Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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Chapter 36, Problem 17PQ
To determine
The thickness of the oil film that will appear green or red.
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One way to determine the index of refraction of a gas is to use an interferometer. As shown below, one of the beams of an interferometer passes through a glass container that has a length of L = 1.8 cm. Initially the
glass container is a vacuum. When gas is slowly allowed into the container, a total of 7571 dark fringes move past the reference line. The laser has a wavelength of 687 nm (this is the wavelength when the light from
the laser is moving through a vacuum).
Laser
Mirror
Glass Container
Beam Splitter
Diffraction Pattern
Mirror
A.) Determine how many wavelengths will fit into the glass container when it is a vacuum. Since the
light passes through the container twice, you need to determine how many wavelengths will fit into a
glass container that has a length of 2L.
number of wavelengths (vacuum) =
B.) The number of dark fringes is the difference between the number of wavelengths that fit in the
container (length of 2L) when it has gas and the number of wavelengths that fit in…
What visible wave length will be reflected in this scenario:
A 700 nm thick soap film floats on a plate. From the air, white light is struck onto the film at a normal incidence. the wavelength of visible light to air is 400 nm to 700 nm. The index of the soap film is 1.74 and 1.58 for the plate.
One way to determine the index of refraction of a gas is to use an interferometer. As shown below, one of the beams of an interferometer passes through a glass container that has a length of L = 1.8 cm. Initially the glass container is a vacuum. When gas is slowly allowed into the container, a total of 6894 dark fringes move past the reference line. The laser has a wavelength of 635 nm (this is the wavelength when the light from the laser is moving through a vacuum).
A.) Determine how many wavelengths will fit into the glass container when it is a vacuum. Since the light passes through the container twice, you need to determine how many wavelengths will fit into a glass container that has a length of 2L.number of wavelengths (vacuum) =
B.) The number of dark fringes is the difference between the number of wavelengths that fit in the container (length of 2L) when it has gas and the number of wavelengths that fit in the container (length of 2L) when it is a vacuum. Use this knowledge to…
Chapter 36 Solutions
Physics for Scientists and Engineers: Foundations and Connections
Ch. 36.2 - Prob. 36.1CECh. 36.3 - Prob. 36.2CECh. 36.4 - Prob. 36.3CECh. 36.5 - Prob. 36.4CECh. 36.5 - Prob. 36.5CECh. 36 - Many circular apertures are adjustable, such as...Ch. 36 - Many of the images we regularly look at are...Ch. 36 - The hydrogen line at 1420.4 MHz corresponds to the...Ch. 36 - Prob. 4PQCh. 36 - Estimate the diffraction-limited resolution of the...
Ch. 36 - Prob. 6PQCh. 36 - Prob. 7PQCh. 36 - Prob. 8PQCh. 36 - Prob. 9PQCh. 36 - Prob. 10PQCh. 36 - Prob. 11PQCh. 36 - Prob. 12PQCh. 36 - Prob. 13PQCh. 36 - Prob. 14PQCh. 36 - Prob. 15PQCh. 36 - Prob. 16PQCh. 36 - Prob. 17PQCh. 36 - Prob. 18PQCh. 36 - Prob. 19PQCh. 36 - Prob. 20PQCh. 36 - Prob. 21PQCh. 36 - Prob. 22PQCh. 36 - Prob. 23PQCh. 36 - Prob. 24PQCh. 36 - Light of wavelength 566 nm is incident on a...Ch. 36 - Prob. 26PQCh. 36 - Prob. 27PQCh. 36 - Prob. 28PQCh. 36 - Prob. 29PQCh. 36 - Prob. 30PQCh. 36 - A light source emits a mixture of wavelengths from...Ch. 36 - Prob. 32PQCh. 36 - Prob. 33PQCh. 36 - Prob. 34PQCh. 36 - Prob. 35PQCh. 36 - Prob. 36PQCh. 36 - Prob. 37PQCh. 36 - Prob. 38PQCh. 36 - Prob. 39PQCh. 36 - Prob. 40PQCh. 36 - Prob. 41PQCh. 36 - Prob. 42PQCh. 36 - Prob. 43PQCh. 36 - Prob. 44PQCh. 36 - CASE STUDY Michelsons interferometer played an...Ch. 36 - CASE STUDY Michelsons interferometer played an...Ch. 36 - Prob. 47PQCh. 36 - Prob. 48PQCh. 36 - Problems 49 and 50 are paired. C Optical flats are...Ch. 36 - Optical flats are flat pieces of glass used to...Ch. 36 - Prob. 51PQCh. 36 - Prob. 52PQCh. 36 - Figure P36.53 shows two thin glass plates...Ch. 36 - Viewed from above, a thin film of motor oil with...Ch. 36 - Newtons rings, discovered by Isaac Newton, are an...Ch. 36 - Prob. 56PQCh. 36 - What is the radius of the beam of an argon laser...Ch. 36 - Prob. 58PQCh. 36 - A diffraction grating with 428 rulings per...Ch. 36 - How many rulings must a diffraction grating have...Ch. 36 - Prob. 61PQCh. 36 - White light is incident on a diffraction grating...Ch. 36 - X-rays incident on a crystal with planes of atoms...Ch. 36 - Prob. 64PQCh. 36 - Prob. 65PQCh. 36 - Prob. 66PQCh. 36 - The fringe width b is defined as the distance...Ch. 36 - The fringe width is defined as the distance...Ch. 36 - Prob. 69PQ
