Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics
Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics
1st Edition
ISBN: 9781305259836
Author: Debora M. Katz
Publisher: Cengage Learning
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Chapter 35, Problem 87PQ

(a)

To determine

The distance between the central maxima and adjacent maxima in the interference pattern in the screen.

(a)

Expert Solution
Check Mark

Answer to Problem 87PQ

The distance between the central maxima and an adjacent maxima in the interference pattern in the screen is 6.03×103m.

Explanation of Solution

Write the relation of the path difference for bright fringe in young’s double slit experiment.

    nλ=dsinθn                                                                                                        (I)

Here, λ is wavelength, the distance between the two consecutive slits is d and n is integer number.

Write the equation of the angle made by the particular fringe from the central maxima.

    tanθn=ynx                                                                                                           (II)

Here, yn is the distance between the particular fringe from the central fringe and x is the distance between slit and screen.

For small angle θ, consider sinθ=tanθ.

Substitute tanθ for sinθ in the equation (I).

    nλ=dtanθntanθn=nλd

Now, substitute nλd for tanθn in the equation (II).

  nλd=ynxyn=nλxd

Substitute, n=1 in the above equation to get the distance between the central maxima and the next consecutive maxima.

    Δy=λxd

Conclusion:

Substitute, 485nm for λ, 2.3m for x and 0.185mm for d in the above equation to find Δy.

    Δy=(485nm)(2.3m)(0.185mm)=(485×109m)(2.3m)(0.185×106m)=6.03×103m

Therefore, the distance between the central maxima and an adjacent maxima in the interference pattern in the screen is 6.03×103m.

(b)

To determine

The distance between the first and the third dark band of the interference pattern.

(b)

Expert Solution
Check Mark

Answer to Problem 87PQ

The distance between the first and the third dark band of the interference pattern is 1.2×102m.

Explanation of Solution

Write the relation of the path difference for dark fringe in young’s double slit experiment.

    (n+12)λ=dsinθn

Here, λ is wavelength, the distance between the two consecutive slits is d. n is integer number and θn is the angle between the dark band and the central fringe.

Substitute, tanθ for sinθ in the above equation for small angle θ.

    (n+12)λ=dtanθntanθn=(n+12)λd

Consider the equation (2).

    tanθn=ynx

Substitute (n+12)λd for tanθn in the above equation.

    (n+12)λd=ynxyn=(n+12)λxd

Calculate the distance between the first and the third fringe.

    y3y1=(3+12)λxd(1+12)λxd=(72)λxd(32)λxd=(73)λx2d=4λx2d

Simplify the above equation.

    y3y1=2λxd

Conclusion:

Substitute 485nm for λ, 2.3m for x and 0.185mm for d in the above equation to find y3y1.

    y3y1=2×(485nm)(109m1nm)(2.3m)(0.185mm)(103m1mm)=2×(485×109m)(2.3m)(0.185×106m)=1.2×102m

Therefore, the distance between the first and the third dark band of the interference pattern is 1.2×102m.

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Chapter 35 Solutions

Physics For Scientists And Engineers: Foundations And Connections, Extended Version With Modern Physics

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