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Section summary

  • Young’s double slit experiment gave definitive proof of the wave character of light.
  • An interference pattern is obtained by the superposition of light from two slits.
  • There is constructive interference when d sin θ = (for m = 0, 1, 1, 2, 2, …) size 12{d`"sin"θ=mλ,`m=0,`1,` - 1,`2,` - 2,` dotslow } {} , where d size 12{d} {} is the distance between the slits, θ size 12{θ} {} is the angle relative to the incident direction, and m size 12{m} {} is the order of the interference.
  • There is destructive interference when d sin θ = m + 1 2 λ (for m = 0, 1, 1, 2, 2, …) size 12{d`"sin"θ= left (m+ { {1} over {2} } right )λ,`m=0,`1,` - 1,`2,` - 2,` dotslow } {} .

Conceptual questions

Young’s double slit experiment breaks a single light beam into two sources. Would the same pattern be obtained for two independent sources of light, such as the headlights of a distant car? Explain.

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Suppose you use the same double slit to perform Young’s double slit experiment in air and then repeat the experiment in water. Do the angles to the same parts of the interference pattern get larger or smaller? Does the color of the light change? Explain.

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Is it possible to create a situation in which there is only destructive interference? Explain.

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[link] shows the central part of the interference pattern for a pure wavelength of red light projected onto a double slit. The pattern is actually a combination of single slit and double slit interference. Note that the bright spots are evenly spaced. Is this a double slit or single slit characteristic? Note that some of the bright spots are dim on either side of the center. Is this a single slit or double slit characteristic? Which is smaller, the slit width or the separation between slits? Explain your responses.

The figure shows a photo of a horizontal line of equally spaced red dots of light on a black background. The central dot is the brightest and the dots on either side of center are dimmer. The dot intensity decreases to almost zero after moving six dots to the left or right of center. If you continue to move away from the center, the dot brightness increases slightly, although it does not reach the brightness of the central dot. After moving another six dots, or twelve dots in all, to the left or right of center, there is another nearly invisible dot. If you move even farther from the center, the dot intensity again increases, but it does not reach the level of the previous local maximum. At eighteen dots from the center, there is another nearly invisible dot.
This double slit interference pattern also shows signs of single slit interference. (credit: PASCO)
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Problems&Exercises

At what angle is the first-order maximum for 450-nm wavelength blue light falling on double slits separated by 0.0500 mm?

0 . 516º size 12{0 "." "516"°} {}

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Calculate the angle for the third-order maximum of 580-nm wavelength yellow light falling on double slits separated by 0.100 mm.

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What is the separation between two slits for which 610-nm orange light has its first maximum at an angle of 30 . size 12{"30" "." 0°} {} ?

1 . 22 × 10 6 m size 12{1 "." "22" times "10" rSup { size 8{ - 6} } `m} {}

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Find the distance between two slits that produces the first minimum for 410-nm violet light at an angle of 45 . size 12{"45" "." 0°} {} .

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Calculate the wavelength of light that has its third minimum at an angle of 30 . size 12{"30" "." 0°} {} when falling on double slits separated by 3 . 00 μm size 12{3 "." "00"`"μm"} {} . Explicitly, show how you follow the steps in Problem-Solving Strategies for Wave Optics .

600 nm

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What is the wavelength of light falling on double slits separated by 2 . 00 μm size 12{2 "." "00"`"μm"} {} if the third-order maximum is at an angle of 60 . size 12{"60" "." 0°} {} ?

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At what angle is the fourth-order maximum for the situation in [link] ?

2 . 06º size 12{2 "." "06"°} {}

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What is the highest-order maximum for 400-nm light falling on double slits separated by 25 . 0 μm size 12{"25" "." 0`"μm"} {} ?

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Find the largest wavelength of light falling on double slits separated by 1 . 20 μm size 12{1 "." "20"`"μm"} {} for which there is a first-order maximum. Is this in the visible part of the spectrum?

1200 nm (not visible)

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What is the smallest separation between two slits that will produce a second-order maximum for 720-nm red light?

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(a) What is the smallest separation between two slits that will produce a second-order maximum for any visible light? (b) For all visible light?

(a) 760 nm

(b) 1520 nm

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(a) If the first-order maximum for pure-wavelength light falling on a double slit is at an angle of 10 . size 12{"10" "." 0°} {} , at what angle is the second-order maximum? (b) What is the angle of the first minimum? (c) What is the highest-order maximum possible here?

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[link] shows a double slit located a distance x size 12{x} {} from a screen, with the distance from the center of the screen given by y size 12{y} {} . When the distance d size 12{d} {} between the slits is relatively large, there will be numerous bright spots, called fringes. Show that, for small angles (where sin θ θ size 12{"sin"θ approx θ} {} , with θ size 12{θ} {} in radians), the distance between fringes is given by Δ y = / d size 12{Δy=xλ/d} {} .

The figure shows a schematic of a double slit experiment. A double slit is at the left and a screen is at the right. The slits are separated by a distance d. From the midpoint between the slits, a horizontal line labeled x extends to the screen. From the same point, a line angled upward at an angle theta above the horizontal also extends to the screen. The distance between where the horizontal line hits the screen and where the angled line hits the screen is marked y, and the distance between adjacent fringes is given by delta y, which equals x times lambda over d.
The distance between adjacent fringes is Δ y = / d size 12{Δy=xλ/d} {} , assuming the slit separation d size 12{d} {} is large compared with λ size 12{λ} {} .

For small angles sin θ tan θ θ ( in radians ) size 12{"sin"θ - "tan"θ approx θ` \( "in"`"radians" \) } {} .

For two adjacent fringes we have,

d sin θ m = size 12{d`"sin"θ rSub { size 8{m} } =mλ} {}

and

d sin θ m + 1 = m + 1 λ size 12{d`"sin "θ rSub { size 8{m+1} } = left (m+1 right )λ} {}

Subtracting these equations gives

d sin θ m + 1 sin θ m = m + 1 m λ d θ m + 1 θ m = λ tan θ m = y m x θ m d y m + 1 x y m x = λ d Δ y x = λ Δ y = d alignl { stack { size 12{d left ("sin"θ rSub { size 8{m+1} } - " sin"θ rSub { size 8{m} } right )= left [ left (m+1 right ) - m right ]λ} {} # d left (θ rSub { size 8{m+1} } - θ"" lSub { size 8{m} } right )=λ {} #"tan"θ rSub { size 8{m} } = { {y rSub { size 8{m} } } over {x} } approx θ"" lSub { size 8{m} } drarrow d left ( { {y rSub { size 8{m+1} } } over {x} } - { {y rSub { size 8{m} } } over {x} } right )=λ {} # d { {Δy} over {x} } =λ drarrow {underline {Δy= { {xλ} over {d} } }} {}} } {}
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Using the result of the problem above, calculate the distance between fringes for 633-nm light falling on double slits separated by 0.0800 mm, located 3.00 m from a screen as in [link] .

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Using the result of the problem two problems prior, find the wavelength of light that produces fringes 7.50 mm apart on a screen 2.00 m from double slits separated by 0.120 mm (see [link] ).

450 nm

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Practice Key Terms 5

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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