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 68PQ

(a)

To determine

The distance from the slit to the viewing screen.

(a)

Expert Solution
Check Mark

Answer to Problem 68PQ

The distance from the slit to the viewing screen is 2.1m.

Explanation of Solution

Write the expression for the minima of single slit diffraction pattern to find sinθm.

  wsinθm=mλsinθm=mλw                          (I)

Here, w is width of the slit and m is an integer (m=±1,±2,±3,.........), θ is the angular position and λ is the wavelength of light.

Substitute m=1 for the first minima in above equation to find sinθ1.

  sinθ1=λw                                (II)

Write the equation for the width of the central maximum which is surrounded by the first minima.

  B=2y1=2xtanθ1                            (III)

Here, B is the width of central maximum and x is the distance from the slit to the screen.

For small angle of θ1, tanθ1sinθ1.

Substitute λw for tanθ1 in above equation to find x.

  B=2x(λw)x=Bw2λ                            (IV)

Conclusion:

Substitute 6.60mm for B, 0.314mm for w and 490.0nm for λ in equation (IV) to find x.

  x=(6.60mm×10-3m1mm)(0.314mm×10-3m1mm)2(490.0nm×10-9m1nm)=2.1m

Therefore, the distance from the slit to the viewing screen is 2.1m.

(b)

To determine

The width of the first side maximum on the viewing screen.

(b)

Expert Solution
Check Mark

Answer to Problem 68PQ

The width of the first side maximum on the viewing screen is 3.32mm.

Explanation of Solution

Substitute m=2 for the second minima in equation (I) to find sinθ2.

  sinθ2=2λw                                (V)

Write the equation for the distance from the center of the central maximum to the second minima.

  y2=xtanθ2                                  (VI)

For small angle of θ2, tanθ2sinθ2.

Substitute 2λw for tanθ2 in above equation to find y2.

  y2=2λxw                                     (VII)

Substitute m=1 for the second minima in equation (I) to find sinθ2.

  y1=λxw                                     (VIII)

Write Δy from Equations (VII) and (VIII).

Δy=y2y1=λxw

Conclusion:

Substitute 490.0nm for λ, 2.1m for x and 0.314mm for w in equation (VII) to find y2.

  Δy=(490.0nm×10-9m1nm)×2.1m(0.314mm×10-3m1mm)=3.32×103m×1mm10-3m=3.32mm

Therefore, the width of the first side maximum on the viewing screen is 3.32mm.

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Chapter 35 Solutions

EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC

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