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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 Equation 37.6, 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.
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Chapter 37 Solutions
Physics for Scientists and Engineers
- 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_forwardFigure 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_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
- A glass lens, ?glass=1.52, has a 119 nm thick antireflective film coating one side, ?film=1.35. White light, moving through the air, is perpendicularly incident on the coated side of the lens. What is the largest wavelength of the reflected light that is totally removed by the coating? Assume that ?air=1.00. largest removed wavelength:arrow_forwardAn anti-glare coating is applied to minimize reflection of a certain wavelength from the glass of a camera lens. If light travels slower in the coating than it does in the lens' glass, what minimum number of wavelengths need to travel across one thickness of the coating? Thank youarrow_forwardInterference effects are produced at point P on a screen as a result of direct rays from a 5.00 x 102 - nm source and reflected rays off a mirror, as shown in Figure P24.67. If the source is L = 1.00 x 102 m to the left of the screen and h = 1.00 cm above the mirror, find the distance y (in millimeters) to the first dark band above the mirror.arrow_forward
- Light of wavelength 589 nm is used to view an object under a microscope. If the aperture of the objective has a diameter of 0.9 cm, what is the limiting angle of resolution? If the light is in the visible range, what would be the maximum limit of resolution for this microscope?arrow_forwardA slit of width 0.45 mm is illuminated with light of wavelength 544 nm, and a screen is placed 110 cm in front of the slit. Find the widths of the first and second maxima on each side of the central maximum. w1 = mm (1st maxima) w2 = mm (2nd maximaarrow_forwardA slit of width 0.56 mm is illuminated with light of wavelength 492 nm, and a screen is placed 130 cm in front of the slit. Find the widths of the first and second maxima on each side of the central maximum.w1 =_______ mm (1st maxima) w2 = ______mm (2nd maxima)arrow_forward
- A source of polychromatic light is used to create a two-slit interference pattern on a screen. In the n=2 block of that pattern, which of the following colours appears at the largest angle relative to the "straight-through" direction? a) red b) blue c) green d) yellow e) violetarrow_forwardA thin layer of a transparent material that has an index of refraction of 1.25 is used as a nonreflective coating on the surface of glass that has an index of refraction of 1.50. What should the minimum thickness of the material be for the material to be nonreflecting for light that has a wavelength of 578 nm?arrow_forward
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning