Concept explainers
The distance between the first two adjacent dark fringes on the screen.
Answer to Problem 83PQ
The distance between the first two adjacent dark fringes on the screen is
Explanation of Solution
Write the expression for the phase difference for the dark fringes for wavelength
Here,
Write the expression for the phase difference for the dark fringes for wavelength
Here,
Divide equation (I) by (II)
Write the expression to calculate the value of
Here,
Write the expression to calculate the value of
Write the expression for the distance between the two adjacent dark fringes on the screen
Here,
Conclusion:
Substitute
Solve equation (VII) for the values of
Similarly,
Substitute
Substitute
Substitute
Therefore, the distance between the first two adjacent dark fringes on the screen is
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Chapter 35 Solutions
Physics for Scientists and Engineers: Foundations and Connections
- 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_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_forwardConsider 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?arrow_forward
- A beam of monochromatic green light is diffracted by a slit of width 0.550 mm. The diffraction pattern forms on a wall 2.06 m beyond the slit. The distance between the positions of zero intensity on both sides of the central bright fringe is 4.10 mm. Calculate the wavelength of the light.arrow_forwardIn Figure P36.10 (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? Figure P36.10arrow_forwardWhen a monochromatic light of wavelength 430 nm incident on a double slit of slit separation 5 m, there are 11 interference fringes in its central maximum. How many interference fringes will be in the central maximum of a light of wavelength 632.8 nm for the same double slit?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 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 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_forwardShow 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)arrow_forward
- A monochromatic beam of light of wavelength 500 nm illuminates a double slit having a slit separation of 2.00 105 m. What is the angle of the second-order bright fringe? (a) 0.050 0 rad (b) 0.025 0 rad (c) 0.100 rad (d) 0.250 rad (e) 0.010 0 radarrow_forwardRed light (wavelength 632.8 nm in air) from a Helium-Neon laser is incident on a single slit of width 0.05 mm. The entire apparatus is immersed in water of refractive index 1.333. Determine the angular width of the central peak.arrow_forwardUsing the result of the problem two problems prior, find the wavelength of light that produces fringes 7.50 mm apart on a screen 2.00 m from double slits separated by 0.120 mm.arrow_forward
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