Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 37, Problem 43AP
(a)
To determine
To show: The condition of sharp view for monochromatic light with plane wave fronts
(b)
To determine
The optimum pinhole diameter for
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A pinhole camera has a small circular aperture of diameter D. Light from distant objects passes through the aperture into an otherwise dark box, falling on a screen located a distance L away. If D is too large, the display on the screen will be fuzzy because a bright point in the field of view will send light onto a circle of diameter slightly larger than D. On the other hand, if D is too small, diffraction will blur the display on the screen. The screen shows a reasonably sharp image if the diameter of the central disk of the diffraction pattern, specified by as shown, is equal to D at the screen. (a) Show that for monochromatic light with plane wave fronts and L >> D, the condition for a sharp view is fulfilled if D2 = 2.44λL. (b) Find the optimum pinhole diameter for 500-nm light projected onto a screen 15.0 cm away.
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 580-nm light is used, the tube is 5.40 cm long, and 152 fringe shifts occur 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 five decimal places.)
4.0
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 600-nm light is used, the tube is 5.00 cm long, and 160 fringe shifts occur as the pressure of the gas in the tube increases to atmospheric pressure. What is the index of refraction of the gas?
Chapter 37 Solutions
Physics for Scientists and Engineers with Modern Physics
Ch. 37.2 - Suppose the slit width in Figure 37.4 is made half...Ch. 37.3 - Cats eyes have pupils that can be modeled as...Ch. 37.3 - Suppose you are observing a binary star with a...Ch. 37.4 - Ultraviolet light of wavelength 350 nm is incident...Ch. 37.6 - A polarizer for microwaves can be made as a grid...Ch. 37.6 - Prob. 37.6QQCh. 37 - Prob. 1PCh. 37 - Prob. 2PCh. 37 - Prob. 3PCh. 37 - In Figure 37.7, show mathematically how many...
Ch. 37 - Prob. 5PCh. 37 - What If? Suppose light strikes a single slit of...Ch. 37 - Prob. 7PCh. 37 - Coherent light of wavelength 501.5 nm is sent...Ch. 37 - Prob. 9PCh. 37 - Prob. 10PCh. 37 - What is the approximate size of the smallest...Ch. 37 - Prob. 12PCh. 37 - Prob. 13PCh. 37 - Prob. 14PCh. 37 - Impressionist painter Georges Seurat created...Ch. 37 - Prob. 16PCh. 37 - Consider an array of parallel wires with uniform...Ch. 37 - Prob. 18PCh. 37 - A grating with 250 grooves/mm is used with an...Ch. 37 - Show that whenever white light is passed through a...Ch. 37 - Light from an argon laser strikes a diffraction...Ch. 37 - Prob. 22PCh. 37 - You are working as a demonstration assistant for a...Ch. 37 - Prob. 24PCh. 37 - Prob. 25PCh. 37 - Prob. 26PCh. 37 - Prob. 27PCh. 37 - Prob. 28PCh. 37 - Prob. 29PCh. 37 - Prob. 30PCh. 37 - Prob. 31PCh. 37 - Prob. 32PCh. 37 - Prob. 33APCh. 37 - Laser light with a wavelength of 632.8 nm is...Ch. 37 - Prob. 35APCh. 37 - Prob. 36APCh. 37 - Prob. 37APCh. 37 - Prob. 38APCh. 37 - Prob. 39APCh. 37 - Prob. 40APCh. 37 - Prob. 41APCh. 37 - Prob. 42APCh. 37 - Prob. 43APCh. 37 - Prob. 44APCh. 37 - Prob. 45APCh. 37 - Prob. 46APCh. 37 - Prob. 47APCh. 37 - Prob. 48APCh. 37 - Two closely spaced wavelengths of light are...Ch. 37 - Prob. 50CPCh. 37 - Prob. 51CPCh. 37 - In Figure P37.52, suppose the transmission axes of...Ch. 37 - Prob. 53CP
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- A beam of 580-nm light passes through two closely spaced glass plates at close to normal incidence as shown in Figure P27.23. For what minimum nonzero value of the plate separation d is the transmitted light bright?arrow_forwardA plano-convex lens has index of refraction n. The curved side of the lens has radius of curvature R and rests on a flat glass surface of the same index of refraction, with a film of index nflim between them, as shown in Figure P36.42. The lens is illuminated from above by light of wavelength . Show that the dark Newtons rings have radii given approximately by r=mRnfilm where r R and m is an integer. Figure P36.42arrow_forwardLight enters a prism of crown glass and refracts at an angle of 5.00 with respect to the normal at the interface. The crown glass has a mean index of refraction of 1.51. It is combined with one flint glass prism (n = 1.65) to produce no net deviation. a. Find the apex angle of the flint glass. b. Assume the index of refraction for violet light (v = 430 nm) is nv = 1.528 and the index of refraction for red light (r = 768 nm) is nr = 1.511 for crown glass. For flint glass using the same wavelengths, nv = 1.665 and nr = 1.645. Find the net dispersion.arrow_forward
- Figure CQ27.4 shows an unbroken soap film in a circular frame. The film thickness increases from top to bottom, slowly at first and then rapidly. As a simpler model, consider a soap film (n = 1.33) contained within a rectangular wire frame. The frame is held vertically so that the film drains downward and forms a wedge with flat faces. The thickness of the film at the top is essentially zero. The film is viewed in reflected white light with near-normal incidence, and the first violet ( = 420 nm) interference band is observed 3.00 cm from the top edge of the film. (a) Locate the first red ( = 680 nm) interference band. (b) Determine the film thickness at the positions of the violet and red bands. (c) What is the wedge angle of the film?arrow_forwardA soap film, 4.5 x 10-7 m thick, is illuminated by white light normal to its surface. The index of refraction of the film is 1.50. What wavelength will be intensified in the reflected beam? 450 nm 540 nm 420 nm 675 nm 600 nmarrow_forwardIn a double-slit experiment, the optical path difference between the rays from two coherent sources at a point P on one side of the central bright band is 7.5 x 10 m and at a point Q on the other side of the central bright band is 1.8 x 10 m. How many bright and dark bands are observed between points P and Q if the wavelength of light used is 6 × 107 m?arrow_forward
- A green laser beam with a wavelength 500 nm passes through a partially reflecting Pyrex plate that transmits 50% of the incident light. The output surface of that plate is d0 =1.6 cm from a screen. A d1 = 8.0 mm thick glass plate that has an index of refraction n = 1.50 is inserted into that 1.6 cm gap . With that glass plate partially filling the gap, how many wavelengths of the light are there between the output surface of the Pyrex plate and the screen?arrow_forwardA slit of width d is placed in front of a lens of focal length 0.5 m and is illuminated normally with light of wavelength 5.89 × 107 m. The first diffraction minima on either side of the central diffraction maximum are separated by 2 × 10-³ m. The width d of the slit is m.arrow_forwardOne 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…arrow_forward
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