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Newton’s rings, discovered by Isaac Newton, are an interference pattern of dark and bright rings formed because of the air gap of increasing thickness w between a spherical surface and an adjoining flat surface (Example 36.2). Figure P36.55 shows a Newton’s rings apparatus with a plano-convex lens with radius of curvature R atop a flat glass slab, both with index of refraction nglass. In the configuration shown, the seventh bright fringe has a radius of 1.10 cm. After the air gap in the apparatus is filled with an unknown liquid, the radius of the seventh fringe decreases to 0.968 cm. What is the index of refraction of the unknown liquid?
FIGURE P36.55
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Chapter 36 Solutions
Physics for Scientists and Engineers: Foundations and Connections
- Coherent light rays of wavelength strike a pair of slits separated by distance d at an angle 1, with respect to the normal to the plane containing the slits as shown in Figure P27.14. The rays leaving the slits make an angle 2 with respect to the normal, and an interference maximum is formed by those rays on a screen that is a great distance from the slits. Show that the angle 2 is given by 2=sin1(sin1md) where m is an integer.arrow_forwardIn a Newtons-rings experiment, a plano-convex glass (n = 1.52) lens having radius r = 5.00 cm is placed on a flat plate as shown in Figure P36.37. When light of wavelength = 650 nm is incident normally, 55 bright rings are observed, with the last one precisely on the edge of the lens. (a) What is the radius R of curvature of the convex surface of the lens? (b) What is the focal length of the lens? Figure P36.37arrow_forwardFigure 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_forward
- In the double-slit arrangement of Figure P36.13, d = 0.150 mm, L = 140 cm, = 643 nm. and y = 1.80 cm. (a) What is the path difference for the rays from the two slits arriving at P? (b) Express this path difference in terms of . (c) Does P correspond to a maximum, a minimum, or an intermediate condition? Give evidence for your answer. Figure P36.13arrow_forwardTable P35.80 presents data gathered by students performing a double-slit experiment. The distance between the slits is 0.0700 mm, and the distance to the screen is 2.50 m. The intensity of the central maximum is 6.50 106 W/m2. What is the intensity at y = 0.500 cm? TABLE P35.80arrow_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
- Optical flats are flat pieces of glass used to determine the flatness of other optical components. They are placed at an angle above the component as shown in Figure P36.49A, and monochromatic light is incident and observed from above, leading to interference fringes. Figure P36.49C shows the results of one of these tests. What is the approximate difference in the gap thickness between the left and right sides of the optical flat and the component? Is it possible to determine from this figure alone which side has the greater gap thickness (left or right)? Figure P36.49 Problems 49 and 50.arrow_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_forwardAn air wedge is formed between two glass plates separated at one edge by a very line wire of circular cross section as shown in Figure P27.25. 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.arrow_forward
- Coherent light rays of wavelength strike a pair of slits separated by distance d at an angle 1 with respect to the normal to the plane containing the slits as shown in Figure P36.9. The rays leaving the slits make an angle 2 with respect to the normal, and an interference maximum is formed by those rays on a screen that is a great distance from the slits. Show that the angle 2 is given by 2=sin1(sin1md) where m is an integer. Figure P36.9arrow_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_forwardProblems 49 and 50 are paired. C Optical flats are flat pieces of glass used to determine the flatness of other optical components. They are placed at an angle above the component as shown in Figure P36.49A, and monochromatic light is incident and observed from above, leading to interference fringes. Parts B and C of Figure P36.49 show the results of tests on two optical components. Which of the two is more flat? Explain. Figure P36.49 Problems 49 and 50.arrow_forward
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