Concept explainers
Consider the following dialogue:
Student 1: "l don’t see why there are minima when there's only a single silt—I think you need two waves to have destructive interference,”
Student 2: "You can model the single slit as many Identical smaller interfering 'slits,' each small enough to act like a point; source. The first minimum occurs where the path length difference from the two ‘slits’ at the edges of the single slit is
Do you agree with student 2’s response to student 1? Discuss your reasoning with your partners.
Want to see the full answer?
Check out a sample textbook solutionChapter 11 Solutions
Tutorials in Introductory Physics
Additional Science Textbook Solutions
Modern Physics
College Physics: A Strategic Approach (3rd Edition)
College Physics (10th Edition)
Essential University Physics: Volume 1 (3rd Edition)
Cosmic Perspective Fundamentals
University Physics (14th Edition)
- Solve the following exercise; document each step of the process.Consider a double-slit Young interferometer: 1) Write down the evolution operator for propagation from the double slits to the observation screen.arrow_forwardSolve the following: (show your complete solution) (a) At what angle is the first minimum for 550-nm light falling on a single slit of width 1.00 µm? (b) Will there be a second minimum?arrow_forwardIn a Young's double-slit experiment, a set of parallel slits with a separation of 0.130 mm is illuminated by light having a wavelength of 550 nm and the interference pattern observed on a screen 3.50 m from the slits. (a) What is the difference in path lengths from the two slits to the location of a fifth order bright fringe on the screen? 1.1 Your response differs from the correct answer by more than 10%. Double check your calculations. um (b) What is the difference in path lengths from the two slits to the location of the fifth dark fringe on the screen, away from the center of the pattern? X umarrow_forward
- The light intensity vs. position graph of a double-slit experiment is shown below. The graph was made with helium-neon laser light of wavelength 640 nm shined through two very narrow slits separated by a small distance. The slits were 2.0 meters away from the probe. What is the spacing between the two slits, in mm? Your answer needs to have 2 significant figures, including the negative sign in your answer if needed. Do not include the positive sign if the answer is positive. No unit is needed in your answer, it is already given in the question statement. Light level allian Position of probe (mm)arrow_forwardMichelson's interferometer played an important role in improving our understanding of light, and it has many practical uses today. For example, it may be used to measure distances precisely. Suppose the mirror labeled 1 in the figure below is movable. If the laser light has a wavelength of 654.0 nm, how many fringes will pass across the detector if mirror 1 is moved just 1.960 (mm)? If you can easily detect the passage of just one fringe, how accurately can you measure the displacement of the mirror(nm)?arrow_forwardMichelson's interferometer played an important role in improving our understanding of light, and it has many practical uses today. For example, it may be used to measure distances precisely. Suppose the mirror labeled 1 in the figure below is movable. If the laser light has a wavelength of 654.0 nm, how many fringes will pass across the detector if mirror 1 is moved just 1.960 mm? fringesIf you can easily detect the passage of just one fringe, how accurately can you measure the displacement of the mirror? nmarrow_forward
- Derive the Fraunhofer pattern for an array of N slits and show that it gives the familiar result for double slits when N = 2. Hints: Choose coordinates so that the slits are centred at x = 0, a, 2a, ... and stay in complex notation for as long as possible.arrow_forwardWhen the width of the slit is made double, how would this effect the size and intensity of the central diffraction band? Justify your answer with the help of diagram.arrow_forwardUse the following information for the Gas Identification part of the lab. The data below are angles that might be recorded by a student using a spectrometer to map bright lines in the emission spectrum of some unknown gas. Be sure to record both the gas ID number and the diffraction grating constant (slit spacing) to be used in the analysis. In the answer field below, convert the first m=1 angle entry above to the corresponding wavelength expressed in nanometres. PLEASE SHOW THE STEPS FOR HOW TO CALCULATEarrow_forward
- For a given wavelength X, what is the minimum slit width for which there will be no diffraction minima? Express your answer in terms of X. Π| ΑΣΦ D = Submit Part B D = Request Answer ☐ What is the minimum slit width so that no visible light exhibits a diffraction minimum? The range of visible light is from 400 nm to 750 nm. Express your answer to two significant figures and include the appropriate units. µÅ Value ***** Units ? ?arrow_forwardHi please help: A screen is placed 2.00 m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.80 cm wide-that is, the two first-order diffraction minima are separated by 1.80 cm. What is the distance between the two second-order minima? Express your answer to three significant figures.arrow_forwardAn air wedge is formed between two glass plates separated at one edge by a very fine wire of circular cross section as shown in the figure below. When the wedge is illuminated from above by 600 nm light and viewed from above, 30 dark fringes are observed. Calculate the diameter d of the wire (in µm). What If? How many dark fringes will be observed if the gap between the glass plates is filled with water?arrow_forward
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningUniversity Physics Volume 3PhysicsISBN:9781938168185Author:William Moebs, Jeff SannyPublisher:OpenStax