The # of bright fringes in a double slit with finite width?

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SUMMARY

The discussion centers on calculating the number of visible bright fringes in a double-slit experiment using laser light with a wavelength of 633 nm, slit separation of 0.125 mm, and slit width of 0.015 mm. The formula for the number of bright fringes is established as 2(d/w) - 1, resulting in 15.67, which rounds down to 15 bright fringes. However, the textbook solution employs m = d/w, yielding 8, leading to 17 bright fringes when including the central fringe. The confusion arises from the overlap of bright fringes with dark fringes, necessitating clarification on visible fringes.

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Homework Statement


Laser light with a wavelength 633 nm is used to illuminate two slits separated by 0.125 mm. The width of each slit is 0.0150 mm. Assuming that only fringes between the first minima in the pattern are counted, how many bright fringes are visible?

lambda = 633nm
d = 0.125mm
w= 0.0150mm

Homework Equations


# of bright fringes visible = 2(d/w)-1

or

m = d/w

The Attempt at a Solution



I tried to use the first formula for the solution

2(d/w)-1

2(0.125/0.015)-1 = 15.67

therefore, 15 bright fringes are visible.

However, the solution manual for my textbook says to use m=d/w
which results in m=8. Therefore there are 2(8)+1 = 17 bright fringes visible. [/B]
I don't see how this is correct because the question asks for the # of visible bright fringes. Since the bright fringes at the end are not visible because they overlap the first order dark fringes. Can anyone clarify this for me?


Sorry if my post is formatted incorrectly... First time using this website.
 
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I would tend to agree with your reasoning since
sin (theta) = lambda / w for the angle to the first single slit minimum
Plugging this into
m = d sin (theta) / lambda does give you d/w for 8 bright fringes on either side
 
J Hann said:
m = d sin (theta) / lambda does give you d/w for 8 bright fringes on either side
Don't forget the central bright fringe.
 

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