6 Figure 35-24a gives intensity I versus position x on the viewing screen for the central portion of a two-slit interference pattern. The other parts of the figure give phasor diagrams for the electric field components of the waves arriving at the screen from the two slits (as in Fig. 35-13a). Which numbered points on the screen best correspond to which phasor diagram? - Central fringe キャと 1 2 3 (a) (6) (c) (d) 4.

6 Figure 35-24a gives intensity I versus position x on the viewing screen for the central portion of a two-slit interference pattern. The other parts of the figure give phasor diagrams for the electric field components of the waves arriving at the screen from the two slits (as in Fig. 35-13a). Which numbered points on the screen best correspond to which phasor diagram? - Central fringe キャと 1 2 3 (a) (6) (c) (d) 4.

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter37: Wave Optics

Section: Chapter Questions

Problem 37.27P: The intensity on the screen at a certain point in a double- slit interference pattern is 64.0% of...

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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 : mm= (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.
Show that the distribution of intensity in a double-slit pattern is given by Equation 36.9. Begin by assuming that the total magnitude of the electric field at point P on the screen in Figure 36.4 is the superposition of two waves, with electric field magnitudes E1=E0sintE2=E0sin(t+) The phase angle in in E2 is due to the extra path length traveled by the lower beam in Figure 36.4. Recall from Equation 33.27 that the intensity of light is proportional to the square of the amplitude of the electric field. In addition, the apparent intensity of the pattern is the time-averaged intensity of the electromagnetic wave. You will need to evaluate the integral of the square of the sine function over one period. Refer to Figure 32.5 for an easy way to perform this evaluation. You will also need the trigonometric identity sinA+sinB=2sin(A+B2)cos(AB2)
The intensity on the screen at a certain point in a double- slit interference pattern is 64.0% of the maximum value. (a) What minimum phase difference (in radians) between sources produces this result? (b) Express this phase difference as a path difference for 486.1-nm light.
A diffraction pattern is produced on a screen 1.40 m from a single slit, using monochromatic light of wavelength 5.00 102 nm. The distance from the center of the central maximum to the first-order maximum is 3.00 mm. Calculate the slit width. Hint: Assume that the first-order maximum is halfway between the first- and second-order minima.
A diffraction pattern is formed on a screen 120 cm away from a 0.400-mm-wide slit. Monochromatic 546.1-nm light is used. Calculate the fractional intensity I/Imax at a point on the screen 4.10 mm from the center of the principal maximum.
The intensity of light in a diffraction pattern of a single slit is described by the equation I=Imaxsin22 where = (a sin )/. The central maximum is at = 0, and the side maxima are approximately at =(m+12) for m = 1, 2, 3, . Determine more precisely (a) the location of the first side maximum, where m = 1, and (b) the location of the second side maximum. Suggestion: Observe in Figure 37.6a that the graph of intensity versus has a horizontal tangent at maxima and also at minima.
Suppose the single slit in Figure 38.4 is 6.00 cm wide and in front of a microwave source operating at 7.50 GHz. (a) Calculate the angle for the first minimum in the diffraction pattern. (b) What is the relative intensity l/lmax at 0 =- 15.0? (c) Assume two such sources, separated laterally by 20.0 cm, are behind the slit. What must be the maximum distance between the plane of the sources and the slit if the diffraction patterns are to be resolved? In this case, the approximation sin tan 8 is not valid because of the relatively small value of a/.
Light of wavelength 587.5 nm illuminates a slit of width 0.75 mm. (a) At what distance from the slit should a screen be placed if the first minimum in the diffraction pattern is to be 0.85 mm from the central maximum? (b) Calculate the width of the central maximum.
In Figure 37.7, show mathematically how many interference maxima are enclosed by the central diffraction maximum in the pattern. Notice that the diagram is generated by using 650-nm light to illuminate two 3.0-m slits separated by 18 m. Figure 37.7 The combined effects of two-slit and single-slit interference. This pattern is produced when 650-nm light waves pass through two 3.0-m slits that are 18 m apart.
When we studied Youngs double-slit experiment, we mostly ignored the dark fringe pattern produced by diffraction. Use Figure 35.21 to describe situations in which that omission makes sense. Think especially about the single slit used in front of the double slit in Youngs experiment (Fig. 35.9).
Assume light of wavelength 650 nm passes through two slits 3.00 m wide, with their centers 9.00 m apart. Make a sketch of the combined diffraction and interference pattern in the form of a graph of intensity versus = (a sin )/. You may use Figure 37.7 as a starting point.
A plane monochromatic light wave is incident on a double slit as illustrated in Figure 37.1. (i) As the viewing screen is moved away from the double slit, what happens to the separation between the interference fringes on the screen? (a) It increases, (b) It decreases, (c) It remains the same, (d) It may increase or decrease, depending on the wavelength of the light. (e) More information is required, (ii) As the slit separation increases, what happens to the separation between the interference fringes on the screen? Select from the same choices.