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### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

Book edition 4th
Author(s) Randall D. Knight
Pages 1240 pages
ISBN 9780133942651

# Shows a light ray that travels from point A to point B. The ray crosses the boundary at position x, making angles and in the two media. Suppose that you did not know Snell’s law. A. Write an expression for the time t it takes the light ray to travel from A to B. Your expression should be in terms of the distances a, b, and w; the variable x; and the indices of refraction n1 and n2 B. The time depends on x. There’s one value of x for which the light travels from A to B in the shortest possible time. We’ll call it . Write an expression (but don’t try to solve it!) from which could be found.C. Now, by using the geometry of the figure, derive Snell’s law from your answer to part b. You’ve proven that Snell’s law is equivalent to the statement that “light traveling between two points follows the path that requires the shortest time.” This interesting way of thinking about refraction is called Fermat’s principle.

See the step by step solution

## Step 1.Concepts and Principles

1. The index of refraction of a medium is defined by the ratio

where c =m/s is the speed of light in vacuum and is the speed of light in the medium.

2.The Ray Model of Light:

Light travels along straight lines, called light rays, at speed v=c/n, where n is the material's index of refraction. A light ray continues forever unless an interaction with matter causes it to reflect, refract, scatter, or be absorbed. Light rays come from objects. Each point on the object sends rays in all directions. The eye sees an object (or an image) when diverging rays are collected by the pupil and focused on the retina. Ray optics is valid when lenses, mirrors, and apertures are larger than ≈ 1 mm.

3.The average speed of a particle is equal to the ratio of the total distance it travels to the total interval during which it travels that distance:

## Step 2.Given Information

• The light ray that travels from point A to point B.

• The ray crosses the boundary at position

• The angle of incidence of the light ray is:

• The angle of refraction of the light ray is:.

• The index of refraction of the medium for which the light ray is incident is:

• The index of refraction of the medium for which the light ray is refracted is:.

• We are not allowed to use Snell's law.

• is the value of xx for which the light travels from A to B in the shortest possible time.

## Part A) Step 1.Given information

find an expression for the time tt taken by the light ray to travel from point A to point B.

## Part A) Step 2.Simplify

The speed of the light ray in medium 1 is found in terms of its index of refraction from Equation (1):

Similarly, the speed of the light ray in medium 2 is:

The time of travel of the light ray from point A to the boundary is found from Equation (2):

where is the distance traveled by the light ray from point A to the boundary. is found in terms of aa and x by applying the

Pythagorean theorem to the yellow-shaded triangle in Figure 1:

Substitute for into Equation (3):

The time of travel of the light ray from the boundary to point B is found from Equation (2):

where is the distance traveled by the light ray from the boundary to point B. is found in terms of x, w, and b by applying the Pythagorean theorem to the green-shaded triangle in Figure 1:

Substitute for into Equation (4):

The time taken by the light ray to travel from point A to point B is then:

t=

## Part B) Step 1.Given Information

determine an expression from which can be found.

## Part B) Step 2.Simplify

The expression from which can be determined is found by differentiating with respect to x and then setting the derivative equal to zero. So,

The value of can determined using the above expression.

## Part C) Step 1.Given Information

derive Snell's law from the geometry of the figure and the answer to part (b).

## Part C)  Step 2.Simplify

From the geometry of Figure 1, we see that the cosine of the angle is:

where is equal to the sine of 's complementary angle . So

The sine of the angle is also found from the geometry of the green-shaded triangle in Figure 1:

Substituting for from Equations (6) and (7) into Equation (5), we obtain Snell's law: