# 4.4 Diffraction gratings  (Page 3/4)

 Page 3 / 4

## Significance

The large distance between the red and violet ends of the rainbow produced from the white light indicates the potential this diffraction grating has as a spectroscopic tool. The more it can spread out the wavelengths (greater dispersion), the more detail can be seen in a spectrum. This depends on the quality of the diffraction grating—it must be very precisely made in addition to having closely spaced lines.

Check Your Understanding If the line spacing of a diffraction grating d is not precisely known, we can use a light source with a well-determined wavelength to measure it. Suppose the first-order constructive fringe of the ${\text{H}}_{\beta }$ emission line of hydrogen $\left(\text{λ}=656.3\phantom{\rule{0.2em}{0ex}}\text{nm}\right)$ is measured at $11.36\text{°}$ using a spectrometer with a diffraction grating. What is the line spacing of this grating?

$3.332\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{m}$ or 300 lines per millimeter

Take the same simulation we used for double-slit diffraction and try increasing the number of slits from $N=2$ to $N=3,4,5...$ . The primary peaks become sharper, and the secondary peaks become less and less pronounced. By the time you reach the maximum number of $N=20$ , the system is behaving much like a diffraction grating.

## Summary

• A diffraction grating consists of a large number of evenly spaced parallel slits that produce an interference pattern similar to but sharper than that of a double slit.
• Constructive interference occurs when $d\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta =m\lambda \phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}m=0,\phantom{\rule{0.2em}{0ex}}±1,\phantom{\rule{0.2em}{0ex}}±2,\phantom{\rule{0.2em}{0ex}}...,$ where d is the distance between the slits, $\theta$ is the angle relative to the incident direction, and m is the order of the interference.

## Problems

A diffraction grating has 2000 lines per centimeter. At what angle will the first-order maximum be for 520-nm-wavelength green light?

$5.97\text{°}$

Find the angle for the third-order maximum for 580-nm-wavelength yellow light falling on a difraction grating having 1500 lines per centimeter.

How many lines per centimeter are there on a diffraction grating that gives a first-order maximum for 470-nm blue light at an angle of $25.0\text{°}$ ?

$8.99\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}$

What is the distance between lines on a diffraction grating that produces a second-order maximum for 760-nm red light at an angle of $60.0\text{°}$ ?

Calculate the wavelength of light that has its second-order maximum at $45.0\text{°}$ when falling on a diffraction grating that has 5000 lines per centimeter.

707 nm

An electric current through hydrogen gas produces several distinct wavelengths of visible light. What are the wavelengths of the hydrogen spectrum, if they form first-order maxima at angles $24.2\text{°},\phantom{\rule{0.2em}{0ex}}25.7\text{°},\phantom{\rule{0.2em}{0ex}}29.1\text{°},$ and $41.0\text{°}$ when projected on a diffraction grating having 10,000 lines per centimeter?

(a) What do the four angles in the preceding problem become if a 5000-line per centimeter diffraction grating is used? (b) Using this grating, what would the angles be for the second-order maxima? (c) Discuss the relationship between integral reductions in lines per centimeter and the new angles of various order maxima.

a. $11.8\text{°}$ , $12.5\text{°}$ , $14.1\text{°}$ , $19.2\text{°}$ ; b. $24.2\text{°}$ , $25.7\text{°}$ , $29.1\text{°}$ , $41.0\text{°}$ ; c. Decreasing the number of lines per centimeter by a factor of x means that the angle for the x -order maximum is the same as the original angle for the first-order maximum.

What is the spacing between structures in a feather that acts as a reflection grating, giving that they produce a first-order maximum for 525-nm light at a $30.0\text{°}$ angle?

An opal such as that shown in [link] acts like a reflection grating with rows separated by about $8\phantom{\rule{0.2em}{0ex}}\text{μm}.$ If the opal is illuminated normally, (a) at what angle will red light be seen and (b) at what angle will blue light be seen?

a. using $\lambda =700\phantom{\rule{0.2em}{0ex}}\text{nm,}\phantom{\rule{0.2em}{0ex}}\theta =5\text{.0}\text{°};$ b. using $\lambda =460\phantom{\rule{0.2em}{0ex}}\text{nm,}\phantom{\rule{0.2em}{0ex}}\theta =3\text{.3}\text{°}$

At what angle does a diffraction grating produce a second-order maximum for light having a first-order maximum at $20.0\text{°}$ ?

