Minimum Time When light waves traveling in a transparent...

Minimum Time When light waves traveling in a transparent medium strike the surface of a second transparent medium, they change direction. This change of direction is called refraction and is defined by Snell's Law of Refraction, $$\frac{\sin \theta_{1}}{v_{1}}=\frac{\sin \theta_{2}}{v_{2}}$$ where $\theta_{1}$ and $\theta_{2}$ are the magnitudes of the angles shown in the figure and $v_{1}$ and $v_{2}$ are the velocities of light in the two media. Show that this problem is equivalent to that in Exercise $42,$ and that light waves traveling from $P$ to $Q$ follow the path of minimum time.

Minimum Time When light waves traveling in a transparent medium strike the surface of a second transparent medium, they change direction. This change of direction is called refraction and is defined by Snell's Law of Refraction,

$$\frac{\sin \theta_{1}}{v_{1}}=\frac{\sin \theta_{2}}{v_{2}}$$
where $\theta_{1}$ and $\theta_{2}$ are the magnitudes of the angles shown in the
figure and $v_{1}$ and $v_{2}$ are the velocities of light in the two media.
Show that this problem is equivalent to that in Exercise $42,$
and that light waves traveling from $P$ to $Q$ follow the path of
minimum time.

Key Concepts

Fermat's Principle

Fermat's Principle, also known as the principle of least time, states that light travels between two points along the path that takes the least time compared to nearby alternatives. This principle underlies many optical phenomena, including refraction and reflection, and provides a unifying explanation for why light follows a particular curved path in heterogeneous media.

Snell's Law

Snell's Law quantitatively describes the relationship between the angles of incidence and refraction as light passes from one medium to another with different speeds. The law, often stated as the ratio of the sine of the angles being equal to the ratio of the velocities in the two media, is a direct consequence of Fermat's Principle and explains the bending of light at interfaces.

Calculus of Variations

Calculus of Variations is a mathematical technique used to find the function or curve that minimizes (or maximizes) a particular quantity, such as time, energy, or distance. In the context of optics, it is applied to derive the path of least time taken by light, thereby providing a rigorous proof that the light’s actual path, which satisfies Snell's Law, minimizes the travel time between two points.

Transcript

00:01 So i'm going to begin just by re -drawing the picture that we have.

00:04 So we have two mediums which are separated by this horizontal line here.

00:10 It starts up here, travels through medium 1, hits medium 1, hits a different angle to get to this point here on medium 2.

00:19 And we have this vertical line here, which will help us with drawing the angle, theta 1, data 2.

00:27 The distance between medium 1, where it starts and medium 2.

00:33 So starting here, this is the distance of g1.

00:36 Over here is the distance of d2.

00:39 And this is x, x distance away from horizontal starting location.

00:46 This is a minus x.

00:48 We also have, it's actually everything we have.

00:52 All right, i'm going to label one additional thing.

00:55 The path that i'm like take to start here goes along this path.

00:58 I'm going to call this path.

00:59 Capital d1 for distance and goes here, changes mediums, goes along this path, which i would call capital d2.

01:08 All right, and we like to minimize the time it takes, whatever path minimizes the time.

01:14 So time is going to have two parts, time one plus time two, time one is the time it takes to get through medium 1.

01:25 And from the problem, we know that it travels at a rate of v1, since distance equals rate times times, time equals distance divided by rate and so t1 will be distance divided by rate so d1 d1 over v1 and then t2 will be d2 this distance here this distance here divided by v2 the rate at which might go through that medium alright so here is our equation for time and we to use calculus we need to make it so there's only single variable.

02:11 Do in mind v1 and b2 are constant.

02:13 So we need to find expressions for g1 and d2 and to do that we can use the pythagorean theorem.

02:19 So in this right triangle up here, we have that d1 squared plus x squared equals big d1 squared.

02:28 And in this right triangle we have a minus x squared plus middle d2 squared equals big t2 squared.

02:37 So taking square roots, we'll give with expressions for d1 and d2, big d1 and big d2.

02:42 So we get big d1, you a square root, it's a positive square root, and this is positive.

02:51 We have, let's see, d1 squared plus x squared.

02:56 We write that with x squared first, x squared plus d1 squared over v1 plus square root, and you minus x squared plus v2 squared.

03:11 Yeah, d2 squared, okay.

03:13 Or a v2.

03:14 All right.

03:15 And this d1, little d1, letter d2, v1 and v2, they're all constants.

03:20 So the only variable in this function is x.

03:22 This is now a function of x.

03:26 So we just need to minimize this function.

03:32 So do that.

03:33 Let's find the derivative.

03:35 And find the critical numbers.

03:39 So t prime of x.

03:40 Differentiate this, remember x is a variable, v1, d1, they're just constant, so.

03:45 To differentiate the square root, we use the power rule as well as a chain rule...

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