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What is the distance on screen C in Fig. 35-11a between adjacent maxima near the center of the interference pat- tern? The wavelength lambda of the light is 546 nm, the slit separation d is 0.12 mm, and the slit-screen separation D is 55 cm. Assume that theta in Fig. 35-11 is small enough to permit use of the approximations sin theta ~~ tan theta ~~ theta, in which theta is expressed in radian measure. |
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Answer» Solution :KEY IDEAS (1) First, let us pick a maximum with a low value of m to ensure that it is near the center of the pattern. Then, from the GEOMETRY of Fig. 35-11a, the maximum.s vertical distance `y_(m)` from the center of the pattern is related to its angle `theta` from the central axis by `tan theta ~~ theta =(y_(m))/(D)` (2) From Eq. 35-14, this angle o for the mth maximum is given by `sin theta ~~ theta =(m lambda)/(d)` Calculations: If we equate our two expressions for angle `theta` and then solve for `y_(m)`, we find `y_(m)=(m lambda D)/(d)""(35-20)` For the next maximum as we move away from the pattern.s center, we have `y_(m+1)=((m+1)lambda D)/(d)""(35-21)` We find the distance between these adjacent maxima by subtracting Eq. 35-20 from Eq. 35-21: `Delta y=y_(m+1)-y_(m)=(lambda D)/(d)` `=((546 xx 10^(-9) m)(55 xx 10^(-2) m))/(0.12 xx 10^(-3)m)` `= 2.50 xx 10^(-3) m ~~ 2.6 mm`. (ANSWER) As long as d and in Fig. 35-11a are SMALL, the separation of the interference fringes is independent of m, that is, the fringes are evenly spaced. |
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