Understanding Total Internal Reflection

In this video, we're going to go over total internal reflection and also how to calculate the critical angle. So, let's say if this represents the boundary between two materials. On top, we have air, and on the bottom, actually it's water instead of glass. So this is the normal line which is perpendicular to the surface and the ray that strikes the boundary between air and water is called the incident ray and the angle between the ray and the normal line is the incident angle or the angle of incidence. Now two things can happen The ray can reflect or it can refract. If it bounces back to the water, this is reflection. The angle of incidence is equal to the angle of reflection. So theta i is equal to theta r. This is the law of reflection. Now it can also refract. It can bend as it passes from water to air. This is the angle of refraction. The equation that relates the angle of incidence and the angle of refraction is Snell's Law. It's n1 sine theta is equal to n2 sine theta 2. The index of refraction for air. is about 1 and the index of refraction for water is 1.33. So as the light ray travels from a material with a high index of refraction to a material with a low index of refraction, it bends away from the normal line. Notice that theta r is greater than the angle of incidence so it bends away from the normal line. Now, let's say instead of using air, we use diamond. Diamond has a much higher index of a fraction than air. It's about 2.42. So, if we have a light ray that travels from water to diamond, it's going to bend towards... the normal line and so the angle every fraction is much less so therefore as n increases theta decreases now for total internal reflection to occur light has to travel from high end value to a low end value not from a low end value to a high end value So total internal reflection will not occur between water and diamond as light travels from water to diamond But it can occur as light travels from water to air The reason for that is because the angle of refraction bends away from the normal line. And so, eventually, total internal reflection can occur if you increase the angle of incidence. So make sure you keep that in mind. Total internal reflection will only occur if light travels from a material with a high index of refraction to a low index of refraction, and not the other way around. So anytime you have an incident ray striking a boundary, two things can happen. It can reflect, or it can refract. But let's focus on the refraction part. So this is the angle of incidence and this is the angle of refraction. And let's say this is air and water. Now what's going to happen if we increase the incident angle? If you increase the incident angle, the angle of refraction will increase as well. As you increase the incident angle, there comes a point where the angle of refraction becomes 90. When the angle of refraction is 90, the incident angle has now become the critical angle. So the critical angle is the incident angle when the refracted angle is 90. Now, if the incident angle exceeds the critical angle, that is when total internal reflection will occur. So now, let's understand why. As the incident angle increases when it exceeds the critical angle, there's no more refraction. Once the refracted angle, once it reaches 90, if you go past that, there's no more refraction. So the only thing that can happen is reflection. Keep in mind, you always have reflection. But once you take away refraction, then only reflection occurs. Therefore, you have total internal reflection. So anytime the incident angle exceeds the critical angle, total internal reflection occurs. so let's try an example using air and water calculate the critical angle what must be the instant angle at which the refracted angle is 90 So if you want to calculate the critical angle, use Snell's Law. n1 sine theta 1 is equal to n2 sine theta 2. Keep in mind the index of refraction for air is 1, and for water it's 1.33. so let's say one is for air, two is for water so n1 is going to be 1, sine theta 1 that's going to be 90 and on the other side n2 is 1.33 times sine theta 2 which is the critical angle so the critical angle is simply 1 divided by n, it's just 1 over n So 1 over 1.33, that's 0.7519. Now to find the critical angle, it's the inverse sine of 1 over n, or inverse sine of 0.7519. So therefore, the critical angle in our example is about 48.75. At this angle, the angle of refraction is 90. If the incident angle exceeds the critical angle, total internal reflection will occur so let's see what happens if we increase the instant angle to 90 by the way keep in mind that the instant angle could be anything the critical angle is simply the incident angle when the refracted angle is 90. So using this equation again, n1 sine theta 1 is equal to n2 sine theta 2. So we're going to use 1 and 90. n2 is going to be 1.33. Actually, let's change it. Let's solve for the new refracted angle if the incident angle exceeds the critical angle. Let's say the incident angle is 50. Keep in mind the critical angle is about 40. 8.75 so let's attempt to solve for the refracted angle if the incident angle exceeds the critical angle so n1 is 1 we're looking for the refracted angle and 2 is the 1.33 and theta 2 is 50 so 1.33 times sine of 50 is 1.02. Now, sine can never be greater than 1. It doesn't happen. If you were to type in the inverse sine of 1.02, you will get an error in the calculator. It just, it doesn't work. So therefore, there is no refracted angle when the incident angle exceeds the critical angle. Refraction doesn't occur anymore. And that's why you have total internal reflection. If you try to solve it, you're going to get an error. Sine cannot be greater than 1. So after 48.75, refraction no longer occurs. So total internal reflection is what will occur So now let's work on some practice problems. Calculate the critical angle between glass and water. So the first thing we need to know is the index of refraction for these two materials. The index of refraction for water, you can look it up, it's 1.33. And for glass, it's about 1.5. So using the equation, n1 sine theta 1 is equal to n2 sine theta 2. which angle do we need to look for and which one is the refracted angle keep in mind for total internal reflection to occur the light has to travel from a high index of refraction to a low index of refraction or to material with a lower index of a fraction so it has to go from glass to water can go from water to glass so therefore the refracted angle the one that's 90 is the one that is associated with the lower index of refraction. So if n2 is 1.33, theta2 has to be 90, which means theta1 is the critical angle that we're looking for. n1 is 1.5. If you try solving it the other way, it won't work. So whenever you're looking for the critical angle, The 90 is associated with the smaller index of a fraction. It's always going to work out that way. So 1.33 times sine 90 is 1.33. Sine 90 is always going to be 1. So sine theta C is 1.33 divided by 1.5, which is 0.887. So therefore, the critical angle is going to be the inverse sine of 0.887. So it's about 62.5 degrees. That's the critical angle for this particular problem. The critical angle for the interface between a solid and air is 40 degrees. What is the index of refraction of the solid? So how can we solve this one? we have solid and we have air. The index of refraction for air is about 1 and for the solid we don't know what it is we're looking for it. So let's draw a picture. So let's put the solid on the bottom and let's say the air is on top. So light is going to travel from the solid into the air. from a high end value to a low end value the light ray is going to bend away from the normal line but since we're dealing with a critical angle the angle of refraction must be 90 and we know the critical angle is 40 So we just got to solve for n. So let's write the equation that we need. n1 sine theta 1 is equal to n2 sine theta 2. Let's say n1 is the index of refraction for air. And then it's going to be multiplied by sine 90. n2 is the index of refraction for the solid, that's what we're looking for, times sine 40. So n2 is simply sine 90, which is 1, divided by sine 40. And so the answer is 1.56. That's the index of refraction of the solid. So anytime you're dealing with a critical angle problem, just make sure that the angle of refraction is 90. That's what you need to plug in for the other angle. So that is it for this video. Thanks for watching, and have a great day.

