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Diffraction gratings are designed with a variety of slit widths d. For a microscopic view and examples of real diffraction gratings, see the figures below. Diffraction Grating (Microscopic View) EDUCATIONAL DIFFRACTION GRATING 100 lines/mm 300 lines/mm 600 lines/mm d Suppose a diffraction grating contains 929 lines per millimeter. What is its slit width in nanometers? [Note: 1 meter = $10^3$ millimeters = $10^9$ nanometers] 

          Diffraction gratings are designed with a variety of slit widths d. For a microscopic view and examples of real diffraction gratings, see the figures below.
Diffraction Grating
(Microscopic View)
EDUCATIONAL DIFFRACTION GRATING
100
lines/mm
300
lines/mm
600
lines/mm
d
Suppose a diffraction grating contains 929 lines per millimeter. What is its slit width in nanometers?
[Note: 1 meter = $10^3$ millimeters = $10^9$ nanometers]
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Diffraction gratings are designed with a variety of slit widths d. For a microscopic view and examples of real diffraction gratings, see the figures below. Diffraction Grating (Microscopic View) EDUCATIONAL DIFFRACTION GRATING 100 lines/mm 300 lines/mm 600 lines/mm d Suppose a diffraction grating contains 929 lines per millimeter. What is its slit width in nanometers? [Note: 1 meter = 10^3 millimeters = 10^9 nanometers]

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University Physics with Modern Physics
Hugh D. Young 14th Edition
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Diffraction gratings are designed with a variety of slit widths, d. For a microscopic view and examples of real diffraction gratings, see the figures below. Diffraction Grating (Microscopic View) EDUCATIONAL DIFFRACTION GRATING 100 lines/mm 300 lines/mm 600 lines/mm Suppose a diffraction grating contains 929 lines per millimeter. What is its slit width in nanometers? [Note: 1 meter = 10^3 millimeters = 10^3 nanometers]
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Transcript

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00:01 Hi friends, here it is given diffraction grating having 100 lines per millimeter.
00:19 So width of the slate will be 1mm 1 .00 that is 10 to the power minus 5 meter.
00:31 Diffraction pattern is obtained on the screen, 198 centimeter from it.
00:37 And the distance of first order dark from central right fences 19 .8 centimeters.
00:44 We have to find the wavelength.
00:48 For an in order dark fringe, we have d sine of theta to be n into lambda...
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