Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 38, Problem 38.4P

A horizontal laser beam of wavelength 632.8 nm has a circular cross section 2.00 nun in diameter. A rectangular aperture is to lie placed in the center of the beam so that when the light falls perpendicularly on a wall 4.50 m away, the central maximum fills a rectangle 110 mm wide and 6.00 mm high. The dimensions are measured between the minima bracketing the central maximum. Find the required (a) width and (b) height of the aperture. (c) Is the longer dimension of the central bright patch in the diffraction pattern horizontal or vertical? (d) Is the longer dimension of the aperture horizontal or vertical? (e) Explain the relationship between these two rectangles, using a diagram.

(a)

Expert Solution
Check Mark
To determine

The width of the aperture.

Answer to Problem 38.4P

The width of the aperture is 51.77μm .

Explanation of Solution

Given info: The wavelength of the laser beam is 632.8nm for λ , the diameter of the beam is 2.00mm , the distance to the wall is 4.50m , the width of the rectangle is 110mm and the height of the rectangle is 6.00mm .

Write the expression for the destructive interference.

sinθ=mλa (1)

Here,

m is the order.

λ is the wavelength of the light.

a is the width of the aperture.

Write the expression for the distance of the minimum from the central maximum.

y=tanθ×L

Here,

θ is the angle with the horizontal.

L is the distance of the screen from the slit.

The tangent is approximately equal to the sine if the angle is very small.

Substitute sinθ for tanθ in above equation.

y=sinθ×LsinθyL (2)

Write the expression for the width of the central maximum.

w=2yy=w2

Here,

y is the distance of the first minimum from the central maximum.

Equate equation (1) and equation (2).

yL=mλa (3)

Substitute w2 for y in the above equation.

w2L=mλa

Substitute 4.50m for L , 632.8nm for λ , 1 for m and 110mm for w in the above equation.

110mm×1m103mm24.50m=1×632.8nm×1m109nmaa51.77×106m×1μm106m=51.77μm

Conclusion:

Therefore, the width of the aperture is 51.77μm .

(b)

Expert Solution
Check Mark
To determine

The height of the aperture.

Answer to Problem 38.4P

The height of the aperture is 949μm .

Explanation of Solution

Given info: The wavelength of the laser beam is 632.8nm for λ , the diameter of the beam is 2.00mm , the distance to the wall is 4.50m , the width of the rectangle is 110mm and the height of the rectangle is 6.00mm .

Write the expression for the height of the central maximum.

h=2yy=h2

Here,

h is the height of the rectangle.

y is the distance of the first minimum from the central maximum.

Substitute h2 for y in the equation (3).

h2L=mλa

Substitute 4.50m for L , 632.8nm for λ , 1 for m and 6.00mm for h in the above equation.

(6.00mm×1m1000mm)24.50m=1×632.8nm×1m109nmaa949×106m×1μm106m=949μm

Conclusion:

Therefore, the height of the aperture is 949μm .

(c)

Expert Solution
Check Mark
To determine

Whether the longer dimension of the central bright patch is horizontal or vertical.

Answer to Problem 38.4P

The longer dimension of the central bright patch is horizontal.

Explanation of Solution

Given info: The wavelength of the laser beam is 632.8nm for λ , the diameter of the beam is 2.00mm , the distance to the wall is 4.50m , the width of the rectangle is 110mm and the height of the rectangle is 6.00mm .

From the given information, the width of the rectangle in the central bright patch is 110mm and the height of the rectangle in the central bright patch is 6.00mm . The value of the height of the rectangle in the central bright patch is smaller than the width of the rectangle in the central bright patch. So, the longer dimension of the central bright patch is 110mm . The width lies horizontally and the height is vertical. Thus, the longer dimension of the aperture is horizontal.

Conclusion:

Therefore, the longer dimension of the central bright patch is horizontal.

(d)

Expert Solution
Check Mark
To determine

Whether the longer dimension of the aperture is horizontal or vertical.

Answer to Problem 38.4P

The longer dimension of the aperture is vertical.

Explanation of Solution

Given info: The wavelength of the laser beam is 632.8nm for λ , the diameter of the beam is 2.00mm , the distance to the wall is 4.50m , the width of the rectangle is 110mm and the height of the rectangle is 6.00mm .

From part (a), the width of the aperture is 51.77μm and from part (b), the height of the aperture is 949μm . The value of the height is larger than the width of the aperture. So, the longer dimension of the aperture is 949μm . The width lies horizontally and the height is vertical. Thus, the longer dimension of the aperture is vertical.

Conclusion:

Therefore, the longer dimension of the aperture is vertical

(e)

Expert Solution
Check Mark
To determine

The relationship between the two rectangles.

Answer to Problem 38.4P

The longer dimension is 18.33 times the smaller dimension of the aperture.

Explanation of Solution

Given info: The wavelength of the laser beam is 632.8nm for λ , the diameter of the beam is 2.00mm , the distance to the wall is 4.50m , the width of the rectangle is 110mm and the height of the rectangle is 6.00mm .

From part (a), the width of the aperture is 51.77μm and from part (b), the height of the aperture is 949μm . From the given information, the width of the rectangle in the central bright patch is 110mm and the height of the rectangle in the central bright patch is 6.00mm .

The smaller distance between aperture edges causes a wider diffraction angle.

Write the expression for the ratio of larger dimension to the smaller dimension of the aperture.

x=longerdimensionsmallerdimension

Substitute 949μm for longerdimension and 51.77μm for smallerdimension in the above equation.

x=949μm51.77μm18.33

Thus, the longer dimension is 18.33 times the smaller dimension of the aperture.

Conclusion:

Therefore, the longer dimension is 18.33 times the smaller dimension of the aperture.

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Chapter 38 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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