Concept explainers
Determine what happens to the double-slit interference pattern if one of the slits is covered with a thin, transparent film whose thickness is
Trending nowThis is a popular solution!
Chapter 3 Solutions
University Physics Volume 3
Additional Science Textbook Solutions
College Physics
University Physics Volume 1
Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)
Tutorials in Introductory Physics
University Physics with Modern Physics (14th Edition)
Essential University Physics: Volume 2 (3rd Edition)
- A beam of monochromatic green light is diffracted by a slit of width 0.550 mm. The diffraction pattern forms on a wall 2.06 m beyond the slit. The distance between the positions of zero intensity on both sides of the central bright fringe is 4.10 mm. Calculate the wavelength of the light.arrow_forwardSuppose Youngs double-slit experiment is performed in air using red light and then the apparatus is immersed in water. What happens to the interference pattern on the screen? (a) It disappears. (b) The bright and dark fringes stay in the same locations, but the contrast is reduced. (c) The bright fringes are closer together. (d) The bright fringes are farther apart. (e) No change happens in the interference pattern.arrow_forwardMonochromatic light is incident on a pair of slits that are separated by 0.200 mm. The screen is 2.50 m away from the slits. a. If the distance between the central bright fringe and either of the adjacent bright fringes is 1.67 cm, find the wavelength of the incident light. b. At what angle does the next set of bright fringes appear?arrow_forward
- What is the angular width of the central fringe of the interference pattern of (a) 20 slits separated by d=2.0103 mm? (b) 50 slits with the same separation? Assume that =600 nm.arrow_forwardA Fraunhofer diffraction pattern is produced on a screen located 1.00 m from a single slit. If a light source of wavelength 5.00 107 m is used and the distance from the center of the central bright fringe to the first dark fringe is 5.00 103 m, what is the slit width? (a) 0.010 0 mm (b) 0.100 mm (c) 0.200 mm (d) 1.00 mm (e) 0.005 00 mmarrow_forwardRed light (wavelength 632.8 nm in air) from a Helium-Neon laser is incident on a single slit of width 0.05 mm. The entire apparatus is immersed in water of refractive index 1.333. Determine the angular width of the central peak.arrow_forward
- Which of the following causes the fringes in a two-slit interference pattern to move farther apart? (a) decreasing the wavelength of the light (b) decreasing the screen distance L (c) decreasing the slit spacing d (d) immersing the entire apparatus in waterarrow_forwardTwo slits 4.0106 m apart are illuminated by light of wavelength 600 nm. What is the highest order fringe in the interference pattern?arrow_forwardAn effect analogous to two-slit interference can occur with sound waves, instead of light. In an open field, two speakers placed 1.30 m apart are powered by a single-function generator producing sine waves at 1200-Hz frequency. A student walks along a line 12.5 m away and parallel to the line between the speakers. She hears an alternating pattern of loud and quiet, due to constructive and destructive interference. What is (a) the wavelength of this sound and (b) the distance between the central maximum and the first maximum (loud) position along this line?arrow_forward
- Light at 633 nm from a helium-neon laser shines on a pair of parallel slits separated by 1.45 105 m and an interference pattern is observed on a screen 2.00 m from the plane of the slits. (a) Find the angle from the central maximum to the First bright fringe. (b) At what angle from the central maximum does the second dark fringe appear? (c) Find the distance from the central maximum to the first bright fringe.arrow_forwardShow 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)arrow_forwardIn Figure P27.7 (not to scale), let L = 1.20 m and d = 0.120 mm and assume the slit system is illuminated with monochromatic 500-nm light. Calculate the phase difference between the two wave fronts arriving at P when (a) = 0.500 and (b) y = 5.00 mm. (c) What is the value of for which the phase difference is 0.333 rad? (d) What is the value of for which the path difference is /4?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 LearningCollege PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning
- University Physics Volume 3PhysicsISBN:9781938168185Author:William Moebs, Jeff SannyPublisher:OpenStaxPhysics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning