Some modern optical devices are made with glass whose index of re...

Question

Some modern optical devices are made with glass whose index of refraction changes with distance from the front surface. FIGURE P16.72 shows the index of refraction as a function of the distance into a slab of glass of thickness L. The index of refraction increases linearly from n₁ at the front surface to n₂ at the rear surface.b. Evaluate your expression for a 1.0-cm-thick piece of glass for which n₁ = 1.50 and n₂ = 1.60.

Accepted Answer

Hey everyone. So this problem is dealing with the refraction of light. Let's see what it's asking us. A linear gradient refractive index or a linear grin is a type of grin material where the refractive index changes linearly with distance. Suppose that the refractive index of a linear grin increases at a constant rate between point A which is at 1.48 and point B which is at 1.56 A and B lie on the horizontal axis and the distance separating the two points is 3.5 centimeters. We're asked to calculate the time needed for an electromagnetic wave to travel between points A and B. Our multiple choice answers are a 2.25 times 10 to the negative 12 seconds. B 1.77 times 10 to the negative 10 seconds. C 3.54 times 10 to the negative nine seconds or D 1.85 times 10 to the negative eight seconds. So we know that the refractive index changes at a constant rate between A and B, it increases between A and B and so we can express this variation by the equation N is equal to and sub A plus and sub B minus and sub A divided by D, the distance between A and B multiplied by X the position at any point in that medium. We can also recall that for refractive indices and electromagnetic waves, the speed varies with relation to the refractive index. So the speed is equal to the speed of light divided by N the fract index. And lastly, the equation that we can use to pull all of this together focusing on what the problem is asking for or time time is equal to our position divided by our speed. And so when we combine these equations, we get DT or the change in time is equal to one divided by C multiplied by N A plus NB minus N A divided by DAB multiplied by X GX. And so from there, the time it takes to travel from point A to point B can be found by integrating. So we have from zero to TDT is equal to, we'll keep one over C as a constant. We'll keep that outside of the integral. And then from point A to point B. So actually write this as X sub A and X sub B um and sub A plus NB minus N A divided by dab multiplied by XDX. So we can use the sum rule and pulling out the constants outside of the integral. And this equation simplifies to an A divided by C of the integral from X A to XV DX plus NB minus N A divided by C multiplied by dab multiplied by the integral from X of A to Xab um XDX. Now this looks like common integrals. We can use our CALC ROS to solve. And so on the left hand side of the equation we have T and that's equal to N A divided by C multiplied by the integral of DX assembly, X from X A of 0 m to XP of 0.035 m. So because we know the distance between point A and point B is 0.3 0.035 m or 3.5 centimeters, we can assign point A to be 0 m and then point B becomes 0.035 m. We're just choosing our reference point along that X axis. OK plus NB minus N A divided by C and dab multiplied by the integral of XDX is simply X squared divided by two. And then again, from point A 0 m to point B 0.035 m. The last thing we need to do here is plug in our known values for all from the problem. So N A or the refractive index at point A is 1.48 divided by C speed of light constant three times 10 to the eighth meters per second. And then X where X is from 0.35 m minus 0 m is simply 0.035 m plus NB 1.56 minus N A 1.48 multiplied again by or sorry, divided by C again, three times 10 to the 8 m per second multiplied by dab that's the distance from A to B 0.035 m and then multiplied by X square divided by two from zero 20.35. The zero obviously goes to zero and then we have 0.035 m squared divided by two, plug that into our calculator and we get a time of 1.77 times 10 to the negative 10 seconds. And so that's our final answer for this problem. When we look at our multiple choice answers, it aligns with answer choice B so B is the correct answer. That's all we have for this one. We'll see you in the next video.

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