Using a multi-slit diffraction grating, you find that for the same grating,constructive interference appears for two different monochromatic sources oflight at the exact same non-zero angle. For the first source (with wavelength664 nm), you observe that this point of interference is the first non-zero pointof constructive interference.What is the smallest wavelength that the second light source could have?Report your answer in nm.

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What is the smallest wavelength that the second light source could have?

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter27: Wave Optics

Section: Chapter Questions

Problem 10OQ: A Fraunhofer diffraction pattern is produced on a screen located 1.00 m from a single slit. If a...

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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 P36.13
In 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?
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 mm
A thin sheet of transparent material has an index of refraction of 1.40 and is 15.0 µm thick. When it is inserted in the light path along one arm of an interferometer, how many fringe shifts occur in the pattern? Assume the wavelength (in a vac- uum) of the light used is 600 nm. Hint: The wavelength will change within the material.
A student performs a double-slit interference experiment. The student records the fact that a third-order dark spot is located at an angle of 20.11 for a given light source. The wavelength of light is then decreased by 40.0nm, and the student records the fact that the second-order dark spot is now located at an angle of 11.17. What is the slit separation?
In a single-slit diffraction experiment, there is a minimum of intensity for orange light (l= 600 nm) and a minimum of intensity for blue-green light (l = 500 nm) at the same angle of 1.00 mrad. For what minimum slit width is this possible?
Problem 2: The width of a rectangular slit is measured in the laboratory by means of its diffraction pattern at a distance of 2 m from the slit. When illuminated normally with a parallel beam of laser light (632.8 nm), the distance between the third mini- ma on either side of the principal maximum is measured. An average of several tries gives 5.625 cm. a. Assuming Fraunhofer diffraction, what is the slit width? b. Is the assumption of far-field diffraction justified in this case? What is the ratio L/Lmin?
A student performs a double-slit interference experiment. The student records the fact that a third-order dark spot is located at an angle of 20.11o for a given light source. The wavelength of light is then decreased by 40.0nm, and the student records the fact that the second-order dark spot is now located at an angle of 11.17 degrees. What is the slit separation?
A thin sheet of transparent material has an index of refraction of 1.40 and is 15.0 μm thick. When it is inserted in the light path along one arm of an interferometer, how many fringe shifts occur in the pattern? Assume the wavelength (in a vacuum) of the light used is 600 nm. Hint: The wavelength will change within the material.
A monochromatic source of light of wavelength λ, is incident on a slit of width a. Obtain an expression to represent the angle of the 4th diffraction minimum. Hint: Think about Huygens construction with the single slit divided into four equal length sections acting as sources of secondary waves. Write down a general expression for the nth minimum angle. If λ =500 nm, a = 4 μm, determine the angle of the 2nd minimum.
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…
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…