for a transmitted medium is that of air, which is 1 and we're looking for that refracted angle so it's the same set up, let me minimise myself. I already have the equation I need to use so I'm just going to write that out, sine of theta 2 equals N1 over N2 sine of theta 1 which is now 1.33 over 1 right the indices of refraction are flipped but the incidence angle is still 33 degrees and this whole things equals about 0.72. So if we use the arc sine on our calculator we find that the refracted angle is 46 degrees from the normal, always from the normal. Alright guys that wraps up our discussion on Snell's Law. Thanks for watching.

**Problem:** A light ray passes from air to glass, with an index of refraction of 1.5. If the light if blue, with a wavelength of 450 nm, what is the wavelength of the light ray after it passes into the glass?

**Example:** Refraction Through Multiple Boundaries

Hey guys, let's do an example, a light ray enters a 25 centimetre deep layer of oil at an angle of 37 degrees to the normal. Below the oil is a layer of water 15 centimetres deep. How far does the light ray move horizontally from the point it enters the oil to the point it hits the bottom of the water? This is actually like a very classic Snell's law problem so the odds are you're going to see it maybe as a homework problem maybe as a test problem so I would study this problem because it's a very very common problem. So what's happening here is simply that an incident light ray is hitting oil right here so refracts going through the oil. Now the oil we're told is 25 centimetres deep right, then that light ray which is refractived once already passes through into the third medium right the first medium is air the second medium is oil the third medium is water right here and it refracts again then it travels 15 centimetres further down, that's the depth of the water at a new refractived angle and eventually hits the bottom of the water. So it's going to cross some horizontal distance through both media. Through the oil it crosses this horizontal distance which I'll call delta X 1 and then through the water it crosses this horizontal distance which I will call delta X 2 sorry that's cut off a little, delta X 2 and quite clearly our total distance travelled X is just delta X 1 plus delta X 2. So at the boundaries this boundary and this boundary we're going to have Snell's law that we have to apply to figure out the refracted angles but after that it's just a geometry problem but buckle up because this is a long geometry problem.

Let's talk about the first boundary, this is the air to oil boundary what we have that is let me draw that solid. We have our incident ray from air which has an index of refraction of 1 hitting this at an angle of 37 degrees and then passes into water which has an index of refraction of sorry passes into oil first oil since it's hot water, an index of refraction of 1.47 and it's going to be refracted at some new angle theta 2 we need to apply Snell's law to find what that new angle is. Snell's law says N1 sine theta 1 equals N2 sine theta 2 so that means that sine of theta 2 is simply N1 over N2 sine theta 1 and that's going to be 1 over 1.47 times sine of the incident angle which is 37 degrees once again it's measured from the normal the problem said to the normal always make sure that the angles that you plug in Snell's law are measured to the normal, and this whole thing equals about 0.41 and so our first refracted angle is if we use our calculator 24 degrees. Let me bubble this, that is the angle at which the light travels through the oil so this angle right here 27 degrees. Let's talk about the second boundary now. In the second boundary we have lights that's coming in at what angle? We don't know that angle but we do know this angle that angle is 24 degrees that we found right here that's just a refracted angle this angle is the alternate interior angle so it too is 24 degrees. So now we can apply Snell's law to find what is the refracted angle which I'll call theta 3 inside the water. So N2 once again for oil was 1.47, N3 the third medium for water is 1.33 I'm going to use this exact same equation I'm just going to use different numbers 2 is now going to become 3 and 1 is now going to become 2 so sine of theta 3 is N2 over N3 sine theta 3. This is going to be 1.47 over 1.33 sine of our incident angle which is 24 degrees and this whole thing equals 0.45. Which means what that if we use our calculators the refracted angle into the water is 27 degrees. So only slightly off from our refracted angle into the oil. Now let me separate some area right here, now we need to do our geometry if we look at our figure we have two triangles we have a triangle inside the oil and a triangle inside the water let's start with the oil.

