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When laser light of wavelength \(632.8 \mathrm{nm}\) passes through a diffraction grating, the first bright spots occur at \(\pm 17.8^{\circ}\) from the central maximum. (a) What is the line density (in lines/cm) of this grating? (b) How many additional bright spots are there beyond the first bright spots, and at what angles do they occur?

Short Answer

Expert verified
The line density of the grating is found by calculating the reciprocal of the grating separation, which is derived from given conditions. Then, the number of additional bright spots and their angles are calculated using the known relationship of m, λ and d.

Step by step solution

01

Determine the grating separation

We are given the wavelength \(\lambda = 632.8 \mathrm{nm} = 632.8 \times 10^{-9} \mathrm{m}\) (note the conversion from nanometers to meters). The angle of the first-order maximum is \(\theta = 17.8^\circ\). We use these values in the grating equation to solve for d: \(m \lambda = d \sin \theta\) becomes \(1 \times 632.8 \times 10^{-9} \mathrm{m} = d \times \sin 17.8^\circ\). Solving this equation gives us the separation between the grating lines, d.
02

Calculate the line density

The line density is the reciprocal of the grating separation d. So once we have found d, we just need to calculate \(1/d\), and then convert to lines/cm (since 1 m = 100 cm).
03

Determine the number of additional bright spots

For there to be another bright spot, there must be an integer m such that \(\sin \theta = m \lambda / d ≤ 1\). We already have the value of d from Step 1, and we know \(\lambda\), so we can compute \(m = d \sin \theta / \lambda\). As m is an integer, we round down to the nearest integer to find the maximum possible m.
04

Determine the angles of the additional bright spots

Once we have our maximum m, we can compute the angles of the bright spots. Each additional bright spot (beyond the first) corresponds to a different m value. We can calculate the angle for each m with the grating equation, rearranging it to \(\theta = \arcsin(m \lambda / d)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Line Density
The line density of a diffraction grating describes the number of lines drawn per unit length, typically measured in lines per centimeter. It is a crucial factor because it determines how light diffracts as it passes through the grating.
To find the line density, first, you must calculate the grating separation (distance between adjacent grating lines) using the grating equation. Once you have this separation, denoted as \(d\), the line density \(L\) is simply the reciprocal of \(d\):
  • \(L = \frac{1}{d}\)
This converts to lines per meter, so to find the line density in lines per centimeter, divide \(L\) by 100. This conversion is important for practical applications as it aligns with standard measurement units.
In experiments requiring precise light diffraction, knowing the line density helps predict where and how many diffraction maxima will occur.
Grating Equation
The grating equation is the fundamental formula used to analyze diffraction patterns produced by a grating. It relates the angle at which a certain maxima appears to the wavelength of light and the grating's line separation.
The equation is expressed as:
  • \(m \lambda = d \sin \theta\)
Here, \(m\) is the order of the diffraction maximum, \(\lambda\) is the wavelength, \(d\) is the grating separation, and \(\theta\) is the angle observed.
The scientific formula shows how the order of maxima and the spacing between the lines on the grating determine the angles at which light will diffract.
  • When \(m = 1\), it refers to the first-order maximum.
  • This variable \(m\) depicts how many wavelengths fit into the path difference created by the grating.
By understanding and applying the grating equation, you can predict where any subsequent maxima will appear.
Order of Maximum
In the context of diffraction patterns, the "order of maximum," denoted as \(m\), refers to the numbering of the diffraction maxima from the central maximum (the directly transmitted light). Each order indicates how many wavelengths of light constructively interfere at that point:
  • \(m = 0\) represents the central (or zero-order) maximum where there's no change in the path difference.
  • \(m = 1\) is the first-order maximum, caused by one wavelength fitting into the path difference.
  • Higher orders follow similarly, such as \(m = 2\) for second-order, and so on.
The central maximum will always be the brightest, while each order thereafter typically decreases in intensity.
Orders are crucial because they indicate how far (in terms of angle) the diffraction maximum is from the central point. Higher orders may only occur if the path difference and the sin of the angle allow multiple wavelengths to fit constructively.
Wavelength
The wavelength, \(\lambda\), is a key property of light influencing how it interacts with a diffraction grating. Wavelength is the distance between successive peaks (or troughs) of a wave, and it is generally specified in nanometers (nm) when dealing with visible light.
Visible light, such as laser light, typically ranges from about 400 nm (violet) to 700 nm (red). The wavelength directly impacts diffraction patterns:
  • The grating equation requires knowledge of the wavelength to calculate angles \(\theta\) and line separation \(d\).
  • Shorter wavelengths bend less, resulting in smaller diffraction angles \(\theta\).
  • Longer wavelengths bend more, resulting in larger angles \(\theta\).
In practical scenarios, understanding the wavelength assists in predicting interference patterns and designing effective optical systems. The example uses a laser light of 632.8 nm, a typical value for red laser pointers.

