Concept explainers
Monochromatic coherent light of amplitude E0 and angular frequency ω passes through three parallel slits, each separated by a distance d from its neighbor. (a) Show that the time-averaged intensity as a function of the angle θ is
(b) Explain how this expression describes both the primary and the secondary maxima. (c) Determine the ratio of the intensities of the primary and secondary maxima. Hint: See Problem 16.
(a)
To show: The time averaged intensity as a function of angle
Answer to Problem 18P
Explanation of Solution
Given info: The amplitude of monochromatic light is
The amplitude of the monochromatic light is,
Here,
The total amplitude of monochromatic light is,
Substitute
Apply the trigonometric identity to the above expression as,
Apply the trigonometric identity to the above expression as,
Add the above result to
As it is known that intensity of monochromatic light is directly proportional to the the square of the electric field that is,
Here,
The resultant field is obtained by square the above expression as,
Substitute
Conclusion:
Thus, the time averaged intensity as a function of angle
(b)
Answer to Problem 18P
Explanation of Solution
Given info: The amplitude of monochromatic light is
From the above expression obtained in part (a), the minimum interference is obtained when
Conclusion:
Thus, the expression describes both the primary and secondary maxima.
(c)
Answer to Problem 18P
Explanation of Solution
Given info: The amplitude of monochromatic light is
Consider the amplitude of the monochromatic light is,
The total amplitude of monochromatic light is,
Substitute
Apply the trigonometric identity to the above expression as,
Apply the trigonometric identity to the above expression as,
Add the above result to
As it is known that intensity of monochromatic light is directly proportional to the square of the electric field that is,
Here,
The resultant field is obtained by square the above expression as,
Substitute
The expression for the intensity for primary maxima is,
The expression for the intensity for secondary maxima is,
Take the ratio of the above two expression as,
Conclusion:
Therefore, the ratio of the intensities of the primary and secondary maxima is
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Chapter 36 Solutions
Physics for Scientists and Engineers
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