The triangle inside the oil looks like this this is the incident sorry this is the ray travelling through the oil which has an angle to the normal of 24 degrees how far horizontally it goes we called delta X 1 how far down it goes is just the depth of the oil which we said was 25 centimetres so all we have to do is use trigonometry to figure out what delta X 1 is since we have the opposite and the adjacent edges we should use tangent tangent of 24 degrees is going to be the opposite edge delta X 1 divided by the adjacent edge which is 25 centimetres so delta X 1 is going to be 25 centimetres times tangent of 24 degrees which is going to be 11.1 centimetres. Now what about through the water well we have another triangle that looks like this. Where the refractive angle inside the water is 27 degrees how far it goes horizontally is delta X 2 right and how far it goes vertically is just the depth of the water which is just 15 centimetres and the hypotenuse of course just that ray that's travelling through the water. So once again we have opposite and adjacent edges we want to use tangent. So tangent of 27 degrees equals the opposite edge over the adjacent edge so delta X 2 is 15 centimetres times tangent of 27 degrees. Which is going to be 7.6 centimetres. So lastly putting this all together, we have that X which was delta X 1 plus delta X 2 equals 11.1 centimetres plus 7.6 centimetres which equals 18.7 centimetres. Alright guys like I said this is a classic problem and it has a butt load of geometry but the only physics involved in it is the application of Snell's law at the two boundaries. Alright that wraps up this problem, thanks for watching guys.

A beam of white light is incident on a triangular glass prism with an index of refraction of about 1.5 for visible light, producing a spectrum. Initially, the prism is in a glass aquarium filled with air, as shown. If the aquarium is now filled with water, with an index of refraction of 1.3, which of the following is true?

1. The spectrum produced has greater separation between red and violet than that produced in air.

2. A spectrum is produced, but the deviation of the beam is opposite to that seen in air.

3. The spectrum has the same separation between red and violet as that produced in air.

4. Violet light will not emerge from the prism.

5. The intensity of the light emerging increases.

6. There is no light seen coming out of the prism.

7. The spectrum produced has less separation between red and violet than that produced in air.

8. The positions of red and violet are reversed in the spectrum.

9. No spectrum is produced.

10. Red light will not emerge from the prism.

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Light of wavelength 862 nm is incident on the face of a silica prism at an angle of θ _{1} = 74.1° (with respect to the normal to the surface). The apex angle of the prism is ∅ = 53.8°. Given: The value of the index of refraction for silica is n = 1.455. Find the angle of refraction at this first surface.

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A narrow beam of light passes through a plate of glass with thickness 1.03 cm and a refractive index 1.46. The beam enters from air at an angle 20.1°. The goal is to calculate the deviation d of the ray as indicated in the figure. Find deviation d.

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Consider a light ray which enters from air to a liquid, where the index of refraction of the liquid is given by n = √2. Denote the wavelength and the frequency in the liquid by ⋋′ and f′ and those in the air by ⋋ and f. Choose the correct pair of ratios:

1. ⋋' / ⋋ = √2, f' / f = √2

2. ⋋' / ⋋ = 1, f' / f = √2

3. ⋋' / ⋋ = 1/2, f' / f = 1

4. ⋋' / ⋋ = 1/√2, f' / f = 1

5. ⋋' / ⋋ = 2, f' / f = √2

6. ⋋' / ⋋ = 1/2, f' / f = √2

7. ⋋' / ⋋ = 1, f' / f = 1

8. ⋋' / ⋋ = 2, f' / f = 1

9. ⋋' / ⋋ = 1/√2, f' / f = √2

10. ⋋' / ⋋ = √2, f' / f = 1

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A ray of light traveling in air is incident on the surface of a block of clear ice (of index 1.309) at an angle of 43.4° with the normal. Part of the light is reflected and part is refracted. Find the angle between the reflected and the refracted light. Answer in units of °.

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Light strikes a glass plate at an angle of incidence of 60 degrees. Part of the beam is reflected and part is refracted. It is observed that the reflected and refracted rays are perpendicular to each other. What is the index of refraction of the glass?

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A submarine is 400 m horizontally from the shore of a freshwater lake and 100 m beneath the surface of the water. A laser beam is sent from the submarine so that the beam strikes the surface of the water 300 m from the shore. A building stands on the shore, and the laser beam hits a target at the top of the building. Recall, the index of refraction of water is 1.33. Find the height of the target above the level of the water.

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As shown at right below, a laser is directed into a water tank (filled with water having an index of refraction n_{2 }at the laser wavelength) at an angle *θ*_{1} from the vertical and strikes the bottom of the tank at point A, a horizontal ditance *r* away from B, the spot directly underneath the place where the light enters the water. The index of refraction of water for red light is 1.331, while the index of refraction of water for green light is 1.336. The angle of refraction θ_{2} would be

a) the same for a green and a red laser.

b) smaller for a red laser than for a green laser.

c) smaller for a green laser than for a red laser.

