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Concept explainers
In the What If? section of Example 36.2, it was claimed that overlapping fringes in a two-slit interference pattern for two different wavelengths obey the following relationship even for large values of the angle θ:
(a) Prove this assertion. (b) Using the data in Example 36.2, find the nonzero value of y on the screen at which the fringes from the two wavelengths first coincide.
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Chapter 36 Solutions
Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 10th + WebAssign Printed Access Card for Serway/Jewett's Physics for Scientists and Engineers, 10th, Multi-Term
- In the double-slit arrangement of Figure P36.13, d = 0.150 mm, L = 140 cm, = 643 nm. and y = 1.80 cm. (a) What is the path difference for the rays from the two slits arriving at P? (b) Express this path difference in terms of . (c) Does P correspond to a maximum, a minimum, or an intermediate condition? Give evidence for your answer. Figure P36.13arrow_forwardCoherent light rays of wavelength strike a pair of slits separated by distance d at an angle 1, with respect to the normal to the plane containing the slits as shown in Figure P27.14. The rays leaving the slits make an angle 2 with respect to the normal, and an interference maximum is formed by those rays on a screen that is a great distance from the slits. Show that the angle 2 is given by 2=sin1(sin1md) where m is an integer.arrow_forwardIn a double-slit experiment, green light (5303 Å) falls on a double slit having a separation of 19.44 µ-m and a width of 4.05 µ-m. The number of bright fringes between the first and the second diffraction minima is (a) 5 (b) 10 (c) 9 (d) 4arrow_forward
- Light of wavelength λ = 610 nm and intensity I0 = 240 W/m2 passes through a slit of width w = 4.8 μm before hitting a screen L = 1.6 meters away. Part (a) Use the small-angle approximation to write an equation for the phase difference, β, between rays that pass through the very top and very bottom of the slit when the rays hit a point y = 46 mm above the central maximum. Part (b) Calculate this phase difference, in radians? Part (c) What is the intensity of the light, in watts per square meter, at this point?arrow_forwardA double-slit interference pattern is created by two narrow slits spaced 0.20 mm apart. The distance between the first and the fifth minimum on a screen 59 cm behind the slits is 6.5 mm. What is the wavelength (in nmnm) of the light used in this experiment?arrow_forwardLight of wavelength 470 nm passes through a double slit, yielding a diffraction pattern whose graph of intensity I versus angular position 0 is shown in the figure. Calculate (a) the slit width and (b) the slit separation. If Im = 7.1 mW/cm2 what are the intensities of the (c) m %3D = 1 and (d) m = 2 interference fringes? m, 9. 0 (degrees) (a) Number Units (b) Number Units (c) Number Units (d) Number Units Intensity (mW/cm²)arrow_forward
- The figure shows the interference pattern that appears on a distant screen when coherent light is incident on a mask with two identical, very narrow slits. Points P and Q are maxima; Point R is a minimum. The wavelength of the light that created the interference pattern is λ=678nm, the two slites are separated by rm d=6 μm, and the distance from the slits to the center of the screen is L=80cm . The difference in path length at a point on the screen is Δs=|s1−s2|, where s1s1 and s2s2 are the distances from each slit to the point. What is ΔsΔs (in nm) at Point P? What is ΔsΔs (in nm) at Point Q? What is ΔsΔs (in nm) at Point R?arrow_forwardThe figure shows a Young's double slit experimental setup. It is observed that when a thin transparent sheet of a Its) Hepre T thickness t and refractive index u is put in front of one of the slits, the a(μµ − 1) - central maximum gets shifted by a distance equal to n fringe widths. If the wavelength of light used is λ, t will be 2nDλ (a) (b) 2Dλ a(μ-1) (c) Dλ a(µ − 1) - D (d) Screen nDλ a(μµ - 1)arrow_forwardIn a double-slit diffraction experiment, two slits of width 12.4 x 10-6 m are separated by a distance of 32.2 x 10-6 m, and the wavelength of the incident light is 671 nm. The diffraction pattern is viewed on a screen 4.63 m from the slits. Assume Ip is the intensity at a point P, a distance y= 69.8 cm on the screen from the central maximum. Find the ratio of the intensity at P and the intensity at the center of the diffraction pattern (I), Ip/Io. i Hint The ratio of the intensities is given by = (sin(a)) ² Ip Io Save for Later cos (B)². Submit Answerarrow_forward
- Consider a 560 nm light falling on a single slit of width 1.1 μm.λ = 560 nmw = 1.1 μm At what angle (in degrees) is the first minimum for the light? θ=??arrow_forwardIn the two-slit interference experiment, the slit widths are each 4.0 μm, their separation is 24.0 μm, the wavelength is 600 nm, and the viewing screen is at a distance of 2.00 m from the slits. Point P lies at distance y =5.0 cm from the center of the pattern. (a) Without diffraction effects taken into account, what is the ratio of IP to the intensity Im at the center of the pattern? (b) With diffraction effects taken into account, what is the ratio of IP to the intensity Im at the center of the pattern?arrow_forward
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningPrinciples of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
- Physics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
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