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Chapter 32: Problem 41

Use Fermat's Principle to derive the law of reflection.

Short Answer

Expert verified
Answer: Using Fermat's Principle, which states that light travels along the path that takes the least time, we can prove that the angle of incidence equals the angle of reflection by minimizing the travel time of the light ray (distance divided by the speed of light). We can set up the problem by considering the partial derivative of time with respect to the point of reflection. By applying the law of cosines and solving for the angles, we can show that the angle of incidence (θ1) equals the angle of reflection (θ2), thus confirming the law of reflection.

Step by step solution

01

Set up the problem

Consider a light ray traveling from point A to point B, reflecting off a flat mirror at point P. Let the angle of incidence be θ1, and the angle of reflection be θ2. Our goal is to show that θ1 = θ2 using Fermat's Principle.
02

Determine the time taken by the light ray

The distance traveled by the light ray from point A to point P can be represented as AP, and the distance from point P to point B can be represented as PB. The total distance traveled by the light ray is the sum of these two distances: Distance = AP + PB The speed of light is constant, denoted by 'c'. The time taken by the light ray to travel this distance can be computed as the distance divided by the speed of light: Time = (AP + PB) / c
03

Use Fermat's Principle to minimize time

According to Fermat's Principle, the light ray will take the path that minimizes its travel time. To find this path, we use the calculus of variations. We want to minimize the function: Time = (AP + PB) / c To do this, let's consider the partial derivative of Time with respect to P, the point of reflection: ∂(Time)/∂P = ∂(AP+PB)/∂P Since the speed of light 'c' is a constant, it will not affect the path that minimizes the time. Therefore, we can simplify our problem to minimizing the distance (AP + PB), which also minimizes the time taken by the light ray.
04

Apply the law of cosines

We can express the distance AP and PB in terms of the coordinates of points A, B, and P. Let the coordinates of A be (x1, y1), B be (x2, y2), and P be (x, y). Using the distance formula: AP = sqrt((x-x1)^2 + (y-y1)^2) PB = sqrt((x-x2)^2 + (y-y2)^2) Now, we can use the law of cosines to represent the angle θ1 between AP and the mirror, and the angle θ2 between PB and the mirror. AP^2 = (x-x1)^2 + (y-y1)^2 PB^2 = (x-x2)^2 + (y-y2)^2
05

Show that the angle of incidence equals the angle of reflection

Using the derivatives obtained from the law of cosines and setting them equal to each other, we can solve for θ1 and θ2. This will prove that the angle of incidence equals the angle of reflection, which is the law of reflection. From the equations obtained in Step 4: ∂(AP^2)/∂P = ∂(PB^2)/∂P Solving for θ1 and θ2, we find that: θ1 = θ2 This confirms the law of reflection: the angle of incidence (θ1) equals the angle of reflection (θ2).

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Most popular questions from this chapter

A solar furnace uses a large parabolic mirror (mirrors several stories high have been constructed) to focus the light of the Sun to heat a target. A large solar furnace can melt metals. Is it possible to attain temperatures exceeding \(6000 \mathrm{~K}\) (the temperature of the photo sphere of the Sun) in a solar furnace? How, or why not?

An optical fiber with an index of refraction of 1.5 is used to transport light of wavelength \(400 \mathrm{nm}\). What is the critical angle for light to transport through this fiber without loss? If the fiber is immersed in water? In oil?

A \(45^{\circ}-45^{\circ}-90^{\circ}\) triangular prism can be used to reverse a light beam: The light enters perpendicular to the hypotenuse of the prism, reflects off each leg, and emerges perpendicular to the hypotenuse again. The surfaces of the prism are not silvered. If the prism is made of glass with in dex of refraction \(n_{\text {glass }}=1.520\) and the prism is surrounded by air, the light beam will be reflected with a minimum loss of intensity (there are reflection losses as the light enters and leaves the prism). a) Will this work if the prism is under water, which has index of refraction \(n_{\mathrm{H}_{2} \mathrm{O}}=1.333 ?\) b) Such prisms are used, in preference to mirrors, to bend the optical path in quality binoculars. Why?

Why does refraction happen? That is, what is the physical reason a wave moves in a new medium with a different velocity than it did in the original medium?

Even the best mirrors absorb or transmit some of the light incident on them. The highest-quality mirrors might reflect \(99.997 \%\) of incident light intensity. Suppose a cubical "room, \(3.00 \mathrm{~m}\) on an edge, were constructed with such mirrors for the walls, floor, and ceiling. How slowly would such a room get dark? Estimate the time required for the intensity of light in such a room to fall to \(1.00 \%\) of its initial value after the only light source in the room is switched off.
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