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As shown at right below, a laser is directed into a water tank (filled with water having an index of refraction n_{2 }at the laser wavelength) at an angle *θ*_{1} from the vertical and strikes the bottom of the tank at point A, a horizontal ditance *r* away from B, the spot directly underneath the place where the light enters the water. The distance *r* is:

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Consider a spherical rain drop in the air. A laser beam is directed at the rain drop from the left and slightly above the central axis (see figures). Which of the following diagrams best indicates the trajectory of the light? (Note, the dotted lines cross in the center of each drop, and therefore indicate the normal to the surface.)

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Light in air is shone on the surface of oil with an index of refraction of 1.25. Beneath the layer of oil is water, which has an index of refraction of 1.33. If the light travels through the water with an angle of 32° from the normal, at what angle (measured from the normal) was the light incident on the oil with?

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A prism with angles of 60°-60° -60° and refractive index of 1.6 is partially immersed in an oil with n=1.2 as shown. A light beam entering from air at the top (P) strikes point Q at the critical angle. Find the incident angle θ.

a. 27.0°

b. 16.5°

c. 18.5°

d. 11.4°

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The index of refraction for red light in water is 1.331 and that for blue is 1.340. If the incident angle of a white light beam is 83°, what is the angle between the red and blue light inside water?

a. 0.22°

b. 0.75°

c. 0.43°

d. 0.36°

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The figure shows the path of a portion of a ray of light as it passes through three different materials. Note: The figure is drawn to scale.

What can be concluded concerning the refractive indices of these three materials?

a. n_{1} < n_{2} < n_{3}

b. n_{1} > n_{2} > n_{3}

c. n_{3} < n_{1} < n_{2}

d. n_{2} < n_{1} < n_{3}

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A light beam shown in the figure makes an angle of 30° with the normal to the air-oil surface. Determine the angle θ.

a. 22.2°

b. 24.7°

c. 40.5°

d. 37.0°

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Light passes from air into some other medium. The light enters with an incident angle of 40^{o} and a wavelength of 425 nm, and passes through the boundary with some refracted angle of 30^{o}. What is the wavelength of the light in the medium? Assume all angles are measured with respect to the normal of the boundary.

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A person in the air above the water in a swimming pool looks straight down into the water (n = 1.33) at a diamond ring that lies on the bottom of the pool. If the image of the ring is 1.20 m below the surface of the water, what is the depth of the water (the distance from the surface of the water to the bottom of the pool)? Hint: Assume θ<<1, so sinθ=θ, and tanθ=θ, for a θ in radians.

(a) 0.60 m

(b) 0.90 m

(c) 1.20 m

(d) 1.60 m

(e) 2.40 m

(f) none of the above answers

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A ray of light traveling in material *a* strikes the interface between materials *a* and *b* and refracts into material *b*. In material *a*, the light makes an angle of 30° with respect to the surface of material *a* and in material *b*, it makes an angle of 60° with respect to the surface of material *b*. Which of the following statements is correct?

(a) The light travels faster in material *b* than in material *a*.

(b) The light travels slower in material *b* than in material *a*.

(c) The speed of the light is the same in both materials.

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Light goes from flint glass into ethanol. The angle of refraction in the ethanol is 24.8°, the index of refraction for flint glass is 1.61, and the index of refraction for ethanol is 1.36. What is the angle of incidence in the glass?

1. 10.8702

2. 12.7958

3. 20.4236

4. 21.3253

5. 18.6127

6. 20.7518

7. 16.378

8. 17.6211

9. 9.6954

10. 12.3777

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A tank holds a layer of oil, 1.65 m thick, which floats on a layer of syrup that is 0.83 m thick as shown in the figure. Both liquids are clear and do not intermix. A ray, which originates at the bottom of the tank on a vertical axis, crosses the oil-syrup interface at a point 0.90 m from the axis. The ray continues and arrives at the oil-air interface, 2.00 m from the axis and at the critical angle. Find the index of refraction of the syrup.

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Light of wavelength 475nm in vacuum enters a plastic as shown in Fig. 1. Measurements of the light in the plastic indicate that its wavelength there is 394 nm.

At what direction does the light travel with respect to the normal in the plastic?

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Light of wavelength 475nm in vacuum enters a plastic as shown in Fig. 1. Measurements of the light in the plastic indicate that its wavelength there is 394 nm. What is the frequency of the light in the plastic?

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