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Problem 12 Medium Difficulty

Light containing wavelengths of 400. nm, 500. nm, and 650. nm is incident from air on a block of crown glass at an angle of $25.0^{\circ} .$ (a) Are all colors refracted alike, or is one color bent more than the others? (b) Calculate the angle of refraction in each case to verify your answer.

Answer

a. $\text { The angle of refraction is different for the colors; because the index of refraction is different for the different frequencies. }$
b.$ \begin{array}{l}{\text { The angle of refraction for } 300 \mathrm{mm}, 400 \mathrm{nm} \text { and } 500 \mathrm{nm} \text { are } 8.14^{\circ}, 10.88^{\circ}} \\ {\text { and } 13.65^{\circ} \text { respectively. }}\end{array}$

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Top Physics 103 Educators
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Video Transcript

part a of our question asked if all the colors reflected are alike or if one color is bent a little bit more than the other colors. So light bends at the interface of two mediums due to changes in speed. So when it moves from one transparent medium to another, this is called refraction and white lie incident. Upon this crown, glass gets divided into components and speeds of light dependent upon the wavelength. Since the refractive index of Crown glass is different from different way, Blakes, the longer wavelength deviate from the original path less than the shorter wavelength. So the answer to our question then would be, uh, they are different, and so we can go ahead and say part. Eh, let's type out. They are different for different, uh, colors because the index of refraction is different. Four different color slash frequencies. That's because color and frequency are related. So that would be the answer to part A. So for part B, it asks us to calculate the angle of refraction if we're given the fact that the angle of incident um, we go back here. The angle of incident is 25 degrees. We're coming out of a medium that has an index of refraction in someone of one and were asked to find Ah, find it for 405 106 115 centimeters of light. And I've written their index of refraction is here for each associated. What? Okay, So when Lambda then when the wavelength is equal to will, say 400 nano meters, let's start with that one. We can use this equation which says that the index of refraction of the incident times the sine of the angle of the incident is equal to the index of refraction. So that's inside i index of refraction, of what's refracted times the sign of that angle. And that's the angle we're trying to find. So in this case, data are is going to be equal to the inverse sine of the ratio of in one times the sign of the incident angle. They die, which we were told this 25 degrees divided by the index of refraction at 400 nanometers, which we called in some 400. So plugging those values into this expression where instead of 400 was 1.531 and someone is one and the angle of refraction is 25 degrees plugging those values. And we find that this angle is equal to 16 degrees weaken box. That in is their solution for that part of the question. Okay, so now we can do the same thing. For when Lambda is equal to 500 nano meters, we're gonna have the same equation. But this time status of our is equal to the inverse sine and above here we also this is inverse sine we forgot our minus one so inverse sine of in one times this sign of the incident angle, they die divided by and said 500 which we have listed as 1.52 playing those values into the six fresh and we find that this is equal to 16.1 degrees making box said it is her solution for part B for the second part of part B and then for the last part of part B, we need a little bit room here. This is for Lambda equal to 650 nano meters set this time, Davis of our it's going to be equal to same equation. We're just gonna use the addicts every fraction for 650 nano meters, which was 1.51 So we have the inverse sine of the ratio of in one climbs the sign of the index of refraction. Uh, the incident and exit incident angle divided by in sub 650 and find that this comes out to equal 16 0.3 degrees. So they're all slightly different Anatomy box. It is the final part of our solution.

University of Kansas
Top Physics 103 Educators
Elyse G.

Cornell University

LB
Liev B.

Numerade Educator

Marshall S.

University of Washington

Aspen F.

University of Sheffield