Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term
Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term
9th Edition
ISBN: 9781305932302
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 38, Problem 4P

(a)

To determine

The width of aperture.

(a)

Expert Solution
Check Mark

Answer to Problem 4P

The width of aperture is 51.8μm.

Explanation of Solution

On looking at the figure P38.4, width of the rectangular patch is more than that of its height.

Write the equation for tangent of angular width of aperture.

    tanθwidth=ywidthL

Here, θwidth is the angular width of aperture, ywidth is half the height of rectangle, and L is the separation between the aperture and wall.

Since θwidth is very small, the following approximation can be used.

    tanθwidthsinθwidth

Write the relation between width of aperture and the wavelength of light used for first order diffraction pattern.

    awidthsinθwidth=(1)λ

Here, awidth is the width of aperture and λ is the wavelength of light used.

Conclusion:

Substitute 110mm2 for ywidth and 4.50m for L in the equation for tanθwidth.

    tanθwidth=(110mm1m103mm2)4.50m=0.0122

Substitute 632.8nm for λ and 0.0122 for sinθwidth in the equation for λ.

    awidth(0.0122)=(1)(632.8nm(109m1nm))

Rewrite the above expression in terms of awidth.

    awidth=632.8×109m0.0122=51,868×109m(1μm106m)=51.8μm

Therefore, the width of aperture is 51.8μm.

(b)

To determine

The height of aperture.

(b)

Expert Solution
Check Mark

Answer to Problem 4P

The height of aperture is 949μm.

Explanation of Solution

On looking at the figure P38.4, width of the rectangular patch is more than that of its height.

Write the equation for tangent of angular width of aperture.

    tanθheight=yheightL

Here, θheight is the angular height of aperture, yheight is the half the height of rectangle, and L is the separation between the aperture and wall.

Since θwidth is very small, the following approximation can be used.

    tanθheightsinθheight

Write the relation between width of aperture and the wavelength of light used for first order diffraction pattern.

    aheightsinθheight=(1)λ

Here, aheight is the height of aperture.

Conclusion:

Substitute 6.00mm2 for ywidth and 4.50m for L in the equation for tanθheight.

    tanθwidth=(6.000mm(1m103mm)2)4.50m=0.000667

Substitute 632.8nm for λ and 0.000667 for sinθheight in the equation for λ.

    awidth(0.000667)=(1)(632.8nm(109m1nm))

Rewrite the above expression in terms of aheight.

    awidth=632.8×109m0.000667=9.49×104m(1μm106m)=949μm

Therefore, the height of aperture is 949μm.

(c)

To determine

Check whether the horizontal or vertical dimension of central bright portion is greater.

(c)

Expert Solution
Check Mark

Answer to Problem 4P

Horizontal dimension of central bright portion is longer than its vertical dimension.

Explanation of Solution

Draw the diagram showing the diffraction pattern on light passing through a circular aperture.

Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term, Chapter 38, Problem 4P

From the diagram, it can be seen that the central bright patch has an ellipse shape. It has greater length in horizontal direction than in vertical direction.

Therefore, the horizontal dimension of central bright portion is longer than its vertical dimension.

(d)

To determine

Check whether the horizontal or vertical dimension of aperture is greater.

(d)

Expert Solution
Check Mark

Answer to Problem 4P

Vertical dimension of aperture is greater.

Explanation of Solution

Refer the diagram shown in part (c). From the diagram, it is understood that to obtain diffraction pattern with greater horizontal dimension its vertical length, the vertical length of aperture must be greater than that of horizontal length. If the horizontal dimension of aperture is greater, the vertical dimension of bright becomes greater than that of the horizontal dimension.

Therefore, the vertical dimension of aperture is greater.

(e)

To determine

Identify the relation between the two rectangles given in question with the help of a diagram.

(e)

Expert Solution
Check Mark

Answer to Problem 4P

The distances between edges of rectangular aperture is inversely proportional to size of central maxima rectangle on the wall.

Explanation of Solution

Refer the figure 1shown in part (c). The size of aperture is inversely proportional to the size of diffraction pattern. Smaller the size of aperture, larger will be the size of diffraction pattern. It is found that the width of aperture is 51.8μm and the height is 949μm. It means that the larger dimension of aperture is around 18.3 times longer than its smaller dimension.

