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The graph shows the variation of intensity as a function of sine theta. The curve has a strong peak at sine theta equals zero, then has small oscillations spreading symmetrically to the left and right of this central peak. The oscillations all appear to be of the same height. Between each oscillation, the curve appears to go to zero, and each zero is labeled. The first zero to the left of the main peak is labeled minus lambda over d and the first zero to the right is labeled lambda over d. The second zero to the left is labeled minus two lambda over d and the second zero to the right is labeled two lambda over d. The third zero to the left is labeled minus three lambda over d and the third zero to the right is labeled three lambda over d.
A graph of single slit diffraction intensity showing the central maximum to be wider and much more intense than those to the sides. In fact the central maximum is six times higher than shown here.

Thus, to obtain destructive interference for a single slit    ,

D sin θ = , for m = 1, –1, 2, –2, 3, (destructive), size 12{D`"sin"θ= ital "mλ",~m="1,"`"2,"`"3,"` dotslow } {}

where D size 12{D} {} is the slit width, λ size 12{λ} {} is the light’s wavelength, θ size 12{θ} {} is the angle relative to the original direction of the light, and m size 12{m} {} is the order of the minimum. [link] shows a graph of intensity for single slit interference, and it is apparent that the maxima on either side of the central maximum are much less intense and not as wide. This is consistent with the illustration in [link] (b).

Calculating single slit diffraction

Visible light of wavelength 550 nm falls on a single slit and produces its second diffraction minimum at an angle of 45.0º size 12{"45" "." 0°} {} relative to the incident direction of the light. (a) What is the width of the slit? (b) At what angle is the first minimum produced?

The schematic shows a single slit to the left and the resulting intensity pattern on a screen is graphed on the right. The single slit is represented by a gap of size d in a vertical line. A ray of wavelength lambda enters the gap from the left, then five rays leave from the gap center and head to the right. One ray continues on the horizontal centerline of the schematic. Two rays angle upward: the first at an unknown angle theta one above the horizontal and the second at an angle theta two equals forty five degrees above the horizontal. The final two rays angle downward at the same angles, so that they are symmetric about the horizontal with respect to the two rays that angle upward. The intensity on the screen is a maximum where the central ray hits the screen, whereas it is a minimum where the angled rays hit the screen.
A graph of the single slit diffraction pattern is analyzed in this example.

Strategy

From the given information, and assuming the screen is far away from the slit, we can use the equation D sin θ = size 12{D`"sin"θ= ital "mλ"} {} first to find D size 12{D} {} , and again to find the angle for the first minimum θ 1 size 12{θ rSub { size 8{1} } } {} .

Solution for (a)

We are given that λ = 550 nm size 12{λ="500"`"nm"} {} , m = 2 size 12{m=2} {} , and θ 2 = 45.0º . Solving the equation D sin θ = size 12{D`"sin"θ= ital "mλ"} {} for D size 12{D} {} and substituting known values gives

D = sin θ 2 = 2 ( 550 nm ) sin 45.0º = 1100 × 10 9 0.707 = 1.56 × 10 6 .

Solution for (b)

Solving the equation D sin θ = size 12{D`"sin"θ= ital "mλ"} {} for sin θ 1 size 12{"sin"θ rSub { size 8{1} } } {} and substituting the known values gives

sin θ 1 = D = 1 550 × 10 9 m 1 . 56 × 10 6 m . size 12{"sin"θ rSub { size 8{1} } = { {mλ} over {D} } = { {1 left ("550" times "10" rSup { size 8{ - 9} } m right )} over {1 "." "56" times "10" rSup { size 8{ - 6} } m} } } {}

Thus the angle θ 1 size 12{θ rSub { size 8{1} } } {} is

θ 1 = sin 1 0.354 = 20.7º. size 12{θ rSub { size 8{1} } ="sin" rSup { size 8{ - 1} } 0 "." "354"="20" "." 7°} {}

Discussion

We see that the slit is narrow (it is only a few times greater than the wavelength of light). This is consistent with the fact that light must interact with an object comparable in size to its wavelength in order to exhibit significant wave effects such as this single slit diffraction pattern. We also see that the central maximum extends 20.7º on either side of the original beam, for a width of about 41º . The angle between the first and second minima is only about 24º ( 45.0º 20.7º ) size 12{"24"°` \( "45" "." 0° - "20" "." 7° \) } {} . Thus the second maximum is only about half as wide as the central maximum.

Section summary

  • A single slit produces an interference pattern characterized by a broad central maximum with narrower and dimmer maxima to the sides.
  • There is destructive interference for a single slit when D sin θ = , (for m = 1, –1, 2, –2, 3, …) size 12{D`"sin"θ= ital "mλ",~m="1,"`"2,"`"3,"` dotslow } {} , where D is the slit width, λ is the light’s wavelength, θ is the angle relative to the original direction of the light, and m is the order of the minimum. Note that there is no m = 0 size 12{m=0} {} minimum.

Conceptual questions

As the width of the slit producing a single-slit diffraction pattern is reduced, how will the diffraction pattern produced change?

Problems&Exercises

(a) At what angle is the first minimum for 550-nm light falling on a single slit of width 1 . 00 μm size 12{1 "." "00"`"μm"} {} ? (b) Will there be a second minimum?

(a) 33 . size 12{"33" "." 4°} {}

(b) No

(a) Calculate the angle at which a 2 . 00 -μm size 12{2 "." "00""-μm"} {} -wide slit produces its first minimum for 410-nm violet light. (b) Where is the first minimum for 700-nm red light?

Questions & Answers

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ninjadapaul
20/(×-6^2)
Salomon
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ninjadapaul
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ninjadapaul
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ninjadapaul
Commplementary angles
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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Differences Between Laspeyres and Paasche Indices
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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Source:  OpenStax, Yupparaj english program physics corresponding to thai physics book #3. OpenStax CNX. May 19, 2014 Download for free at http://legacy.cnx.org/content/col11657/1.1
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