A problem using Young's equation from the double-slit experiment

In summary, the question discusses the use of coherent light with wavelength 400 nm passing through two very narrow slits separated by 0.200 mm, resulting in an interference pattern observed on a screen 4.00 m away. The question asks for the width of the central interference maximum, and the conversation reveals that the approach of doubling the distance of the first maximum from the center was incorrect. Instead, the width of the central maximum can be estimated by finding the distance between the first minima, which can be calculated using the equation d \cdot \sin(\theta) = (m + \frac{1}{2}) \cdot \lambda.
  • #1
erik-the-red
89
1
Question:

Coherent light with wavelength 400 nm passes through two very narrow slits that are separated by 0.200 mm and the interference pattern is observed on a screen 4.00 m from the slits.

A

What is the width (in mm) of the central interference maximum?

[tex]y_m = R \cdot \frac{m \cdot \lambda}{d}[/tex]

I thought I would get the width of the central interference maximum by doubling [tex]y_m = 4.00 \cdot \frac{4.00 \cdot 1 \cdot 4.00 \cdot 10^(-9)}{.00200} = .016[/tex].

That's not correct, though.

Why was my approach incorrect?
 
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  • #2
for central maximum doesn't m=0?
 
  • #3
But, does [tex]y_0 = 0[/tex] make sense physically?
 
  • #4
Your substitution doesn't look right to me

YOu appear to have one too many 4.0s in there.

400 nm is either 400 x 10^-9 or 4 x 10^-7 - you appear to have used 4.0 x 10^-9

2mm is 0.0002m, not 0.002

[Edited to correct should read 0.2 = 0.0002 not 0.002]

Incidentally, I don't know whether that's why you've gone wrong, because either using the extra 4 or not, I can see the answer being 0.016. 4x4 is 16 so divided by 2 should give you an 8 - using an extra 4 would give you a 32 in your answer.
 
Last edited:
  • #5
rsk said:
2mm is 0.0002m, not 0.002

2 [mm] = 0.002 [m].
 
  • #6
radou said:
2 [mm] = 0.002 [m].

My typing mistake - he's got 0.2 mm in the original, which gives 0.0002
 
  • #7
erik-the-red said:
Why was my approach incorrect?
It looks like you found the distance of the first maximum (m = 1) from the center and then doubled it. (There are typos in your expression.) That gives you the distance between the two first non-central maxima, which is not the same thing as the width of the central maximum. A good estimate of the width of the central maximum would be the distance between the first minima. (For double slit interference, the minima are centered between adjacent maxima.)
 
  • #8
Thanks for all the responses. Indeed, my tex equation had numerous typos, I apologize for those.

I was finally able to do the problem by using [tex]d \cdot \sin(\theta) = (m + \frac{1}{2}) \cdot \lambda[/tex] and its corresponding constructive counterpart.

I used the equation [tex]y_m = R \cdot \sin(\theta_m)[/tex], and I got my answers for part A and part B (not shown).
 

Related to A problem using Young's equation from the double-slit experiment

1. What is Young's equation and how is it used in the double-slit experiment?

Young's equation is a mathematical formula that describes the interference pattern created by light passing through two adjacent slits. It is used in the double-slit experiment to calculate the location of bright and dark fringes on a screen placed behind the slits, based on the wavelength of the light and the distance between the slits.

2. How do we know that Young's equation accurately describes the double-slit experiment?

Young's equation has been extensively tested and confirmed through numerous experimental observations. It has also been derived from fundamental principles of wave optics, providing a theoretical basis for its accuracy in describing the double-slit experiment.

3. Can Young's equation be applied to other types of waves besides light?

Yes, Young's equation can be applied to any type of wave, as long as the wave's wavelength is small compared to the distance between the two slits. This includes water waves, sound waves, and even quantum particles like electrons.

4. How does changing the distance between the slits affect the interference pattern in the double-slit experiment?

The distance between the slits plays a crucial role in determining the location and intensity of the bright and dark fringes on the screen. As the distance between the slits increases, the spacing between the fringes also increases, resulting in a more spread-out interference pattern.

5. Are there any limitations or assumptions to using Young's equation in the double-slit experiment?

Young's equation assumes that the light is passing through two equally sized and equally spaced slits, and that the light is monochromatic (having a single wavelength). It also does not take into account any secondary effects such as diffraction or scattering, which may affect the observed interference pattern.

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