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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_forwardProblem 2: Consider light that has its third minimum at an angle of 23.6° when it falls on a single slit of width 3.55 μm. Randomized Variables 9 = 23.6° w = 3.55 um D Find the wavelength of the light in nanometers. λ=1 sin() cos() cotan() asin() atan() acotan() tanh() cosh() O Degrees Hints: 2% deduction per hint. Hints remaining: 2 Submit tan() JU acos() E sinh() cotanh() Radians Hint ( + 7 8 9 4 5 6 1 0 VO BACKSPACE Feedback 2 3 All content © 2022 Expert TA, LLC DEL HOME END I give up! Feedback: 2% deduction per feedback. CLEARarrow_forward
- The limit to the eye’s visual acuity is related to diffraction by the pupil.D = 2.85 mmdh = 1.25 m a. What is the angle between two just-resolvable points of light for a 2.85 mm diameter pupil in radians, assuming an average wavelength of 550 nm? θmin = b. Take your result to be the practical limit for the eye. What is the greatest possible distance in km a car can be from you if you can resolve its two headlights, given they are 1.25 m apart? L= c. What is the distance between two just-resolvable points held at an arm’s length (0.800 m) from your eye in mm? da =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_forwardProblem 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_forward
- You 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_forwardProblem 1: In a double slit experiment the first minimum for 415 nm violet light is at an angle of 42°. Randomized Variables 2 = 415 nm e = 42 ° Find the distance between the two slits in micrometers. d= 8 9 5 6 sin() cos() tan() 7 HOME cotan() asin() acos() E A 4 atan() acotan() sinh() 1 2 3 cosh() tanh() cotanh() END O Degrees O Radians Vol BACKSPACE DEL CLEAR +arrow_forward
- Find the angular radius of the tenth bright fringe in a Michelson interfer- ometer when the central-path difference (2d) is (a) 1.50 mm and (b) 1.5 cm. The orange light of a krypton arc is 6057.8 ˚Aand that the interferometer is adjusted in each case so that the first bright fringe forms a maximum at the center of the pattern. Ans: (a) 4.885◦ , (b) 1.542◦arrow_forwardanswer 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_forward
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