The Bragg equation (10.10.13) is similar in form to the grating equation (10.8.10) with \(\theta_{\phi}=\theta_{2}\) However, for the grating the angles \(\theta_{\alpha}, \theta_{\theta}\) are measured from the normal, whereas in the Bragg case the angle \(\theta\) is measured, by convention, from the diffracting planes in question (rather than from the normal to the planes). What is the relation between the two situations? Why not use the cosine in place of the sine in one equation, since the angles appear to be complements?

Short Answer

Expert verified
Answer: Sine is used in both the Bragg equation and the grating equation because it represents the projection of the distance covered by the light on the normal between the diffracting planes or slits, which is important for maintaining constructive interference. Using cosine would not provide the same information and would not follow the same mathematical principles as sine, so it cannot be used as a substitute.

Step by step solution

01

Write down the Bragg equation and grating equation

The Bragg equation is given by: nλ = 2d sin(θ) And the grating equation is given by: nλ = d(sin(θ_α) + sin(θ_θ)) Here, n is the order of diffraction, λ is the wavelength of the light, d is the distance between the diffracting planes or slits.
02

Analyze the angles involved in both equations

For the Bragg equation, the angle θ is measured from the diffracting planes. In the grating equation, the angles θ_α and θ_θ are measured from the normal to the planes. However, the angle θ_φ in the grating equation equals θ_2, so we can say that θ_α = θ_2.
03

Explain the relation between the two situations

Since angle θ in the Bragg equation is measured from the diffracting planes, we can say that it is complementary to the angle θ_α in the grating equation because the sum of the two angles is indeed 90°. Mathematically, this is represented by: θ + θ_α = 90°
04

Determine the reason for using sine instead of cosine

One might think that we should use cosine in one equation since the angles in question appear to be complementary. However, cosine cannot be used here because of the context of these equations. In the plane of diffraction, the distance covered by the light must be an integer multiple of the wavelength to produce constructive interference, which is the core principle of both Bragg and grating equations. The sine function is used because it represents the projection of the distance covered by the light on the normal between the diffracting planes or slits while keeping the equations consistent. Using cosine would not provide the same information and would not follow the same mathematical principles as sine, thus it cannot be used as a substitute.

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Most popular questions from this chapter

A radio-frequency antenna (such as a TV antenna) may be thought of semiquantitatively as an aperture with dimensions comparable to the physical dimensions (usually of the order of a wavelength). How far away from the antenna, expressed in wavelengths, must one be for the Fraunhofer case to be valid? Compare with the case for a pinhole of diameter \(0.1 \mathrm{~mm}\) for visible wavelengths.

For the \(N\)-slit grating at normal incidence, devise elementary arguments, not involving integrals or a formal summation like \((10.7 .5)\), to show that (a) the principal interference maxima occur for $$ b \sin \theta_{0}=m \lambda $$ where \(m\) is an integer; \((b)\) the nulls adjacent to a particular principal maximum are displaced in angle from the maximum by \(\Delta \theta_{0}\) such that $$ \left(\frac{N}{2} b \cos \theta_{*}\right) \Delta \theta_{\varphi}=\pm \frac{\lambda}{2} $$ (c) hence, the resolving power is \(m N\).

A particular transparent diffraction grating is ruled with 5,500 lines per centimeter, and \(2.5 \mathrm{~cm}\) of it is used in a simple spectrometer having a source slit, collimating lens, and telescope. At what angles are the sodium D lines \((\lambda=589.0\) and \(589.6 \mathrm{~nm})\) found if the grating is illuminated at normal incidence? What is the dispersion in minutes of arc per nanometer and the theoretical resolution at each of these angular positions?

Review the elementary arguments for discussing single-slit Fraunhofer diffraction by pairing up portions of the wavefront in the aperture that are \(180^{\circ}\) out of phase to show that (a) a null is observed when the slit may be divided into two equal portions such that $$ \frac{\delta}{2}=\frac{a}{2} \sin \theta=\frac{\lambda}{2} $$ Prob. 10.4.1 (b) Similarly, nulls result when the aperture is divided up into \(2 n\) portions such that $$ \frac{a}{2 n} \sin \theta=\frac{\lambda}{2} $$ where \(n\) is an integer; \((c)\) the secondary maxima occur approximately when the aperture is divided up into \(2 n+1\) portions such that $$ \frac{a}{2 n+1} \sin \theta=\frac{\lambda}{2} $$ and that the intensity of these maxima, relative to the principal maximum, is approximately $$ \frac{1}{2}\left(\frac{1}{2 n+1}\right)^{2} . $$ How do you justify the factor of \(\frac{1}{2}\) ?

Show that when a circular aperture is illuminated by a plane wave whose amplitude varies with radius as \(T(\rho)\), from (9.4.19), the diffraction pattern is given by $$ \psi(u)=2 \pi \breve{C}^{\prime} \int_{0}^{a} T(\rho) J_{c}\left(\frac{u \rho}{a}\right) \rho d \rho, $$ which reduces to \((10.5 .4)\) when \(T(\rho)=1\).

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