The separation between the slits is d = 8.96
What is fringe width?
- Fringe width is the distance between two consecutive bright spots (maximas, where constructive interference take place)
- Or two consecutive dark spots (minimas, where destructive interference take place).
Fringe width is given by β = λL/d
In the first case fringe width is β1 = λLA /d = 625 x 10-9 x 0.36 / ( 1.4 x 10-5 ) = 0.016071428 m
The total width of the screen is 0.2 m . So, on one side of the central maximum, the width is 0.1 m
No. of fringes in this 0.1m = 0.1 / 0.016071428 = 6.222
So, since there is a bright fringe after every fringe width, the number of bright fringes on one side of central maximum is 6.
In the second case fringe width is β1 = λLAB /d = 625 x 10-9 x 0.25 / ( 1.4 x 10-5 ) = 0.011160714 m
The total width of the screen is 0.2 m . So, on one side of the central maximum, the width is 0.1 m
No. of fringes in this 0.1m = 0.1 / 0.011160714 = 8.96
So, since there is a bright fringe after every fringe width, the number of bright fringes on one side of central maximum is 8. The ninth one will not be seen since the screen is less a little less in width.
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<u>The complete question is -</u>
In a setup like that in Figure 27.7, a wavelength of 625 nm is used in a Young's double-slit experiment. The separation between the slits is d = 1.4 × 10-3 m. The total width of the screen is 0.20 m. In one version of the setup, the separation between the double slit and the screen is LA = 0.36 m, whereas in another version it is LB = 0.25 m. On one side of the central bright fringe, how many bright fringes lie on the screen in the two versions of the setup? Do not include the central bright fringe in your counting. --Tm = 3 (Bright fringe) ++m = 0 (Bright fringe) -m = 3 (Bright fringe) Figure 27.7
