Concept explainers
(a)
To show: The point where
(a)
Answer to Problem 48AP
Explanation of Solution
Given info: The equation of the intensity of the light in the diffraction pattern is
The formula to calculate the intensity of the light is,
Here,
The value of
Substitute
Conclusion
Therefore, the point where
(b)
To draw: Plot
(b)
Answer to Problem 48AP
The graph between
Figure (1)
Explanation of Solution
Given info: The equation of the intensity of the light in the diffraction pattern is
The equation of
The solution of both the equation to coincide at a point is ,
So the solution of the transcendental equation is
(c)
To show: The angular full width at half maximum of the central diffraction maximum is
(c)
Answer to Problem 48AP
Explanation of Solution
Given info: The equation of the intensity of the light in the diffraction pattern is
The formula to calculate the phase angle is,
Rewrite the above equation for
If the value of
The path covered by the light is symmetric so the phase angle is double the initial value.
Substitute
Conclusion
Therefore, the angular full width at half maximum of the central diffraction maximum is
(d)
The number of steps involved to solve the transcendental equation ϕ = 2 sin ϕ
.
(d)
Answer to Problem 48AP
Explanation of Solution
Given info: The equation of the intensity of the light in the diffraction pattern is
The equation of
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The solution of the transcendental equation
Conclusion
Therefore, the number of steps involved to solve the transcendental equation
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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_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 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_forward
- (a) Find the angle of the third diffraction minimum for 633-nm light falling on a slit of width 20.0 m. (b) What slit width would place this minimum at 85.0°?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_forwardIf 570 nm light falls on a slit 4.20×10-2 mm wide, what is the full angular width of the central diffraction peak?arrow_forward
- 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?arrow_forwardIn a double slit experiment, if the separation between the two slits is 0.050 mm and the distance from the slits to a screen is 2.5 m, find the spacing between the first-order and second-order bright fringes when coherent light of wavelength 600 nm illuminates the slits. A) 1.5 cm B) 3.0 cm C) 4.5 cm D) 6.0 cm E) 9.0 cmarrow_forwardWhat must be the ratio of the slit width to the wavelength for a single slit to have the first diffraction minimum at u = 45.0°?arrow_forward
- Consider a double-slit diffraction experiment with slits of width 0.260 mm. Monochromatic light of wavelength 435 nm is used. If the value of the parameter for a point that is an angular distance of 0.0120 rad from the center of the central diffraction peak is 31.76 rad, what must be the slit separation (in mm)?arrow_forward350 nm of light falls on a single slit of width 0.20 mm. What is the angular width of the central diffraction peak?arrow_forwardIn a single-slit diffraction experiment, monochromatic light of wavelength 505 nm is passed through a slit 0.320 mm wide, and the diffraction pattern is observed on a screen 7.14 m from the slit. The intensity at the center of the pattern is ?0. What is the ratio of the intensity at the center of the pattern to the intensity at a point 3.09 mm from the center of the diffraction pattern (?/?0)?arrow_forward
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