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10.What is different about the angles ofrefraction of different colors of lightas compared to the one wavelength laser light?11. Does the index ofrefraction ofa pie...

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10.What is different about the angles ofrefraction of different colors of lightas compared to the one wavelength laser light?11. Does the index ofrefraction ofa piece of glass depend on the wavelength of light passing through it? Why?

10.What is different about the angles ofrefraction of different colors of lightas compared to the one wavelength laser light? 11. Does the index ofrefraction ofa piece of glass depend on the wavelength of light passing through it? Why?



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$\bullet$$\bullet$ A block of glass has a polarizing angle of $60.0^{\circ}$ for red light and $70.0^{\circ}$ for blue light, for light traveling in air and reflecting from the glass. (a) What are the indexes of refraction for red light and for blue light? (b) For the same angle of incidence, which color is refracted more on entering the glass?

All right, so we have our two wavelengths of light. The red light here is going to have a wavelength of 700 nm. It's the longer wavelength Blue Light has a wavelength of 400 nm. And they're going to be entering a different medium. They're going to quote from air to glass where the index of refraction is higher. So the lower wavelength or the shorter wavelength I should say, is going to bend more than the longer wavelength. So the blue light is going to bend more and spread out closer to the normal. Then the red light red light won't bend quite as much and so they will spread out. And the the angle of incidence, which was the same for them both. Um When they came in, the angle of refraction is going to be different. It's going to be smaller for the blue light than it is for the red light. If they were going the other way and they were going from glass to air, it would actually be um kind of the opposite where the blue light is still bent more, but it spent away from the normal line instead of towards it, so the light will speed up as it goes from glass to air instead of slowing down. I think it goes from there to glass.

In this question, we have red light with a wavelength of 6.8 Times 10 to the -7 m. Um enters glass with an index of refraction. So the index refraction of the glass is 1.5 83 and it enters from air, so R. N. One will just be one and it has an angle of incidence. Um So our angle theta here Equals 38°.. We want to know the angle of refraction. So what angle this light bends to as it enters the glass? Um So we can do that by using Snell's Law where we have end of the air. Times the sine of the angle in the air equals, and in the glass times the sine of the angle in the glass and A. Is just one soul. Just write that as one. We can solve this for the angle on the glass. So data G just equals the inverse sine of the sine of theta. A. In the air divided by N. G. When we put the numbers over on the left end of this equation, um and put that into our calculator, We get an angle of 22.9°.. And it's important here that when we go from that less tense, the more dense media that the um light should be bending towards the normal. So we should get the smaller angle like we got here. The second part of this question asks us to find the speed of light in the glass and the speed of light in the glass. It's just the speed of light outside of the glass divided by the index of refraction. Inside the glass. The speed of light is just three times 10 to the 8th m/s and Was just 1.583. No units on the index of refraction. Mhm. And this gives us a speed of 1.8 Um nine times 10 to the 8th meters per second. Likewise, we then want to find the wavelength of the light in the glass, and the wavelength of light will decrease exactly like the speed of lightning, decrease the frequency of the light will stay the same, but the wavelength decreases um by dividing by that index of refraction. Um So we had that Initial wave length is 6.7 times 10 to the -7 m. We divide that by 1.583, and we get a final wave length Inside the glass of 4.3 times 10 To the -7 m.

