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The index of refraction of benzene is 1.31 The critical angle in degrees for total internal reflection; at a benzene- air interface; is about:...

Question

The index of refraction of benzene is 1.31 The critical angle in degrees for total internal reflection; at a benzene- air interface; is about:

The index of refraction of benzene is 1.31 The critical angle in degrees for total internal reflection; at a benzene- air interface; is about:



Answers

Total Internal Reflection
The index of refraction of benzene is 1.8. What is the critical angle for a light ray traveling in benzene toward a flat layer of air above the benzene?

Yeah, this problem covers the concept of slate's slow and from states law Sinai upon sign our equals The effective and next of the medium to upon the effective and next of the medium one. Okay. Now from the streets where we can write sign off the critical angle, TheTA C upon sign off 90° equals the refractive index of the air. That is one Upon the refractive index of the benching that is 1.8 are scientific to see equals one upon 1.8. Or you can buy it the angle TheTA Sequels signing was off one of 1 1.8 or the critical angle for engineering professors 33.7° or in two significant visits, we can write the critical angle is almost equals two 34 degrees.

In this question given. And the index of refraction of Benzene is 1.8. Want to find a critical angle when the light re travelers traveling in the Benzene travels to what the banks in uh boundary. Okay. So like this. Right? So we're using mm King using Snell's law. Yeah. He says that. And one side data one equals two and two assigned to. So data one is the critical anger and two is 1.0 and one is 1.8 and then data tune is 90 degrees. Okay, this is what happens at um critical angle king. So we have signed C. Is equal to mm hmm 11.8. So did you see is oxide 1/1 0.8. The answer is the default degrees. Okay. So this is not answer for this question and that's all.

We're going to start by drawing a diagram. This line right here is showing us the surface of the benzene, so everything below that line is benzene. I'll go ahead and label it be, and everything above that line is air. So I will label that A. I'll also draw a dashed vertical line, so this is showing us the normal to that surface of benzene. So in our problem, we have a beam of light that is traveling through the air. And once it hits the surface of that benzene, the light is refracted or it's bent. So if I draw that you'll see it's not traveling in the same angle relative to the normal as it waas when it was traveling through the air. So let's look at what is given to us what is known in this problem. One thing that's known is this angle right here will call that angle a the angle that the light is traveling as it's going through the air. We know that that angle is equal to 43 degrees. Another thing that's known is this angle down here will call it. They'd be the angle that the light is traveling once it's refracted and that's equal to 27 degrees. There's actually 1/3 variable that's known, although it's not explicitly stated. And that is the index of refraction of the air will call that a and the index of refraction of air is approximately equal to one. What we're trying to find out is the index of refraction of the Benzie, and so we'll call that and be, and that is what we're looking for. So in order to solve this problem, where you're gonna use what's known as Snell's Law and Smells, Law states that and a times the sign I've played a is equal to and be times the sign have they to be. So we'll start by rearranging this equation since we're solving for N B will divide both sides by sign if they to be so. NB is equal to n a times the Sinus eat a divided by sign what they to be. No, we already talked about how, uh, the index of refraction of air is just equal to one. So this term is one simplifies things for us really nicely. That means that N B is equal to sign. They'd a divided by the sine hey, Toby. So we can go ahead and plug in the values that are given to us. Uh, data is 43 degrees, and fate of B is 27 degrees. So if you put that into your calculator, make sure you have it set to degree mode instead of radian load, you'll find that, uh, the index of refraction of benzene is equal to one 0.5. So that is our answer.

In this question we have to find the critical angle beyond which there will be a total internal reflection for area and glass interface. The refractive index of air medium is usually one. And the refractive index of glasses given to us 1.5. Now we know for total internal reflection. Alliance would pass from denser medium to real medium. So in this case the refractive index of the instant medium is nothing. But the refractive index of the glass which is given to us 1.5 the refractive index of the transmitted William is nothing. But they prefer to index of air which we know is equal to one. And now we are given the condition for in total internal reflection. So we know for total internal deflection. If 3 to I is equal to 3 to see then 3 30 will becomes 90 degree that in play that if angle of incidence is equal to critical angle, then Angle of front national become 90°. That is the condition for total internal reflection then according to scale so you can try it from Oh snails law and I shine teeter. I is equal to N. T. Scientific 30. So here the effectiveness of the emission medium is reflectiveness of the glass medium, which is N. G shine tita. I can written a scientist to see equals N. T. S. N. A. And scientific 30 is signed 90 degree. So by simplifying this, we will get scientific to see is equal to one by N. G. S. 1.5 or 3 to see which becomes Sign Universe of one by 1.5. So this will give us Tito see Equals 41 81°. And this is the fellow of the critical angle for a total internal reflection.


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