Therefore, the distances between edges of rectangular aperture is inversely proportional to size of central maxima rectangle on the wall.

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Chapter 38 Solutions

Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term

Ch. 38 - Prob. 4OQCh. 38 - Prob. 5OQCh. 38 - Prob. 6OQCh. 38 - Prob. 7OQCh. 38 - Prob. 8OQCh. 38 - Prob. 9OQCh. 38 - Prob. 10OQCh. 38 - Prob. 11OQCh. 38 - Prob. 12OQCh. 38 - Prob. 1CQCh. 38 - Prob. 2CQCh. 38 - Prob. 3CQCh. 38 - Prob. 4CQCh. 38 - Prob. 5CQCh. 38 - Prob. 6CQCh. 38 - Prob. 7CQCh. 38 - Prob. 8CQCh. 38 - Prob. 9CQCh. 38 - Prob. 10CQCh. 38 - Prob. 11CQCh. 38 - Prob. 12CQCh. 38 - Prob. 1PCh. 38 - Prob. 2PCh. 38 - Prob. 3PCh. 38 - Prob. 4PCh. 38 - Prob. 5PCh. 38 - Prob. 6PCh. 38 - Prob. 7PCh. 38 - Prob. 8PCh. 38 - Prob. 9PCh. 38 - Prob. 10PCh. 38 - Prob. 11PCh. 38 - Coherent light of wavelength 501.5 nm is sent...Ch. 38 - Prob. 13PCh. 38 - Prob. 14PCh. 38 - Prob. 15PCh. 38 - Prob. 16PCh. 38 - Prob. 17PCh. 38 - Prob. 18PCh. 38 - What is the approximate size of the smallest...Ch. 38 - Prob. 20PCh. 38 - Prob. 21PCh. 38 - Prob. 22PCh. 38 - Prob. 23PCh. 38 - Prob. 24PCh. 38 - Prob. 25PCh. 38 - Prob. 26PCh. 38 - Consider an array of parallel wires with uniform...Ch. 38 - Prob. 28PCh. 38 - Prob. 29PCh. 38 - A grating with 250 grooves/mm is used with an...Ch. 38 - Prob. 31PCh. 38 - Prob. 32PCh. 38 - Light from an argon laser strikes a diffraction...Ch. 38 - Show that whenever white light is passed through a...Ch. 38 - Prob. 35PCh. 38 - Prob. 36PCh. 38 - Prob. 37PCh. 38 - Prob. 38PCh. 38 - Prob. 39PCh. 38 - Prob. 40PCh. 38 - Prob. 41PCh. 38 - Prob. 42PCh. 38 - Prob. 43PCh. 38 - Prob. 44PCh. 38 - Prob. 45PCh. 38 - Prob. 46PCh. 38 - Prob. 47PCh. 38 - Prob. 48PCh. 38 - Prob. 49PCh. 38 - Prob. 50PCh. 38 - Prob. 51PCh. 38 - Prob. 52PCh. 38 - Prob. 53APCh. 38 - Prob. 54APCh. 38 - Prob. 55APCh. 38 - Prob. 56APCh. 38 - Prob. 57APCh. 38 - Prob. 58APCh. 38 - Prob. 59APCh. 38 - Prob. 60APCh. 38 - Prob. 61APCh. 38 - Prob. 62APCh. 38 - Prob. 63APCh. 38 - Prob. 64APCh. 38 - Prob. 65APCh. 38 - Prob. 66APCh. 38 - Prob. 67APCh. 38 - Prob. 68APCh. 38 - Prob. 69APCh. 38 - Prob. 70APCh. 38 - Prob. 71APCh. 38 - Prob. 72APCh. 38 - Prob. 73APCh. 38 - Light of wavelength 632.8 nm illuminates a single...Ch. 38 - Prob. 75CPCh. 38 - Prob. 76CPCh. 38 - Prob. 77CPCh. 38 - Prob. 78CPCh. 38 - Prob. 79CP
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