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33. Geometric Optics

Total Internal Reflection


Total Internal Reflection

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Hey, guys. In this video, we're gonna talk about a phenomenon called Total Internal Reflection. Okay, let's get to it. Remember, guys that at a boundary like can undergo both reflection and transmission okay, can undergo a combination of the two, depending on how reflective or how transparent the medium is. Okay, according to Snell's law, wind passing from a high refractive index to a low, the angle increases away from the normal. Okay, so the light ray moves away from the normal. If traveling from ah high refractive index to a low refractive index, that means that there is a maximum amount of deflection. There's a maximum amount of refraction that can occur right. If we look here for a small incident angle, we have a relatively small refracted angle. If we look here for a larger incident angle, we have a larger refracted angle. But at some critical angle, the refracted ray is at 90 degrees from the normal. It cannot refract any more than that, so if you go out to an angle greater than the critical angle, no refraction, which means no transmission occurs. Notice that a component of this light for every instance is also reflected. So when you're talking about angles passed this critical angle, all you have is reflected light. None of the light is transmitted. This is referred to as total internal reflection. There exists some critical angle pass, which light can no longer transmit. Transmission is not possible past this angle. All that's left is reflection. It doesn't matter how transparent the medium is, like glass glasses. Very transparent. But if you're going from an index that's larger to an index that's lower, even across glass, none of that will pass. Okay, This critical angle, at which total internal reflection occurs, is given by a formula that's a result of Snell's law. It's the arc sine or the inverse Sign off into divided by in one. Okay, clearly, just based on trigonometry, the above equation on Lee works when traveling from a large index to a small index. If into is larger than in one sine inverse has no output sign. Can Onley get as big asses one. So if this number is larger than one, meaning the numerator is larger than the denominator, your calculator is just gonna output an error because that's not possible. Okay, A very, very common application of total internal reflection is fiber optics. What fiber optics are our cables that allow the transmission of light across very, very vast distances without losing any energy in the light. None of that light bleeds out through transmission. Right. Normally, at a boundary like this, you would get a little bit of light leaking out here. You get a little bit of light leaking out here. You get a little bit of light leaking out across a long distance. You would lose a lot of the brightness of that light. But if you have an angle larger, then the critical angle then none of that transmission occurs. And all of your light stays inside the fiber optic cable and you don't lose anything, no matter how far you transmit the light. Okay, let's do a quick example. Using total internal reflection, I'm gonna minimize myself for this of fiber optic cable has a refractive index of 1.45 If you want the fiber optic cable to be able to carry a signal through both air and water, what should the minimum angle of the light passing through the cable be okay? So we have two scenarios We have the fiber optic cable in the air and we have a fiber optic cable in the water. I'm just going to separate these two and we'll do them separately in the air. What is the critical angle? This is just our application of Snell's law, and the question is, what's in two and what's in one and one is the medium that the light is traveling in. First in two is the medium that it's moving into. Okay, it's moving from the fiber optic material to air. The index of refraction of air is one. The fiber optic refractive index is 145 So this angle is 43 6 degrees. Okay, let's let me not do that. Let me bubble it. Okay, so as long as your angle is larger than 43.6 degrees, light will not be able to pass from the fiber optic into the air. But what about water? In this case, the critical angle, which is once again the inverse sign of into over in one hour. Original medium is the fiber optic the medium were passing into is water water has a refractive index of 133 and the fiber optic once again has a refractive index of 1.45 Plugging this into your calculator, you get 66 5 degrees. Okay, so the question is, what is the minimum angle? Well, if you have an angle less than 43.6 degrees, you're always gonna be able to transmit light through that fiber optic cable. That light is always gonna bleed out because it can transmit into air and it could definitely transmit into water. What if you're at, let's say, 50 degrees? Well, then for 50 degrees light cannot pass into air, but light can still pass in tow water because that angles less than the critical angle. If you want your fiber optic cable to be able to carry that information without losing anything through both air and water, it has to be larger than 0.5 degrees. Alright, guys, that wraps up our discussion on total internal reflection. Thanks for watching