Question

(a) A narrow beam of light containing yellow (580 nm) and green (550 nm) wavelengths goes from polystyrene to air, striking the surface at a $30.0^\circ$ incident angle. What is the angle between the colors when they emerge? (b) How far would they have to travel to be separated by 1.00 mm?

Final Answer

- $0.043^\circ$
- $2.00 \textrm{ m}$

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Video Transcript

This is College Physics Answers with Shaun Dychko. A beam of light consisting of yellow and green travels through polystyrene and then emerges into air and has an angle of incidence of 30 degrees and when it is… when it goes into air, the angle of refraction will be different for the different colours and this is because the index of refraction for yellow and for green are different in polystyrene. The index of refraction for these two colours in the air are same. It’s always 1.00 in air but the fact that they are different within the polystyrene cause the angle of refraction to be different for two different colours and so the question is asking us for this little bit of angle in here that is different for the two rays. So, we call it delta theta and then part B of this question will ask us for how far would the beams have to travel through the air in order for them to be separated horizontally by distance of one milli-metre. Okay. We’ll use Snell’s law to figure out what these angles of refractions are for yellow and for green and Snell’s law says that the incidence medium, the first medium, that’s its travelling through index of refraction and gets multiplied by sine of the angle of incidence,

*theta one*and that’s goanna equal the index of refraction in the second medium, we call it*n two*times sine*theta two*, the index… or the angle of refraction and then we can solve for*theta two*by dividing both sides by*n two*and then taking inverse sine on both sides and so*theta two*, the angle of refraction is the inverse sign of the first index of refraction times sine of the incident angle divided by the second index of refraction. Now, the*delta theta*, this difference in the angle of refractions for the two colours is goanna be the difference between the angle of refraction for green minus the angle of refraction for yellow and so we will calculate each of these now. So,*theta g*is goanna be inverse sine of 1.493 times sine of 30 degrees divided by one and that’s because the incident angle is 30 degrees and its true for both the colours cause they are in a single beam mixed together initially in the polystyrene, so they both have the same incident angle but the index of refraction for the two colours is different. Its 1.493 in the case of green colour and its 1.492 in the case of yellow and that’s the only difference between these two formulae and we end up with different angles. So, for green its 48.2881 degrees, is the angle of refraction and for yellow, it’s a little bit less 48.2451 degrees and so the difference in the angles is the difference between two numbers which is 0.043 degrees. Now, the next question is how far would the beam have to travel through air in order for the green and the yellow to be separated by one millimetre. So, this is*delta x*, the difference in the x-values for green and for ‘y’… and for yellow. So,*x*value meaning the amount they travel horizontally which is back to where they started. So, we are goanna take*x g*minus*x y*to find this*delta x*, umm… and we are goanna take the distance to be hypotenuse of this triangle here. Its not strictly clear whether they mean the distance travelled through air, meaning the hypotenuse which is what I think it means but they could also mean the distance from the surface. So, if this is the polystyrene there maybe what they are asking for is how far from the surface do the beams have to travel in order to be separated by a milli-metre. So, in this case, this distance here, maybe that’s what they are asking for. I don’t really know so I am goanna give two answers to the questions. So, well from this triangle we can see that sine of*theta g*, the angle of refraction for green is going to be the horizontal displacement of the beam*x g*divided by the hypotenuse*d g*and we are going to say the distance that the green travels along this hypotenuse is approximately the same as for yellow. I mean clearly from this picture, they are not the same but the picture is not drawn to scale and so they are pretty much going to be same in reality because this is meant to be only one millimetre compared to a much much larger by many orders of magnitude distance along the hypotenuse. So, that’s an approximation we have to make otherwise, we would have to solve using numerical methods or something. So, we are goanna say that the distance… the green travels is the same as the distance the yellow travels. So, sine*theta g*equals the opposite*x g*over the hypotenuse*d g*and we will solve this for*x g*by multiplying both sides by*d g*and we have*x g*equals*d g*times sine*theta g*and for ‘y’ for yellow, its essentially the same, except this different angle*theta y*and we will have*x y*as the opposite and then the hypotenuse is goanna be*d y*, what we are goanna approximate that to be the same as*d g*, the same as distance the green travels and so then in their*delta x*formula, we can substitute for the distance horizontally that the green travels and the horizontal distance that the yellow travels and this is a*d g*here and so we factor out the*d g*and then… and then you get*d g*times sine angle of refraction for green minus sine of the angle of refraction for yellow, approximately equals this one milli-metre and then divide both sides by the bracket and you get the*d g*which is what we want to find the distance that the beam travels through the air in order to be separated by milli-metre equals one millimetre divided by the sine of the*theta g*minus sine*theta y*. So, its one millimetre divided by sine of 48.2881 degrees minus sine of 48.2451 degrees and that gives 2001.7 millimetres which is about two metres. So, this distance along the hypotenuse is three orders of magnitude more than the separation of one millimetre. So, that’s why the separation of the green distance and the yellow distance being the same is acceptable. Now, what if they mean distance away from the surface. Well, it that’s the case then we can say that the cosine of*theta g*is the adjacent over the hypotenuse and we already found the hypotenuse. So, we can multiply both sides by the hypotenuse here and solve for the adjacent, which is this distance from the surface perpendicular distance. So, this distance from the surface is 2001.7 millimetres, length of the hypotenuse, times cosine of 48.2881 degrees which gives this many millimetres and we multiply by one metre for every 1000 millimetre and we get 1.33 metres.