Snell’s Law Suppose a light source at A is in a medium in whic...

Question

Snell’s Law Suppose a light source at A is in a medium in which light travels at a speed v₁ and that point B is in a medium in which light travels at a speed v₂ (see figure). Using Fermat’s Principle, which states that light travels along the path that requires the minimum travel time (Exercise 55), show that the path taken between points A and B satisfies (sinΘ₁/v₁ = (sin Θ₂) /v₂ .

Accepted Answer

Hi everyone. Let's take a look at this practice problem. This problem says an ant needs to cross a garden to reach a sugar cube. The first part of the garden is covered in sand, where the ant moves at 2 centimeters per second, and the second part is a patch of grass where the ant's speed decreases to 1 centimeter per second. To minimize the travel time, what should be the ratio of the sign of the angle at which the ant enters the grass theta 1 to the sign of the angle at which it moves through the grass at theta 2. Below the problem, we're given a diagram of what was described in the question. We're also given 4 possible choices as our answers. For choice A, we have 2 to 1. For choice B, we have 1 to 2, for choice C, we have 4 to 1, and for choice D, we have 1 to 4. Now we need to find the ratio of the sign of the angle that the ant enters the grass to the sign of the angle at which the ant moves through the grass. So we're gonna start by defining a couple quantities on our diagram. So, the first quantity we're going to find is the total distance that the ant travels in the sand, which we'll label as D1. We'll also call the total distance that the ant travels in the grass as D2. Now, we were given the speeds of the ant in these two media, and the problems, so we want to relate our speeds to these distances. So recall that speed is equal to our distance divided by time. So, if we call it the speed that the ant travels in the sand as S1, then that's going to be equal to D1 divided by T1. Where T1 here is the time that the ant travels in the sand. We can solve this expression for T1 by multiplying both sides by T1 and dividing both sides by S1. And doing so we'll get T1 is equal to D1 divided by S1. Likewise, we can define the speed that the ant travels through the grass as S2. And that's going to be equal to D2 divided by T2, where T2 is the time that the ant travels through the grass, and we can solve this for T2, again by multiplying both sides by T2 and dividing both sides by S2, and we will get T2 is equal to D2 divided by S2. Now, S1 is going to be equal to. 2 centimeters per second, since that is the speed at which the ant travels through the sand that was given to the problem, and S2 is going to be equal to 1 centimeter per second, since that is the speed given in the problem that the ant travels with through the grass. So we can write our total travel time T as being equal to T1 plus T2, and we can write these 2 times in terms of the distances. So this is going to be equal to D1 divided by S1 plus D2 divided by S2. And we can substitute in our values for S1 and S2. So this is going to be equal to D1 divided by 2, plus D2 divided by 1, which is just D2. So we're going to now define a couple more quantities on our diagram. So we're gonna call the horizontal distance that the ant travels in the sand as X, we call the vertical distance that it travels as A. We're going to call the vertical distance that the ant travels in the grass as B. And we're gonna label the total horizontal distance that the ant travels as D. And that means that the horizontal distance that the ant travels in the grass is going to be D minus X. So now we can rewrite D1 and D2 in terms of these new variables. So, using Pythagorean's theorem, we'll see that D1 is equal to the square root of the quantity of X2 plus A2. And D2 is going to be equal to the square root of the quantity of D minus X in quantity squared, plus B2 in quantity. So this will allow us to write our time T as a function of X. Wet of X is equal to 1/2 multiplied by the square root of quantity of X2 plus a 2 in quantity plus the square root of the quantity of D minus X in quantity squad plus B2 in quantity. And so here, this is a function of X since our variables A, B, and D are all constants, since the vertical distance that the ant travels in both the sand and the grass are constant, as well as the total horizontal distance. And so, what we want to do is minimize our travel time. So, what we're going to do is take a derivative of our function T here with respect to X. So, we're going to calculate T of X. And that's going to be equal to the derivative with respect to X of the quantity of 1/2 multiplied by the square root of the quantity of X2 plus a 2 in quantity, plus the square root of the quantity of D minus X in quantity squared, plus B2 in quantity. So taking the derivative, this is going to be equal to 1/2 multiplied by the derivative with respect to X of the square root of the quantity of X2 plus A2. And when I take that derivative, I will get 1 half again, multiplied by the quantity of X2 plus A2 in quantity raised to the minus 1/2. So here we've just rewritten the square root as raising something to the, to the 12 power. And we still need to apply our chain rule. So applying our chain rule here, we'll multiply this by the derivative with respect to X of the quantity of X2 plus A2, which is going to be 2 X. Then we'll have plus the derivative with respect to x of the square root of the quantity of D minus X in quantity squad plus B2 in quantity. And taking that derivative, we will have 1/2, multiplying the quantity of D minus X in quantity squared, plus B2 in quantity raised to the minus 12 power. And again, we have to apply our chain rule, so we need to multiply this by the derivative with respect to X of the quantity of D minus X in quantity squad plus B2. In doing so, we will get 2 multiplied by the quantity of D minus X in quantity multiplied by -1. So simplifying this expression, this is going to be equal to. X divided by the quantity of 2 multiplied by the square root of the quantity of X2 plus a 2 in quantity minus the quantity of D minus X in quantity, divided by the square root. Of the quantity of D minus X in quantity squared plus B2 in quantity. So, if we look at our derivative here, if we set X equal to 0, we see that this first derivative will actually be negative. And if we set X equals to D, we see that our first derivative here is going to be positive. So since our derivative is changing from a negative value to a positive value, that means that we have a minimum located somewhere in between those two X values. So what we're going to do to find the X value for that minimum is set our first derivative here equal to 0 and solve for X. And so we're going to have X. Divided by the quantity of 2 multiplied by the square root of the quantity of X2 plus a 2 in quantity minus the quantity of D minus X in quantity divided by the square root of the quantity of D minus X in quantity squared plus B2 in quantity is equal to 0. So moving the second term to the other side of the equation, we're going to get X divided by the quantity of 2 multiplied by the square root of the quantity of X2 plus a 2 in quantity is equal to the quantity of D minus X in quantity divided by the square root of the quantity of D minus X in quantity squared plus B2 in quantity. And here at this point, we notice that we can rewrite. This in terms of our sin of theta 1 and sin of theta 2. So, based upon our diagram, we see that sign of data one. is equal to x divided by D1, which is square root of the quantity of X2 plus A2. And sine of theta 2 is equal to the quantity of D minus X. Divided by D2, which is the square root. Of the quantity of D minus X in quantity squad plus B2 in quantity. So rewriting our expression in terms of sin theta 1 and sin of theta 2, we will get 1 half multiplied by sine of theta 1 is equal to sin of theta 2. Now we're looking for the ratio of sin of the 1 to sin of the 2. So we'll divide both sides of this equation by sin of the 2 and multiply both sides by 2. So doing so, we will get sin of the 1 divided by sin of the 2. is equal to 2 divided by 1. So that means that our ratio of our sin of the 1 to sin of theta 2 is going to be 2 to 1. So this here is the answer that we are looking for. And if we look at our answer choices, we see that the correct answer choice corresponds to answer choice A. So I hope that this explanation has been useful, and I'll see you in the next video.

Key Concepts

Fermat's Principle

Snell's Law

Refraction

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