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- A 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_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_forwardTo save money on making military aircraft invisible to radar, an inventor decides to coat them with a nonreflective material having an index of refraction of 1.20, which is between that of air and the surface of the plane. This, he reasons, should be much cheaper than designing Stealth bombers. (a) What thickness should the coating be to inhibit the reflection of 4.00-cm wavelength radar? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?arrow_forward
- Figure P36.53 shows two thin glass plates separated by a wire with a square cross section of side length w, forming an air wedge between the plates. What is the edge length w of the wire if 42 dark fringes are observed from above when 589-nm light strikes the wedge at normal incidence? FIGURE P36.53arrow_forwardBoth sides of a uniform film that has index of refraction n and thickness d are in contact with air. For normal incidence of light, an intensity minimum is observed in the reflected light at λ2 and an intensity maximum is observed at λ1, where λ1 > λ2. (a) Assuming no intensity minima are observed between λ1 and λ2, find an expression for the integer m in Equations 27.13 and 27.14 in terms of the wavelengths λ1 and λ2. (b) Assuming n = 1.40, λ1 = 500 nm, and λ2 = 370 nm, determine the best estimate for the thickness of the film.arrow_forwardIn Figure P36.10 (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? Figure P36.10arrow_forward
- Why is the following situation impossible? A piece of transparent material having an index of refraction n = 1.50 is cut into the shape of a wedge as shown in Figure P36.40. Both the top and bottom surfaces of the wedge are in contact with air. Monochromatic light of wavelength = 632.8 nm is normally incident from above, and the wedge is viewed from above. Let h = 1.00 mm represent the height of the wedge and = 0.500 m its length. A thin-film interference pattern appears in the wedge due to reflection from the top and bottom surfaces. You have been given the task of counting the number of bright fringes that appear in the entire length of the wedge. You find this task tedious, and your concentration is broken by a noisy distraction after accurately counting 5 000 bright fringes. Figure P36.40arrow_forwardInterference fringes are produced using Lloyds mirror and a source S of wavelength = 606 nm as shown in Figure P36.41. Fringes separated by y = 1.20 mm are formed on a screen a distance L = 2.00 m from the source. Find the vertical distance h of the source above the reflecting surface. Figure P36.41arrow_forwardA thin film of oil (no=1.50) with varying thickness floats on water (nw=1.33). When it is illuminated from above by white light, the reflected colors are as shown in the figure. In air, the wavelength of yellow light is 580nm. What is the oil's thickness t at point B?arrow_forward
- The Michelson interferometer can be used to measure the index of refraction of a gas by placing an evacuated transparent tube in the light path along one arm of the device. Fringe shifts occur as the gas is slowly added to the tube. Assume 610-nm light is used, the tube is 5.40 cm long, and 168 bright fringes pass on the screen as the pressure of the gas in the tube increases to atmospheric pressure. What is the index of refraction of the gas? Hint: The fringe shifts occur because the wavelength of the light changes inside the gas-filled tube. (Give your answer to at least five decimal places.)arrow_forwardMonochromatic light (l = 500 nm) is incident on a soap bubble (n = 1.40). What is the wavelength of the light (in nm) in the bubble film?arrow_forwardMonochromatic light with wavelength 620 nm passes through a circular aperture with diameter 7.4 mm. The resulting diffraction pattern is observed on a screen that is 4.5 m from the aperture. What is the diameter of the Airy disk on the screen?arrow_forward
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