If it bounces back to the water, this is reflection. The angle of incidence is equal to the angle of reflection. So theta i is equal to theta r.

This is the law of reflection. Now it can also refract. It can bend as it passes from water to air.

This is the angle of refraction. The equation that relates the angle of incidence and the angle of refraction is Snell's Law. It's n1 sine theta is equal to n2 sine theta 2. The index of refraction for air. is about 1 and the index of refraction for water is 1.33.

So as the light ray travels from a material with a high index of refraction to a material with a low index of refraction, it bends away from the normal line. Notice that theta r is greater than the angle of incidence so it bends away from the normal line. Now, let's say instead of using air, we use diamond. Diamond has a much higher index of a fraction than air.

It's about 2.42. So, if we have a light ray that travels from water to diamond, it's going to bend towards... the normal line and so the angle every fraction is much less so therefore as n increases theta decreases now for total internal reflection to occur light has to travel from high end value to a low end value not from a low end value to a high end value So total internal reflection will not occur between water and diamond as light travels from water to diamond But it can occur as light travels from water to air The reason for that is because the angle of refraction bends away from the normal line.

And so, eventually, total internal reflection can occur if you increase the angle of incidence. So make sure you keep that in mind. Total internal reflection will only occur if light travels from a material with a high index of refraction to a low index of refraction, and not the other way around.

So anytime you have an incident ray striking a boundary, two things can happen. It can reflect, or it can refract. But let's focus on the refraction part.

So this is the angle of incidence and this is the angle of refraction. And let's say this is air and water. Now what's going to happen if we increase the incident angle? If you increase the incident angle, the angle of refraction will increase as well. As you increase the incident angle, there comes a point where the angle of refraction becomes 90. When the angle of refraction is 90, the incident angle has now become the critical angle.

So the critical angle is the incident angle when the refracted angle is 90. Now, if the incident angle exceeds the critical angle, that is when total internal reflection will occur. So now, let's understand why. As the incident angle increases when it exceeds the critical angle, there's no more refraction.