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Most popular questions from this chapter

Quasars, an abbreviation for quasi-stellar radio sources, are distant objects that look like stars through a telescope but that emit far more electromagnetic radiation than an entire normal galaxy of stars. An example is the bright object below and to the left of center in Fig. \(\mathrm{P} 36.58 ;\) the other elongated objects in this image are normal galaxies. The leading model for the structure of a quasar is a galaxy with a supermassive black hole at its center. In this model, the radiation is emitted by interstellar gas and dust within the galaxy as this material falls toward the black hole. The radiation is thought to emanate from a region just a few light-years in diameter. (The diffuse glow surrounding the bright quasar shown in Fig. \(\mathrm{P} 36.58\) is thought to be this quasar's host galaxy.) To investigate this model of quasars and to study other exotic astronomical objects, the Russian Space Agency has placed a radio telescope in a large orbit around the earth. When this telescope is \(77,000 \mathrm{~km}\) from earth and the signals it receives are combined with signals from the ground-based telescopes of the VLBA, the resolution is that of a single radio telescope \(77,000 \mathrm{~km}\) in diameter. What is the size of the smallest detail that this arrangement can resolve in quasar \(3 \mathrm{C} 405,\) which is \(7.2 \times 10^{8}\) light-years from earth, using radio waves at a frequency of \(1665 \mathrm{MHz}\) ? (Hint: Use Rayleigh's criterion.) Give your answer in lightyears and in kilometers.

In the 1920s Clinton Davisson and Lester Germer accidentally observed diffraction when electrons with \(54 \mathrm{eV}\) of energy were scattered off crystalline nickel. The diffraction peak occurred when the angle between the incident beam and the scattered beam was \(50^{\circ}\). (a) What is the corresponding angle \(\theta\) relevant for Eq. (36.16)\(?\) (b) The planes in crystalline nickel are separated by \(0.091 \mathrm{nm}\), as determined by x-ray scattering experiments. According to the Bragg condition, what wavelength do the electrons in these experiments have? (c) Given the mass of an electron as \(9.11 \times 10^{-31} \mathrm{~kg},\) what is the corresponding classical speed \(v_{\mathrm{cl}}\) of the diffracted electrons? (d) Assuming the electrons correspond to a wave with speed \(v_{\mathrm{cl}}\) and wavelength \(\lambda,\) what is the frequency \(f\) of the diffracted waves? (e) Quantum mechanics postulates that the energy \(E\) and the frequency \(f\) of a particle are related by \(E=h f,\) where \(h\) is known as Planck's constant. Estimate \(h\) from these observations. (f) Our analysis has a small flaw: The relevant wave velocity, known as a quantum phase velocity, is half the classical particle velocity, for reasons explained by deeper aspects of quantum physics. Re-estimate the value of \(h\) using this modification. (g) The established value of Planck's constant is \(6.6 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\). Does this agree with your estimate?

A laser beam of wavelength \(\lambda=632.8 \mathrm{nm}\) shines at normal incidence on the reflective side of a compact disc. (a) The tracks of tiny pits in which information is coded onto the CD are \(1.60 \mu \mathrm{m}\) apart. For what angles of reflection (measured from the normal) will the intensity of light be maximum? (b) On a DVD, the tracks are only \(0.740 \mu \mathrm{m}\) apart. Repeat the calculation of part (a) for the DVD.

The VLBA (Very Long Baseline Array) uses a number of individual radio telescopes to make one unit having an equivalent diameter of about \(8000 \mathrm{~km}\). When this radio telescope is focusing radio waves of wavelength \(2.0 \mathrm{~cm}\), what would have to be the diameter of the mirror of a visible-light telescope focusing light of wavelength \(550 \mathrm{nm}\) so that the visible-light telescope has the same resolution as the radio telescope?

A wildlife photographer uses a moderate telephoto lens of focal length \(135 \mathrm{~mm}\) and maximum aperture \(f / 4.00\) to photograph a bear that is \(11.5 \mathrm{~m}\) away. Assume the wavelength is \(550 \mathrm{nm}\). (a) What is the width of the smallest feature on the bear that this lens can resolve if it is opened to its maximum aperture (b) If, to gain depth of field, the photographer stops the lens down to \(f / 22.0\), what would be the width of the smallest resolvable feature on the bear?

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