Young's Experiment (destructive interference)

In summary: Indeed I use a different notation myself, but that really doesn't change the logic too much. Just use ##\frac{wy}{L} = (n - \frac{1}{2}) \lambda## then.
  • #1
roam
1,271
12

Homework Statement



I'm having some trouble understanding the following solved problem:

In Young’s experiment, narrow double slits 0.20 mm apart diffract monochromatic light onto a screen 1.5 m away. The distance between the 5th minima on either side of the zeroth-order maximum is measured to be 34.73 mm. Determine the wavelength of the light.

Solution:

Position of mth minima relative to central maximum is:

##(m-\frac{1}{2}) \lambda = a \ sin \theta =\frac{ay_m}{L} \implies y_m = \frac{L(m-\frac{1}{2})\lambda}{a}##

Distance on the screen between the 5th minima on either side of the central maximum is:

##\Delta y = y_5-y_{-5} = \frac{(4.5-(-4.5))L \lambda}{a} = \frac{9L \lambda}{a}##

##\therefore \ \lambda = \frac{\Delta y a}{9L} = 514 \ nm##

I don't understand why they have used ##(m-\frac{1}{2}) \lambda## as the condition for destructive interference, instead of ##(m+\frac{1}{2}) \lambda##?

Homework Equations



For destructive interference my textbook uses:

##d \ sin \theta_{dark} = (m+\frac{1}{2}) \lambda##

(this is an approximation)

The Attempt at a Solution



If I use -1/2, then 5th minima turn out to be are 9λ apart, just like the model answer.

However when I use +1/2 I get a totally different solution:

##\therefore \ \lambda = \frac{\Delta y a}{(5.5-(-5.5)) L}= \frac{\Delta y a}{11 L} =421 \ nm##

So, why do they use -1/2? And which method is correct? :confused:

Any help is greatly appreciated.
 
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  • #2
I think both you and your book might be wrong on this one. I'm getting ##1029nm## as the answer.

I used: ##\frac{wx_n}{L} = (n - \frac{1}{2}) \lambda##

Where ##n## represents a nodal line number.
 
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  • #3
Zondrina said:
I think both you and your book might be wrong on this one. I'm getting ##1029nm## as the answer.

I used: ##\frac{wx_n}{L} = (n - \frac{1}{2}) \lambda##

Where ##n## represents a nodal line number.

No, that's wrong because you are not given y, you are given ##\Delta y##.
 
  • #4
My textbook ("Optics" by Hecht) uses "m+1/2", while my other book ("Introduction to Optics" by Pedrotti) uses "m-1/2" for destructive interference. The two methods clearly produce different results! :confused:

Why in this particular situation, do we need "-1/2" and not "+1/2"?
 
  • #5
roam said:
No, that's wrong because you are not given y, you are given ##\Delta y##.

No, you are not given ##\Delta x##.

##\Delta x## represents the distance between two consecutive nodal or antinodal fringes.

##x_n## represents the distance from a minimum to the right bisector (central maximum).

The distance between the 5th minima on either side of the zeroth-order maximum is measured to be 34.73 mm

They are not talking about the distance between two consecutive node/antinodal fringes, rather the distance from the 5th minima on either side of the fringe pattern to the right bisector.

Also, because the fringe pattern is symmetric, I believe it wouldn't matter whether you used ##(n - 1/2)## or ##(n + 1/2)##.

It depends on which direction the author likes to measure I believe.
 
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  • #6
Zondrina said:
No, you are not given ##\Delta x##.

##\Delta x## represents the distance between two consecutive nodal or antinodal fringes.

##x_n## represents the distance from a minimum to the right bisector (central maximum).

You are using a different notation. I never defined ##\Delta y## as the distance between consecutive fringes! It is the distance between the position of 5th minima on one side, to the 5th minima on the other.

##y## itself is the linear positions of fringes measured along the screen from the bisector you mentioned.
 
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  • #7
roam said:
You are using a different notation. I never defined ##\Delta y## as the distance between consecutive fringes! It is the distance between the position of 5th minima on one side, to the 5th minima on the other.

##y## itself is the linear positions of fringes measured along the screen from the bisector you mentioned.

Indeed I use a different notation myself, but that really doesn't change the logic too much.

Just use ##\frac{wy}{L} = (n - \frac{1}{2}) \lambda## then.
 

1. What is Young's Experiment and how does it demonstrate destructive interference?

Young's Experiment is a classic experiment used to demonstrate the wave nature of light. It involves a beam of light passing through a double slit and producing an interference pattern on a screen. Destructive interference occurs when two waves of equal amplitude and opposite phase overlap, resulting in a cancellation of the waves and a dark fringe on the screen.

2. What are the key components of Young's Experiment?

The key components of Young's Experiment are a coherent light source, a double slit, and a screen. A coherent light source produces waves of a single frequency and wavelength, while the double slit creates two coherent sources of light. The screen is used to observe the interference pattern produced by the overlapping waves.

3. How does the distance between the slits affect the interference pattern in Young's Experiment?

The distance between the slits, also known as the slit separation, affects the interference pattern by changing the spacing between the bright and dark fringes on the screen. A smaller slit separation results in a wider spacing between fringes, while a larger slit separation results in a narrower spacing. This is because a smaller slit separation leads to a greater difference in path length between the two waves, causing a larger phase difference and more pronounced destructive interference.

4. How does the wavelength of light affect the interference pattern in Young's Experiment?

The wavelength of light also affects the interference pattern by changing the spacing between fringes. A shorter wavelength results in a smaller spacing between fringes, while a longer wavelength results in a larger spacing. This is because the wavelength determines the distance between wave crests, and a shorter wavelength leads to a smaller difference in path length between the two waves and a smaller phase difference.

5. What other factors can affect the interference pattern in Young's Experiment?

Other factors that can affect the interference pattern in Young's Experiment include the width of the slits, the distance between the slits and the screen, and the angle at which the light passes through the slits. A wider slit or a greater distance between the slits and screen can result in a broader interference pattern, while changing the angle of incidence can affect the position of the fringes on the screen.

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