Concept explainers
The intensity of light in a diffraction pattern of a single slit is described by the equation
where ϕ = (πa sin θ)/λ. The central maximum is at ϕ = 0, and the side maxima are approximately at
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- As a single crystal is rotated in an x-ray spectrometer (Fig. 3.22a), many parallel planes of atoms besides AA and BB produce strong diffracted beams. Two such planes are shown in Figure P3.38. (a) Determine geometrically the interplanar spacings d1 and d2 in terms of d0. (b) Find the angles (with respect to the surface plane AA) of the n = 1, 2, and 3 intensity maxima from planes with spacing d1. Let = 0.626 and d0 = 4.00 . Note that a given crystal structure (for example, cubic) has interplanar spacings with characteristic ratios, which produce characteristic diffraction patterns. In this way, measurement of the angular position of diffracted x-rays may be used to infer the crystal structure. Figure P3.38 Atomic planes in a cubic lattice.arrow_forwardThe structure of the NaCl crystal forms reflecting planes 0.541 nm apart. What is the smallest angle, measured from these planes, at which X-ray diffraction can be observed, if X-rays of wavelength 0.085 nm are used?arrow_forwardThe full width at half-maximum (FWHM) of a central diffraction maximum is defined as the angle between the two points in the pattern where the intensity is one-half that at the center of the pattern. (See figure (b).) (a) Does the intensity drop to one-half the maximum value when sin²α = a²/2? (b) Is a = 1.39 rad (about 80°) a solution to the transcendental equation of (a)? (c) Is the FWHM AÐ = 2sin¹(0.442 A/a), where a is the slit width? Calculate the FWHM of the central maximum for slit width (d) 1.17 A, (e) 5.03 A, and (f) 11.7 A. 20 20 Relative intensity 15 10 0.8 0.6 a=2 0.4 0.2 5 05 8 (degrees) (a) 10 15 20 20 Relative intensity 1.0 0.8 0.6 -A0- 0.4 0.2 a= 52 20 15 10 5 0 5 10 15 20 (degrees) (b)arrow_forward
- Problem 7: Consider light falling on a single slit, of width 1.05 μm, that produces its first minimum at an angle of 33.6°.Randomized Variables θ = 33.6°w = 1.05 μm Calculate the wavelength of the light in nanometers.arrow_forwardYou measure three segments of the distance between a diffraction slit an the screen on which the pattern forms: x1 = (14.7 ± 0.1) cm, x2 = (9.9 ± 0.3) cm, and x3 = (17.2 ± 0.3) cm. What is the uncertainty of the total distance x1 + x2 + x3?arrow_forwardAn electric current through an unknown gas produces several distinct wavelengths of visible light. Consider the first order maxima for the wavelengths 403 nm, 428 nm, 511 nm, and 682 nm of this unknown spectrum, when projected with a diffraction grating of 5,000 lines per centimeter.Randomized Variablesλ1 = 403 nmλ2 = 428 nmλ3 = 511 nmλ4 = 682 nm Part (a) What would the angle (in degrees) be for the 403 nm line? Part (b) What would the angle (in degrees) be for the 428 nm line? Part (c) What would the angle (in degrees) be for the 511 nm line? Part (d) What would the angle (in degrees) be for the 682 nm line? Part (e) Using this grating, what would be the angle (in degrees) of the second-order maximum of the 403 nm line?arrow_forward
- answer in radianarrow_forwardThe diffraction grating is a way of separating or dispersing light of different wavelengths, producing a spectrum of light. The grating interferes light constructively in particular directions: dsinθm=mλdsinθm=mλ For a particular angle, we calculate the wavelength. The grating constant (or line density) is 500 lines per mm -- every millimeter has 500 lines scratched onto it, equally spaced. The quantity d is the distance between the lines, and λ is the light wavelength. A meter stick shows the spots, and ym is position on the meter stick of the mth-order light beam. (Negative order is the same as positive order.) Calculate the light wavelength, λ in nm, given this information: The grating constant is 500 lines/mm. L = 31.5 cm y1 = 56.9 cm y-1 = 40.8 cm y0 isn't specified because of computer issues. (It's the average of y1 and y-1.)arrow_forwardThe diffraction grating is a way of separating or dispersing light of different wavelengths, producing a spectrum of light. The grating interferes light constructively in particular directions: dsinθm=mλdsinθm=mλ For a particular angle, we calculate the wavelength. The grating constant (or line density) is 500 lines per mm -- every millimeter has 500 lines scratched onto it, equally spaced. The quantity d is the distance between the lines, and λ is the light wavelength. In the previous problem, calculate y2, where one of the second-order spots appears on the meter stick. Either that, or show that y2 can't be determined.arrow_forward
- In Fig the central diffraction maximum contains exactly seven interference fringes, and in this case d/a = 4. (a) What must the ratio d/a be if the central maximum contains exactly five fringes? (b) In the case considered in part (a), how many fringes are contained within the first diffraction maximum on one side of the central maximum?arrow_forwardProblem 18: Consider a single slit that produces its first minimum at 54° for 590 nm light. Randomized Variablesθ1 = 54 °θ1 = 54 °θ2 = 67 °λ1 = 590 nm Part (a) What is the width of the single slit, w, in nanometers?Numeric : A numeric value is expected and not an expression.w = __________________________________________ Part (b) Find the wavelength, in nanometers, of light that has its first minimum at 67°.Numeric : A numeric value is expected and not an expression.λ2 = __________________________________________arrow_forwardHow much diffraction spreading does a light beam undergo? One quantitative answer is the full width at half maximum of the central maximum of the single-slit Fraunhofer diffraction pattern. You can evaluate this angle of spreading in this problem. (a) as shown, define φ = πa sin φ/λ and show that at the point where I = 0.5Imax we must have φ = √2 sin φ. (b) Let y1 = sin φ and y2 = φ = /√2. Plot y1 and y2 on the same set of axes over a range from φ = 1 rad to φ = π/2 rad. Determine φ from the point of intersection of the two curves. (c) Then show that if the fraction λ/a is notlarge, the angular full width at half maximum of the central diffraction maximum is θ = 0.885λ/a. (d) What If? Another method to solve the transcendental equation φ = √2 sin φ in part (a) is to guess a first value of φ, use a computer or calculator to see how nearly it fits, and continue to update your estimate until the equation balances. How many steps(iterations) does this process take?arrow_forward
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