Suppose that light reflects off a mirror to get from point A...

Suppose that light reflects off a mirror to get from point $A$ to point $B$ as indicated in the figure. Assuming a constant velocity of light, we can minimize time by minimizing the distance traveled. Find the point on the mirror that minimizes the distance traveled. Show that the angles in the figure are equal (the angle of incidence equals the angle of reflection). (FIGURE CAN'T COPY)

Key Concepts

Fermat's Principle

This principle states that light takes the path that requires the least time when traveling from one point to another. In problems involving reflection, where the speed of light is constant in a given medium, minimizing the path length is equivalent to minimizing the travel time. This concept is foundational in understanding why the law of reflection holds true.

Law of Reflection

The law of reflection asserts that the angle of incidence is equal to the angle of reflection. This occurs because the path that light follows, according to Fermat's Principle, is the one that minimizes travel time, and geometric analysis of the reflected path leads directly to this equality of angles.

Method of Images

The method of images involves reflecting one of the points across the mirror to create an image point. By connecting the original point to this image, the straight-line path intersects the mirror at the optimal reflection point. This method provides a geometric demonstration of why the angles of incidence and reflection are equal and helps in finding the specific point on the mirror that minimizes the distance traveled.

Transcript

00:01 So there's a mirror laying flat like this, and light is going to bounce off of that mirror and hit two different points on two different sides here.

00:12 So this point is two units high, and this point is one unit up.

00:21 So then with that in mind, the goal is to figure out what this point is here that's going to minimize the distance.

00:33 Traveled by this light on that diagonal path.

00:38 We also know that it's four units across the entire base, which means these different triangles have units of x and then four minus x as those base values.

00:51 So to minimize the distance, we need to minimize this hypotenuse from the first triangle and the hypotenuse from the second triangle, because that's really what makes the distance.

01:04 Of the light, right? it's this diagonal piece and then this other diagonal piece.

01:09 So the distance for that first diagonal piece, that first hypotenuse would be the square root of x squared plus two squared, just from that pythagorean theorem.

01:22 And then the second hypotenuse there, again, pythagorean theorem would be the square root of 1 squared plus 4 minus x squared.

01:37 So now that it's in terms of one variable, the goal is to take that first derivative and find some critical numbers.

01:45 Now this first derivative is ugly because of all these square roots that we have here, but it's just going through a chain rule for both of these factors.

01:57 So bringing the one half in front and then going through the terms inside, going through the chain rolls from inside, and bringing those out.

02:06 So this is just a simplified form of that first derivative, which then i can let equal to zero to the solve for x.

02:13 So i'm going to graph this first derivative to see where x or where the graph crosses zero, where that x value is.

02:22 And as i do that, this first derivative is zero when x crosses at 2 .667.

02:30 And that went on for a long time...

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