Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781337026345
Author: Katz
Publisher: Cengage
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Chapter 38, Problem 98PQ

A Fermat’s principle of least time for refraction. A ray of light traveling in a medium with speed v1 leaves point A and strikes the boundary between the incident and transmitted media a horizontal distance x from point A as shown in Figure P38.98. The refracted ray travels with speed v2 in the second medium, eventually reaching point B. The horizontal distance between points A and B is L. a. Calculate the time t required for the light to travel from A to B in terms of the parameters labeled in the figure. b. Now take the derivative of t with respect to x. What is the condition for which the ray of light will take the shortest time to travel from A to B?

Chapter 38, Problem 98PQ, A Fermats principle of least time for refraction. A ray of light traveling in a medium with speed v1

Figure P38.98

(a)

Expert Solution
Check Mark
To determine

The time required for the light ray to travel from the points A to B.

Answer to Problem 98PQ

The time taken for the light to travel from point A to B is t=x2+y12v1+y22+(Lx)2v2.

Explanation of Solution

Write the expression for the time taken for the light ray in the second medium.

    t2=d2v2

Here, v2 is the speed of the light in the second medium, d2 is the distance travelled by the light in the second medium, t2 is the time taken for the light ray in the second medium.

Write the expression for distance travelled by the light in the second medium.

    d2=y22+(Lx)2

Here, y2 is the vertical distance of the light in the second medium, L is the distance between the point A to the point B, and x is the horizontal distance from point A to the transmitted media as shown in the Figure-(1).

Write the expression for the time taken for the light ray in the first medium.

    t1=d1v1

Here, v1 is the speed of the light in the first medium, d1 is the distance travelled by the light in the first medium, t1 is the time taken for the light ray in the first medium.

Write the expression for distance travelled by the light in the first medium.

    d1=x2+y12

Here, y1 is the vertical distance of the light in the first medium, and x is the horizontal distance from point A to the transmitted media as shown in the Figure-(1).

Write the equation for total time taken for the light is.

    t=t1+t2                           (I)

Conclusion:

Substitute d1v1 for t1 and t2=d2v2 for t2 in equation (I).

    t=d1v1+d2v2                            (II)

Substitute x2+y12 for d1 and y22+(Lx)2 for d2 in the above equation.

    t=x2+y12v1+y22+(Lx)2v2                   (III)

Therefore, the time taken for the light to travel from point A to B is     t=x2+y12v1+y22+(Lx)2v2

(b)

Expert Solution
Check Mark
To determine

The required condition to travel for getting the shortest time to reach the point A to point B.

Answer to Problem 98PQ

The required condition for getting the minimum time to reach the point A to point B is.    v2v1=(Lxx)((x2+y12)(y22+(Lx)2)).

Explanation of Solution

Take the derivative of t with respect to x in the equation (III).

    t=x2+y12v1+y22+(Lx)2v2dtdx=ddx(x2+y12v1+y22+(Lx)2v2)

From the formula xn=nxn1 simplify the above equation.

    dtdx=1v1(12(x2+y12)12(2x))+1v2(12(y22+(Lx)2)122(Lx)(1))=1v1(12((x2+y12))(2x))+1v2(12((y22+(Lx)2))2(Lx)(1))=xv1(x2+y12)Lxv2(y22+(Lx)2)            (IV)

The light ray in which its shortest time taken is.

    dtdx=0

Conclusion:

Substitute 0 for dtdx in equation (IV).

    dtdx=xv1(x2+y12)Lxv2(y22+(Lx)2)0=xv1(x2+y12)Lxv2(y22+(Lx)2)Lxv2(y22+(Lx)2)=xv1(x2+y12)Lxx=v2(y22+(Lx)2)v1(x2+y12)

Simplify the above equation.

    Lxx=v2(y22+(Lx)2)v1(x2+y12)Lxx=(v2v1)((y22+(Lx)2)(x2+y12))v2v1=(Lxx)((x2+y12)(y22+(Lx)2))

Therefore, the required condition for getting the minimum time to reach the point A to point B is. v2v1=(Lxx)((x2+y12)(y22+(Lx)2)).

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Chapter 38 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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