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

In summary: I'm not sure why the textbook answer would give you 2(8)+1 = 17 bright fringes visible.In summary, when using a laser light with a wavelength of 633 nm to illuminate two slits separated by 0.125 mm with a width of 0.0150 mm each, the number of visible bright fringes can be calculated using the equation m = d/w, which results in 8 bright fringes on either side. This includes the central bright fringe. However, the solution manual for the textbook suggests using the equation 2(d/w)-1, which results in 15 bright fringes. It is unclear why the solution manual gives this answer as it does not take into account overlapping with the
  • #1
Plana
1
0

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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  • #3
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
 
  • #4
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.
 

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

1. What is the relationship between the width of the double slit and the number of bright fringes observed?

The width of the double slit has a direct impact on the number of bright fringes that can be observed. As the width of the slits increases, the number of bright fringes also increases. This is because a wider slit allows for more light to pass through, resulting in more interference patterns and bright fringes.

2. Can the number of bright fringes be altered by changing the distance between the double slit and the screen?

No, the distance between the double slit and the screen does not affect the number of bright fringes observed. This is because the distance only affects the spacing between the fringes, not the actual number of fringes.

3. What happens to the number of bright fringes when the wavelength of light is changed?

The number of bright fringes remains the same when the wavelength of light is changed. This is because the number of fringes is determined by the distance between the slits and the screen, not the wavelength of light.

4. How does the number of bright fringes change when the distance between the slits is altered?

The number of bright fringes increases as the distance between the slits decreases. This is because a shorter distance between the slits results in a wider interference pattern, leading to more bright fringes being observed on the screen.

5. Is there a maximum number of bright fringes that can be observed in a double slit with finite width?

No, there is no maximum number of bright fringes in a double slit with finite width. Theoretically, as the width of the slit decreases, the number of fringes would increase infinitely. However, in practice, the diffraction limit of light would eventually limit the number of fringes that can be observed.

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