**Concept:** Total Internal Reflection

Hey guys, in this video we're going to talk about a phenomenon called total internal reflection OK let's get to it. Remember guys that at a boundary light can undergo both reflection and transmission OK it can undergo a combination of the two depending on how reflective or how transparent the medium is OK according to Snell's law when passing from a high refractive index to a low. The angle increases away from the normal so the light ray moves away from the normal if travelling from a high refractive index to a lower refractive index that means that there is a maximum amount of deflection there is 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 past 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 past 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. Glass is 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 OK this critical angle. At which total internal reflection occurs is given by a formula that's resultant of Snell's law it's the arc sign or the inverse sign of N2 divided by N1 OK clearly just based on trigonometry the above equation only works when travelling from a large index to a small index. If N2 is larger than N1.Sin inverse has no output sin can only get as big as 1 so if this number is larger than one meaning the numerator is larger than the denominator your calculator is just going to output an error because that's not possible OK. A very common application of total internal reflection is fiber-optics with fibre-optics are cables that allow the transmission of light across very very fast 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 and across a long distance you would lose a lot of the brightness of that light but if you have an angle larger than 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 OK let's do a quick example using total internal reflection I'm going to minimize myself for this a 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 OK So we have two scenarios we have the fiber optic cable in the air and we have a fiber optic cable in the water and I am just going to separate these two and we will 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 N2 and what's N1,N1 is the medium that the light is travelling in the first N2 is the medium that it's moving into OK it's moving from the fiber optic material to air the index of refraction of air is 1 of the fiber optic refractive index is 1.45 so this angle is 43.6 degrees. OK, Folks let me not do that let me. Bubble it. OK so. Long as your angle as your 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 N2 over N1 our original medium is the fiber-optic the medium we're passing into is water, water has a refractive index of 1.33.And fibre optic once again has a refractive index of 1.45 putting this into your calculator you get 66.5 degrees. OK So the question is what is the minimum angle Well if you have an angle less than 43.6 degrees you're always going to be able to transmit light through that fiber optic cable that light is always going to bleed out because it can transmit into air and it can definitely transmit into water what if your at say 50 degrees well then for 50 degrees lights cannot pass into air but light can still pass into water because that angle is 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 66.5 degrees. All right guys that wraps up our discussion on total internal reflection Thanks for watching.

**Problem:** Most pools have an underwater light for swimming at night. If the underwater light is 1 m below the surface, for what area of the surface of the water are you able to see the light? Note that the refractive index of water is 1.33.

Consider a prism with the shape shown in the diagram. Its index of refraction is labeled by n_{p}, and it is submerged in a special liquid which has an index of refraction n _{ℓ}, where the difference n_{p} − n_{ℓ} = 0.59. The light ray is perpendicularly incident from the liquid into the prism as shown in the diagram. Notice that the incident angle at both points A and B is 45°. Find the index of refraction of the prism for a critical angle of 45°.

Watch Solution

A fiber optic cable (n _{fiber} = 1.56) is submergedd in water (n _{water} = 1.35). What is the critical angle for light to stay inside the cable?

Watch Solution

Light travels from water (n = 1.33) into glass (n = 1.5). Which of the following is the maximum incident angle with which light can still pass through the boundary?

A) 27.5°

B) 62.5°

C) 48.4°

D) No such angle exists

Watch Solution

A ray of light traveling in a block of glass (n = 1.52) is incident on the top surface of the block at an angle of 62.8° with respect to the normal. If a layer of oil is placed on the top surface of the glass, the ray is totally reflected. What is the maximum possible index of refraction of the oil?

Watch Solution

Determine the maximum angle *θ* for which the light rays incident on the end of the light pipe shown in the figure above are subject to total internal reflection along the walls of the pipe. The pipe of diameter 4.27 *μ*m has an index of refraction of 1.22 and the outside medium is air.

1. 65.0828

2. 27.2747

3. 67.1814

4. 63.1211

5. 48.5904

6. 54.5798

7. 36.0065

8. 71.9894

9. 69.4603

10. 44.3353

Watch Solution

A ray in glass is incident onto a water-glass interface, at an angle of incidence equal to half the critical angle for that interface. The indices of refraction for water and the glass are 1.33 and 1.64, respectively. The angle that the refracted ray in the water makes with the normal is closest to:

a) 29°

b) 39°

c) 34°

d) 44°

e) 24°

Watch Solution

The speed of light in a material is 0.41 *c*. What is the critical angle of a light ray at the interface between the material and a vacuum?

a) 19°

b) 22°

c) 17°

d) 24°

Watch Solution