EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC
EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC
1st Edition
ISBN: 9781337684668
Author: Katz
Publisher: VST
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Chapter 35, Problem 83PQ
To determine

The distance between the first two adjacent dark fringes on the screen.

Expert Solution & Answer
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Answer to Problem 83PQ

The distance between the first two adjacent dark fringes on the screen is 2.80×102m.

Explanation of Solution

Write the expression for the phase difference for the dark fringes for wavelength λ1.

    dsinθ=(2n1+1)λ12                                                                                     (I)

Here, θ is the angle, d is the slit width, λ1 is the first monochromatic wavelength and n1 is an integer.

Write the expression for the phase difference for the dark fringes for wavelength λ2.

    dsinθ=(2n2+1)λ22                                                                                  (II)

Here, λ2 is the second monochromatic wavelength and n2 is an integer.

Divide equation (I) by (II)

dsinθdsinθ=(2n1+1)λ12(2n2+1)λ22(2n1+1)λ12=(2n2+1)λ22(2n1+1)λ1=(2n2+1)λ2(2n1+1)(2n2+1)=λ2λ1 (III)

Write the expression to calculate the value of n1.

    n1=(2n1+1)(n1)                                                                                 (IV)

Here, n1 is an integer.

Write the expression to calculate the value of n1.

    n2=(2n2+1)(n2)                                                                               (V)

Write the expression for the distance between the two adjacent dark fringes on the screen

    Δs=λ1D2d((2n1+1)(2n1+1))                                                                (VI)

Here,D is the distance between the slit and the screen.

Conclusion:

Substitute 560.0nm for λ2 and 400.0nm for λ1 in equation (III) to calculate the value of (2n1+1)(2n2+1)

    (2n1+1)(2n2+1)=560.0nm400.0nm(2n1+1)(2n2+1)=75                                                                              (VII)

Solve equation (VII) for the values of n1 and n2.

    2n1+1=7n1=(712)n1=3

Similarly,

    2n2+1=5n2=(51)2n2=2

Substitute 3 for n1 in equation (V) to calculate the value of n1.

    n1=((2×3)+1)(3)=10

Substitute 2 for n1 in equation (VI) to calculate the value of n2.

    n2=((2×2)+1)(2)=7

Substitute 1.00m for D, 0.100mm for d, 400.0nm for λ1, 10 for n1 and 3 for n1 in equation (VI) to calculate the value of Δs

    Δs=(400.0nm)(109m1nm)(1.00m)(2)(0.100mm×103m1mm)(((2×10)+1)((2×3)+1))=(14.0)(2.00×103m)=2.80×102m

Therefore, the distance between the first two adjacent dark fringes on the screen is 2.80×102m

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Chapter 35 Solutions

EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC

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