Concept explainers
Coherent light of wavelength 501.5 nm is sent through two parallel slits in an opaque material. Each slit is 0.700 μm wide. Their centers are 2.80 μm apart. The light then falls on a semicylindrical screen, with its axis at the midline between the slits. We would like to describe the appearance of the pattern of light visible on the screen. (a) Find the direction for each two-slit interference maximum on the screen as an angle away from the bisector of the line joining the slits. (b) How many angles are there that represent two-slit interference maxima? (c) Find the direction for each single-slit interference minimum on the screen as an angle away from the bisector of the line joining the slits. (d) How many angles are there that represent single-slit interference minima? (e) How many of the angles in part (d) are identical to those in part (a)? (f) How many bright fringes are visible on the screen? (g) If the intensity of the central fringe is Imax, what is the intensity of the last fringe visible on the screen?
(a)
Answer to Problem 38.12P
Explanation of Solution
Given information: The wavelength of light is
The condition for double slit interference maxima is,
Here
Further solve equation (1) as;
Substitute
For different value of
For
Substitute
Thus for
For
Substitute
Thus for
For
Substitute
Thus for
For
Substitute
Thus for
For
Substitute
Thus for
For
Substitute
Thus for
For
Substitute
The value of
Conclusion:
Therefore, there are
(b)
Answer to Problem 38.12P
Explanation of Solution
The calculation in part (a) shows that for zero order there is one angle and for first, second, third, fourth and fifth order there are each two direction for a single that represent the two slit interference maxima
The possible angles are
Conclusion:
Therefore, there are
(c)
Answer to Problem 38.12P
Explanation of Solution
Given Info: The condition for the interference minima in single slit interference minima is,
Here,
Substitute
For different value of
For
Substitute
Thus for
For
Substitute
The value of
Thus up to second order the single slit interference is possible.
Conclusion:
Therefore, the possible directions are
(d)
Answer to Problem 38.12P
Explanation of Solution
The calculation in part (c) shows that for first order only the single slit interference minima is possible.
The possible angles are
So there are total
Conclusion:
Therefore, there are
(e)
Answer to Problem 38.12P
Explanation of Solution
The calculation in part (a) and part (c) shows that the angles
So there are total
Conclusion:
Therefore, there are
(f)
Answer to Problem 38.12P
Explanation of Solution
The calculation in part (a) and part (c) shows that the angles
So, for the position at
Thus there are
So, there are
Conclusion:
Therefore, there are
(g)
Answer to Problem 38.12P
Explanation of Solution
The formula to calculate the intensity at any angle is,
Here,
The last fringe occurs for the fifth order so the value of
Substitute
Conclusion:
Therefore, the intensity at the last fringe is
Want to see more full solutions like this?
Chapter 38 Solutions
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
- 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_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_forwardCoherent 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_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_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
- For 600-nm wavelength light and a slit separation of 0.12 mm, what are the angular positions of the first and third maxima in the double slit interference pattern?arrow_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_forwardA 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_forward
- Monochromatic light is incident on a pair of slits that are separated by 0.200 mm. The screen is 2.50 m away from the slits. a. If the distance between the central bright fringe and either of the adjacent bright fringes is 1.67 cm, find the wavelength of the incident light. b. At what angle does the next set of bright fringes appear?arrow_forwardBoth 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.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
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningUniversity Physics Volume 3PhysicsISBN:9781938168185Author:William Moebs, Jeff SannyPublisher:OpenStaxPhysics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning
- Physics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningGlencoe Physics: Principles and Problems, Student...PhysicsISBN:9780078807213Author:Paul W. ZitzewitzPublisher:Glencoe/McGraw-Hill