Light is incident on a diffraction grating with 7200 slits/cmand the pattern is viewed on a screen located 2.5 m fromthe grating. The incident light beam consists of twowavelengths, d1 = 4.4 × 10-7 m and A2 = 6.8 × 10-7 m.Calculate the linear distance between the first-order brightfringes of these two wavelengths on the screen.

Given Data: The number of silts is, N=7200 silts/cm The distance between the screen and the silts…

Answered: Light is incident on a diffraction…

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter35: Diffraction And Interference

Section: Chapter Questions

Problem 82PQ

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A 5.08-cm-long rectangular glass chamber is inserted into one arm of a Michelson interferometer using a 633-nm light source. This chamber is initially filled with air (n=1.000293) at standard atmospheric pressure but the air is gradually pumped out using a vacuum pump until a near perfect vacuum is achieved. How many fringes are observed moving by during the transition?
Eight slits equally separated by 0.149 mm is uniformly illuminated by a monochromatic light at =523 nm. What is the intensity of a secondary maxima compared to that of the principal maxima?
A monochromatic light of unknown wavelength is incident on a slit of width 20 m. A diffraction pattern is seen at a screen 2.5 m away where the central maximum is spread over a distance of 10.0 cm. Find the wavelength.
Consider a single-slit diffraction pattern for =589 nm, projected on a screen that is 1.00 m from a slit of width 0.25 mm. How far from the center of the pattern are the centers of the first and second dark fringes?
Figure 3.14 shows two 7.50-cm-long glass slides illuminated by pure 589-nm wavelength light incident perpendicularly. The top slide touches the bottom slide at one end and rests on some debris at the other end, forming a wedge of air. How thick is the debris, if the dark bands are 1.00 mm apart? Figure 3.14 (a) The rainbow-color bands are produced by thin-film interference in the air between the two glass slides. (b) Schematic of the paths taken by rays in the wedge of air between the slides. (c) If the air wedge is illuminated with monochromatic light, bright and dark bands are obtained rather than repeating rainbow colors.
In a double-slit experiment, the fifth maximum is 2.8 cm from the central maximum on a screen that is 1.5 m away from the slits. If the slits are 0.15 mm apart, what is the wavelength of the light being used?
An effect analogous to two-slit interference can occur with sound waves, instead of light. In an open field, two speakers placed 1.30 m apart are powered by a single-function generator producing sine waves at 1200-Hz frequency. A student walks along a line 12.5 m away and parallel to the line between the speakers. She hears an alternating pattern of loud and quiet, due to constructive and destructive interference. What is (a) the wavelength of this sound and (b) the distance between the central maximum and the first maximum (loud) position along this line?
A film of oil on water will appear dark when it is very thin, because the path length difference becomes small compared with the wavelength of light and there is a phase shift at the top surface. If it becomes dark when the path length difference is less than one-fourth the wavelength, what is the thickest the oil can be and appear dark at all visible wavelengths? Oil has an index of refraction of 1.40.
A single slit of width 0.1 mm is illuminated by a mercury light of wavelength 576 nm. Find the intensity at a 10° angle to the axis in terms of the intensity of the central maximum.
Show that the distribution of intensity in a double-slit pattern is given by Equation 36.9. Begin by assuming that the total magnitude of the electric field at point P on the screen in Figure 36.4 is the superposition of two waves, with electric field magnitudes E1=E0sintE2=E0sin(t+) The phase angle in in E2 is due to the extra path length traveled by the lower beam in Figure 36.4. Recall from Equation 33.27 that the intensity of light is proportional to the square of the amplitude of the electric field. In addition, the apparent intensity of the pattern is the time-averaged intensity of the electromagnetic wave. You will need to evaluate the integral of the square of the sine function over one period. Refer to Figure 32.5 for an easy way to perform this evaluation. You will also need the trigonometric identity sinA+sinB=2sin(A+B2)cos(AB2)
Both 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.
The interference pattern of a He-Ne laser light (=632.9nm) passing through two slits 0.031 mm apart is projected on a screen 10.0 m away. Determine the distance between the adjacent bright fringes.

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