This is problem eight Chapter 23 for this problem were given the wavelength Lambda equal to 6/79 0 meters on the cover meters, multiplying about Tim seven of nine meters her one now. So when animator is equal to 10 to 9 meters had a multiply across with me at six under 70 times 10 to the neck of nine meters from our wavering. The first part of this question asked, Find the frequency for this light as a trump's through air, glass and water. So the frequency is determined by the source and it's gonna be constant. Throughout those materials in air, we have a speed of light. It's gonna be equal to Lambda. I'm just gonna be equal to 3.0 times 10% here. We're gonna sell for F by dividing both sides by Lambda. So yeah, uh, sort of equal. See over lambda. Let's go ahead and plug in the valleys for seeing we have this in meters per second. And for Lambda we have this and meters. So that's what we converted two meters at the outset so that M cancels out with them. And all that's left is in for seconds, which is hurts. This comes out to be 4.5 times 10 to the 14 parents. Since frequency is determined by the source, that frequency is gonna be equal to frequency of that light as a trumpster air charms through through glass and as it travels to war was that same corner. What changes for the light as it travels through the different materials is the wavelength. To find the wavelength as the light travels through, say, the glass, we're gonna need the refractive index of the glass, and that's given as 1.6. So this is defined to be the speed of light over the velocity of light in that medium, which is be a bus. Let's go ahead and saw for V. Gauss. We won't supply both sides of the expression by the glass. We get the glass and glass is equal to see. Now let's go ahead and divide both sides by n again V glass as equal to see over and glass. But velocity is just lambda a glass that's frequency glass, but frequency grasses just after we just leave it without the subscript here. Okay, here we can software land the glass as being seat over and glass. Let's go ahead and plug in all the values. C is gonna be speed of right f is the quantity we just calculated. And And glasses 1.6 hurt is inverse seconds in for seconds. Customs within for seconds. And what we're left with these leaders. So put the same. We get before a 0.219 10 7 meters. So let's go ahead and comfort this to, uh, banana meters since that nine, since a negative nine meters is equal to one at a meter. So when we simplify this, get lambda glass is gonna be equal to 419 90 meters. Okay, so this is for a glass. We're gonna repeat this calculation for water, so end of water is gonna fall the same the same pattern. Let's go ahead and do it right here on the side is gonna be equal to 1.33 There's gonna be equal to the speed of light in vacuum over the speed of light in water. We repeat the procedure we're gonna solve for the velocity of light and water. So this is gonna be the same algebra as the one for the glass. We're gonna multiply both sides here and then divide by N water. There's gonna be seat over and water, and this is gonna be equal to glammed, uh, water wavelength in the water. Time to the frequency of the water, which is just f. Okay, here, let's go ahead of salt. For Lam Dewar, we divide both sides by F way, plug in from the specific values. So for the speed of light, that's unchanged. Nothing changes but the refractive index of water here. So I'll just go ahead and copied this from before and just change this denominator. So the denominator is gonna be 1.33 And when we saw for this, because not too 5.4 10 10 days of seven meters, which is probable four No, no meters. So the conversion falls. What we did here in this problem, when it's tapped into the negative seven, we could go ahead and multiplied by this conversion factor, and it shows the decimal place to to the right. Okay, so this concludes problem number eight

Part A of our question wants us to figure out the minimum incident angle a ray of light under under ghost, total internal reflection that's needed. So we need to find this minimum incident angle or this critical angle, considering the fact that the index refraction of the glass in sub d is 1.78 and index refraction of air is 1.0 Okay, so to, uh, To do this, we're gonna use Snell's law of refraction to find the minimum angle. So Snell's law tells us that in one times this sign of the angle one which we're gonna call data one is equal to into times this sign data too. But if things are completely internally reflected the minimum angle that that would happen, that is, when faded, too is equal to 90 degrees. If they tatoos equal to 90 degrees in the sign of 90 degrees is equal. The won their fourth data One would now become the critical angle which we could call. Say this up, see. So for part, eh, let's go ahead and solve for that critical angle, though, so data sub sea is going to be equal to the inverse sine, uh, the index of refraction to to the, uh, the ratio of the index of refraction to to index of refraction one. But here the index every fraction, too, is the air, right, cause that's the second medium and the index of refraction. One is the glass and sub G. So taking the inverse sine of these two, we find that this is equal to 34 0.2 degrees. We can go ahead and box it. And as our answer for part a part B says, if a layer of water is placed over the glass, what is the minimum angle of incident on the glass water interface that will result in total internal reflection to the water air interface? Okay, so we want total internal reflection to the water air interface. So for cart be, we want the critical angle, which here is going to be equal to the look. Let's re write that again. Using cells. Snell's law the inverse side in two times the sign data to which is index of refraction for water times the sign of the angle state and W divided by and Sigi. But since we wanted to be totally internally reflected in the air. We know that in W times, the sign of fated W has to be equal to in sub a right cause. Insulate times sign of today. Which data is 90 degrees, right? Cause we want totally internal effect reflected. That means that this sign is equal to in Sunday, Divided by and Sigi. So again, this is equal to 34.2 degrees. That could be boxing as our answer for part B. Okay, Parsi says. Does the thickness of the water layer or the glass affect the results? Well, the results are affected by Snell's law, and Snell's law on Lee has the index of refraction in the angle of incident, and there is no thickness into a dependence there, so we can say no. According to Snell's law, the thickness of the medium does not matter so we can box it in as our solution for part C. And then finally, Part D Part D says, Does the index of refraction of the inverting layer affect the results? Well, the, uh, excuse me intervening later, the intervening layer is the water. But if you look at part one or excuse me part B the index of refraction of the water times the sine of the angle of incidence of the water is just gonna be equal to the index of refraction of air. Since that's the third medium, and we want it totally internally reflected. And since that's always gonna be true, that index, every fraction of water is not going to depend on the final solution. The final solution is just gonna depend on the index of refraction of air. So we can say, uh, no, according to be part B on Lee, the index of refraction, of air and glass matters. And that committee boxed in is our solution for a party in the final solution to our question.


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