Change the chapter
Students in a lab group are given a plastic cube with a hollow cube-shaped space in the middle that fills about half the volume of the cube. The index of refraction of the plastic is known. The hollow space is filled with a gas, and the students are asked to collect the data needed to find the index of refraction of the gas. What information would you need to collect, and how would you use this information in order to deduce the index of refraction of the gas in the cube?
Question by OpenStax is licensed under CC BY 4.0.
Final Answer
data needed: $\textrm{l}, \theta_1, $\Delta d$, $\textrm{n}_c$.
Solve the formula shown in the video numerically for $\Delta d$.
Solution Video

OpenStax College Physics for AP® Courses Solution, Chapter 25, Problem 9 (Test Prep for AP® Courses) (5:51)

Sign up to view this solution video!

View sample solution
Video Transcript

This is College Physics Answers with Shaun Dychko. We're given a plastic cube with a known index of refraction we’ll call <i>n subscript c</i> for Cube and it has a known height we’ll call that <i>l</i>. And inside of it is a cube full of gas with an unknown index of refraction <i>n subscript g</i> and this Cuban side has a height which is half that of the outer Cube. So we'll call this inner Cube <i>l</i> over two for its height and we're going to have a ray of light that goes at a known angle of incidence will call it <i>Theta one</i> here and then we'll figure out by how much this Ray is displaced here compared to if the cube was not there compared to this straight line that would go here if the gas was not present to displace the ray. So we're going to measure Delta <i>d</i>, <i>Theta one</i>, index refraction of the cube and the cubes height and knowing that we'll be able to figure out the index of refraction of the gas. So how we're going to do that is first consider this picture where we have in the plastic compared to this point here, the beam will get displaced distance <i>d1</i> by the time it gets to the top of the gas Cube. And then in the gas is going to get displaced an amount <i>d2</i>. And then in the plastic again, it's going to get displaced an amount <i>d3</i> here. And so this leg of the triangle is <i>l</i> over four, that's the distance here and that first triangle above the gas Cube and then within the gas Cube, we have a distance of <i>l</i> over two. And then in the plastic again, we have a distance <i>l</i> over four. And so we're going to compare this total distance of <i>d1</i> plus <i>d2</i> plus <i>d3</i> with the distance, if you had no gas <i>ng</i> for no gas. And so, you know, it's going to go add an incident angle <i>Theta One</i> have the entire distance <i>l</i> the tire height of the plastic Cube and we can figure out <i>dng</i> by saying that the tangent of this angle <i>Theta One</i> is the opposite over the adjacent. And so that's <i>d</i> no gas divided by <i>l</i> and we'll multiply both sides by <i>l</i> . To solve for distance of with no gas <i>l</i> times <i>Tan Theta One</i>. And then we're going to take the difference between that and the total distance when there is gas. So we have these portions <i>d1</i> and <i>d3</i> within the plastic and those parts are straightforward to calculate and in each case is going to be <i>Theta One</i> is going to be the opposite <i>d1</i> or <i>d3</i> and then tangent of that is going to be that horizontal displacement divided by <i>l</i> over four. And so <i>d1</i> is <i>l</i> over four times <i>Tan Theta One</i> and now this part for calculating <i>d2</i> is a bit more involved. So this triangle has faded two here. This is an angle of refraction when it goes through the into the gas and we know that from Snell's law that the index refraction of the cube times sine <i>Theta One</i> is going to equal the index of refraction of the gas times sine <i>Theta two</i> . And we can solve for <i>Theta two</i> by dividing both sides by <i>ng</i>. And then taking the inverse sine of both sides. So <i>Theta two</i> inverse sine of <i>nc</i> sine <i>Theta One</i> over <i>ng</i> and that is something that we can plug into the tangent. So we're going to go <i>d2</i> is <i>l</i> over two times tangent of <i>Theta two</i> . But <i>Theta two</i> is all of this and so that's where that comes from. So the total with the gas is <i>D1</i> plus <i>d2</i> plus <i>d3</i> , but<i>d3</i> equals <i>d1</i> . Because once this beam returns back to the same plastic material of the cube, it's going to return to the same angle here <i>Theta One</i> that is started with up here. And so <i>d3</i> down here is going to equal <i>d1</i> up there. And so that's two <i>d1</i> plus <i>d2</i> and so that's two times <i>l</i> over four times <i>Tan Theta One</i> plus <i>l</i> over two times Tan of inverse sine <i>nc</i> sine <i>Theta One</i> over <i>ng</i> and this Delta <i>d</i> that we can measure is going to be the difference between the distance with no gas <i>l</i> times Tan <i>Theta One</i> minus all of this. And then we can factor out the <i>l</i> over two here and we get Delta <i>d</i> then is <i>l</i> over two times tan <i>Theta One</i> minus tan of the inverse sine of <i>nc</i> times sine of <i>Theta One</i> over <i>ng</i>. Now this formula we really hard to I don't think you can solve it analytically, you need to use a spreadsheet and solve this numerically to figure out what <i>ng</i> if you keep plugging in different possible possibilities for <i>ng</i> what <i>ng</i> would give you the Delta <i>d</i> that you measure given all these other things that you've also measured and so we solve this numerically and the data we need is the height of the cube, the incident angle, the displacement of the beam with no gas compared to with gas, and the index of refraction of the cube material.