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Ch 33: The Nature and Propagation of Light

Chapter 33, Problem 33

The indexes of refraction for violet light λ = 400 nm2 and red light λ= 700 nm2 in diamond are 2.46 and 2.41, respectively. A ray of light traveling through air strikes the diamond surface at an angle of 53.5° to the normal. Calculate the angular separation between these two colors of light in the refracted ray.

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Hi, everyone. In this problem, we are being asked to calculate the angular separation between the yellow and the red color. In a refracted ray, we will have a fly glass with a refractive index between 1.45 and two and is surrounded by air. A narrow beam of white light strikes the vertically placed piece of flame glass at an angle of 62 degrees. With the normal in air. We're being asked to calculate the angular separation between the yellow and the red color in the reflected rate when the yellow has a wavelength of nanometer and the red has a wavelength of 700 nanometer. And the index of refraction for the red is 1.42 and for the yellow is 1.44 inside of the flint glass. The options given for the angular separations are a 0.5 degrees B 0.3 degrees C 0.4 degrees and lastly D 0.6 degrees. So the red and the yellow light will bend to different angles because they have different indexes of reflection. In order for us to determine the degrees at which the yellow wrap and yellow light is bent. We will have to use Snell's law at the boundary layer. So Snell slot itself will use N one multiplied by sine of theta one for the first medium to be equals to N two multiplied by sine of theta two for the second medium. So we will apply the snail slot twice the first time using the index of refraction for the red light. And the second one using the index of the refraction for the yellow light. The index of reflection for air as it is surrounded by air is one, so N of air is one. So in this case, let's start with the red light. So for the rat, we still want to use N one multiplied by sine of theta one equals to N two multiplied by sine of theta two. N one is going to be the air which is going to be one, the index of reflection is going to be one and then sign of theta one is going to be the angle of incidence which is going to be sine of degrees which will then be equals to N two multiplied by sine of theta two where N two is the end of red, which is 1.42. So 1.42 multiplied by sine of theta two. So next, we want to rearrange this so that we will have sine of the two to then be sine of 62 degrees divided by 1.42. And what we are interested to find is the two which in this case, we have to utilize the in first of sine or s sign to the power of negative one of sine of 62 degrees divided by 1.42. In order to get theta two or in this case, the inverse of sine is a. So we can utilize axent as well. So solving this equation, so then theta two which will then equals to the theta of the red light which will then equals to 38. degrees. Awesome. So now looking into the yellow light for the yellow light, we still want to use the same smell slaw. So N one multiplied by sine of theta one equals to N two multiplied by sine of theta two and one is one multiplied by sine of theta one which is sine of 62 degrees and two is the N of yellow which is 1.44 multiplied that by sign of the two. So rearranging this, we will then get sine of theta two to then be sine of 62 degrees divided by 1.44. And we want to use the arc sin or the inverse of the sin to then get data two which will be signed to the power of negative one of, of sine 62 degrees divided by 1.44. That will then give us the two or the data of yellow to equals to 37.8 degrees. So from these two values, we can then calculate the angle between the two initially coincident rays which is going to then be delta theta which will be the theta of red, minus yellow or yellow minus red. In absolute value, either one will work which will then be 38.4 degrees minus 37.8 degrees, which will give us a 0.6 degrees value of delta theta or in this case, the angular separation between the yellow and the red color in the refracted ray is going to be 0. degrees. So that will be it for this video. If you guys still have any sort of confusion, please make sure to check out our adolescent videos on similar topics. But other than that option D with the angular separation of 0.6 degrees will be the answer to this practice problem. Thank you.