(a) Find the maximum number of lines per centimeter a diffraction grating can have and produce a maximum for the smallest wavelength of visible light. (b) Would such a grating be useful for ultraviolet spectra? (c) For infrared spectra?

a. 26,300 lines/cm; b. yes; c. no

(a) Show that a 30,000 line per centimeter grating will not produce a maximum for visible light. (b) What is the longest wavelength for which it does produce a first-order maximum? (c) What is the greatest number of line per centimeter a diffraction grating can have and produce a complete second-order spectrum for visible light?

The analysis shown below also applies to diffraction gratings with lines separated by a distance d . What is the distance between fringes produced by a diffraction grating having 125 lines per centimeter for 600-nm light, if the screen is 1.50 m away? ( Hin t : The distance between adjacent fringes is $\text{Δ}y=x\lambda \text{/}d,$ assuming the slit separation d is comparable to $\text{λ}.$ ) $1.13\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-2}\phantom{\rule{0.2em}{0ex}}\text{m}$

#### Questions & Answers

A round diaphragm S with diameter of d = 0.05 is used as light source in Michelson interferometer shown on the picture. The diaphragm is illuminated by parallel beam of monochromatic light with wavelength of λ = 0.6 μm. The distances are A B = 30, A C = 10 . The interference picture is in the form of concentric circles and is observed on the screen placed in the focal plane of the lens. Estimate the number of interference rings m observed near the main diffractive maximum.
Jyoti Reply
A Pb wire wound in a tight solenoid of diameter of 4.0 mm is cooled to a temperature of 5.0 K. The wire is connected in series with a 50-Ωresistor and a variable source of emf. As the emf is increased, what value does it have when the superconductivity of the wire is destroyed?
Rupal Reply
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in the wave equation y=Asin(kx-wt+¢) what does k and w stand for.
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derivation of lateral shieft
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hi
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total binding energy of ionic crystal at equilibrium is
All Reply
How does, ray of light coming form focus, behaves in concave mirror after refraction?
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Refraction does not occur in concave mirror. If refraction occurs then I don't know about this.
Sushant
What is motion
Izevbogie Reply
Anything which changes itself with respect to time or surrounding
Sushant
good
Chemist
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Chemist
No
Sushant
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Chemist
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Chemist
Not so more
Sushant
Radioactive substance
DHEERAJ
These substance create harmful radiation like alpha particle radiation, beta particle radiation, gamma particle radiation
Sushant
But ask anything changes itself with respect to time or surrounding A Not any harmful radiation
DHEERAJ
explain cavendish experiment to determine the value of gravitational concept.
Celine Reply
Cavendish Experiment to Measure Gravitational Constant. ... This experiment used a torsion balance device to attract lead balls together, measuring the torque on a wire and equating it to the gravitational force between the balls. Then by a complex derivation, the value of G was determined.
Triio
For the question about the scuba instructor's head above the pool, how did you arrive at this answer? What is the process?
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as a free falling object increases speed what is happening to the acceleration
Success Reply
of course g is constant
Alwielland
acceleration also inc
Usman
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jay
normal distributiin of errors report
Dennis
normal distribution of errors
Dennis
acceleration also increases
Jay
there are two correct answers depending on whether air resistance is considered. none of those answers have acceleration increasing.
Michael
Acceleration is the change in velocity over time, hence it's the derivative of the velocity with respect to time. So this case would depend on the velocity. More specifically the change in velocity in the system.
Big
photo electrons doesn't emmit when electrons are free to move on surface of metal why?
Rafi Reply
What would be the minimum work function of a metal have to be for visible light(400-700)nm to ejected photoelectrons?
Mohammed Reply
give any fix value to wave length
Rafi
40 cm into change mm
Arhaan Reply
40cm=40.0×10^-2m =400.0×10^-3m =400mm. that cap(^) I have used above is to the power.
Prema
i.e. 10to the power -2 in the first line and 10 to the power -3 in the the second line.
Prema
there is mistake in my first msg correction is 40cm=40.0×10^-2m =400.0×10^-3m =400mm. sorry for the mistake friends.
Prema
40cm=40.0×10^-2m =400.0×10^-3m =400mm.
Prema
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Prema
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