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Figure a is a photograph of an opal pendant reflecting various colours. Figure b is the photograph of a butterfly.
(a) This Australian opal and (b) butterfly wings have rows of reflectors that act like reflection gratings, reflecting different colors at different angles. (credit b: modification of work by “whologwhy”/Flickr)

Applications of diffraction gratings

Where are diffraction gratings used in applications? Diffraction gratings are commonly used for spectroscopic dispersion and analysis of light. What makes them particularly useful is the fact that they form a sharper pattern than double slits do. That is, their bright fringes are narrower and brighter while their dark regions are darker. Diffraction gratings are key components of monochromators used, for example, in optical imaging of particular wavelengths from biological or medical samples. A diffraction grating can be chosen to specifically analyze a wavelength emitted by molecules in diseased cells in a biopsy sample or to help excite strategic molecules in the sample with a selected wavelength of light. Another vital use is in optical fiber technologies where fibers are designed to provide optimum performance at specific wavelengths. A range of diffraction gratings are available for selecting wavelengths for such use.

Calculating typical diffraction grating effects

Diffraction gratings with 10,000 lines per centimeter are readily available. Suppose you have one, and you send a beam of white light through it to a screen 2.00 m away. (a) Find the angles for the first-order diffraction of the shortest and longest wavelengths of visible light (380 and 760 nm, respectively). (b) What is the distance between the ends of the rainbow of visible light produced on the screen for first-order interference? (See [link] .)

A vertical line on the left is labeled grating and one on the right is labeled screen. They are a distance x equal to 2 meters apart. Four arrows radiate from the grating to the screen. The first and second from the top make angles theta R and theta V respectively with the central axis. The points where they fall on the screen are at distances yR and yV respectively from the central axis. Rainbows are formed on the screen between the first and second arrow and between the third and fourth arrow.
(a) The diffraction grating considered in this example produces a rainbow of colors on a screen a distance x = 2.00 m from the grating. The distances along the screen are measured perpendicular to the x -direction. In other words, the rainbow pattern extends out of the page.
(b) In a bird’s-eye view, the rainbow pattern can be seen on a table where the equipment is placed.

Strategy

Once a value for the diffraction grating’s slit spacing d has been determined, the angles for the sharp lines can be found using the equation

d sin θ = m λ for m = 0 , ± 1 , ± 2 , ... .

Since there are 10,000 lines per centimeter, each line is separated by 1/10,000 of a centimeter. Once we know the angles, we an find the distances along the screen by using simple trigonometry.

Solution

  1. The distance between slits is d = ( 1 cm ) / 10 , 000 = 1.00 × 10 −4 cm or 1.00 × 10 −6 m . Let us call the two angles θ V for violet (380 nm) and θ R for red (760 nm). Solving the equation d sin θ V = m λ for sin θ V ,
    sin θ V = m λ V d ,

    where m = 1 for the first-order and λ V = 380 nm = 3.80 × 10 −7 m . Substituting these values gives
    sin θ V = 3.80 × 10 −7 m 1.00 × 10 −6 m = 0.380 .

    Thus the angle θ V is
    θ V = sin −1 0.380 = 22.33 ° .

    Similarly,
    sin θ R = 7.60 × 10 −7 m 1.00 × 10 −6 m = 0.760 .

    Thus the angle θ R is
    θ R = sin −1 0.760 = 49.46 ° .

    Notice that in both equations, we reported the results of these intermediate calculations to four significant figures to use with the calculation in part (b).
  2. The distances on the secreen are labeled y V and y R in [link] . Notice that tan θ = y / x . We can solve for y V and y R . That is,
    y V = x tan θ V = ( 2.00 m ) ( tan 22.33 ° ) = 0.815 m

    and
    y R = x tan θ R = ( 2.00 m ) ( tan 49.46 ° ) = 2.338 m .

    The distance between them is therefore
    y R y V = 1.523 m .

Questions & Answers

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hi
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hii
Alvin
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what is the value of speed of light
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1.79×10_¹⁹ km per hour
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কাজের একক কী
Jasim
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Jasim
friction ka direction Kaise pata karte hai
Rahul Reply
friction is always in the opposite of the direction of moving object
Punia
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b) symmetric of time only
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Alvin
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mallam Reply
according to de Broglie any matter particles by attaining the higher velocity as compared to light'ill show the wave nature and equation of wave will applicable on it but in practical life people see it is impossible however it is practicaly true and possible while looking at the earth matter at far
Manikant
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Manikant
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Jerry
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Practice Key Terms 1

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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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