Once the refracted angle, once it reaches 90, if you go past that, there's no more refraction. So the only thing that can happen is reflection. Keep in mind, you always have reflection. But once you take away refraction, then only reflection occurs. Therefore, you have total internal reflection.

So anytime the incident angle exceeds the critical angle, total internal reflection occurs. so let's try an example using air and water calculate the critical angle what must be the instant angle at which the refracted angle is 90 So if you want to calculate the critical angle, use Snell's Law. n1 sine theta 1 is equal to n2 sine theta 2. Keep in mind the index of refraction for air is 1, and for water it's 1.33. so let's say one is for air, two is for water so n1 is going to be 1, sine theta 1 that's going to be 90 and on the other side n2 is 1.33 times sine theta 2 which is the critical angle so the critical angle is simply 1 divided by n, it's just 1 over n So 1 over 1.33, that's 0.7519.

Now to find the critical angle, it's the inverse sine of 1 over n, or inverse sine of 0.7519. So therefore, the critical angle in our example is about 48.75. At this angle, the angle of refraction is 90. If the incident angle exceeds the critical angle, total internal reflection will occur so let's see what happens if we increase the instant angle to 90 by the way keep in mind that the instant angle could be anything the critical angle is simply the incident angle when the refracted angle is 90. So using this equation again, n1 sine theta 1 is equal to n2 sine theta 2. So we're going to use 1 and 90. n2 is going to be 1.33. Actually, let's change it. Let's solve for the new refracted angle if the incident angle exceeds the critical angle.

Let's say the incident angle is 50. Keep in mind the critical angle is about 40. 8.75 so let's attempt to solve for the refracted angle if the incident angle exceeds the critical angle so n1 is 1 we're looking for the refracted angle and 2 is the 1.33 and theta 2 is 50 so 1.33 times sine of 50 is 1.02. Now, sine can never be greater than 1. It doesn't happen. If you were to type in the inverse sine of 1.02, you will get an error in the calculator. It just, it doesn't work.

So therefore, there is no refracted angle when the incident angle exceeds the critical angle. Refraction doesn't occur anymore. And that's why you have total internal reflection.

If you try to solve it, you're going to get an error. Sine cannot be greater than 1. So after 48.75, refraction no longer occurs. So total internal reflection is what will occur So now let's work on some practice problems.

Calculate the critical angle between glass and water. So the first thing we need to know is the index of refraction for these two materials. The index of refraction for water, you can look it up, it's 1.33.

And for glass, it's about 1.5. So using the equation, n1 sine theta 1 is equal to n2 sine theta 2. which angle do we need to look for and which one is the refracted angle keep in mind for total internal reflection to occur the light has to travel from a high index of refraction to a low index of refraction or to material with a lower index of a fraction so it has to go from glass to water can go from water to glass so therefore the refracted angle the one that's 90 is the one that is associated with the lower index of refraction. So if n2 is 1.33, theta2 has to be 90, which means theta1 is the critical angle that we're looking for. n1 is 1.5. If you try solving it the other way, it won't work.

So whenever you're looking for the critical angle, The 90 is associated with the smaller index of a fraction. It's always going to work out that way. So 1.33 times sine 90 is 1.33.

Sine 90 is always going to be 1. So sine theta C is 1.33 divided by 1.5, which is 0.887. So therefore, the critical angle is going to be the inverse sine of 0.887. So it's about 62.5 degrees. That's the critical angle for this particular problem.

The critical angle for the interface between a solid and air is 40 degrees. What is the index of refraction of the solid? So how can we solve this one? we have solid and we have air.

The index of refraction for air is about 1 and for the solid we don't know what it is we're looking for it. So let's draw a picture. So let's put the solid on the bottom and let's say the air is on top. So light is going to travel from the solid into the air. from a high end value to a low end value the light ray is going to bend away from the normal line but since we're dealing with a critical angle the angle of refraction must be 90 and we know the critical angle is 40 So we just got to solve for n.

So let's write the equation that we need. n1 sine theta 1 is equal to n2 sine theta 2. Let's say n1 is the index of refraction for air. And then it's going to be multiplied by sine 90. n2 is the index of refraction for the solid, that's what we're looking for, times sine 40. So n2 is simply sine 90, which is 1, divided by sine 40. And so the answer is 1.56. That's the index of refraction of the solid. So anytime you're dealing with a critical angle problem, just make sure that the angle of refraction is 90. That's what you need to plug in for the other angle.

So that is it for this video. Thanks for watching, and have a great day.

Transcript for:Understanding Total Internal Reflection

Transcript for:
Understanding Total Internal Reflection