General questions about Fraunhofer diffraction

[bcjochim07]

Ok, so my book explains single-slit diffraction like this:

For a single slit with width a, for every point on the wave front there is another point a/2 distance away that it can be paired with.

1) If the distance (a2)*sin(theta) is half a wavelength, the interference is destructive.
So equivalently the condition for this destructive interference is sin(theta)= lambda


That's fine with me, however, here's the part that is a little confusing to me:

We can extend this idea to find other angles of perfect away. destructive interference. Suppose each wavelet is paired with another wavelet from a point a/4 away. Replacing a/2 with a/4in the above equation, the condition for destructive interference becomes a*sin(theta)= 2lambda.

So the general condition for dest. interference is a*sin(theta)=p*lambda , p=1,2,3...

But, you cannot replace the a/2 with anything with an odd denominator, such as a/3. Could someone explain the whole logic of picking a point a/2 or a/4 away?

Also, in my lab section, we produced double-slit diffraction and traced it on paper. We observed two less noticeable diffraction minima on each side. We used the positions of these to calculate the width of our slit (a) with the equation asin(theta)=p*lambda. Why did these diffraction minima occur, and would their p value in the above general condition be 1?

 

[eljose79]

I can provide some clarification on the concepts of single-slit diffraction and the general condition for destructive interference.

First, let's start with the explanation of single-slit diffraction given in your book. This explanation is based on the Huygens-Fresnel principle, which states that every point on a wavefront can be considered as a source of secondary waves. In the case of a single slit, these secondary waves interfere with each other to produce a diffraction pattern. When considering the points on the wavefront, each point has a corresponding point a/2 distance away that it can be paired with. This means that when the distance (a/2)sin(theta) is equal to half a wavelength, the secondary waves from these two points will interfere destructively, resulting in a dark spot in the diffraction pattern. This is the condition for destructive interference in a single-slit diffraction pattern.

Now, let's address the part that is confusing to you. The book mentions that we can extend this idea to find other angles of perfect destructive interference by considering points a/4 distance away instead of a/2. This means that when the distance (a/4)sin(theta) is equal to half a wavelength, destructive interference will occur. This can be seen as a generalization of the condition for destructive interference in a single slit, where the distance a/2 is replaced by any distance a/p, where p is an integer. This leads to the general condition for destructive interference in a single slit: a*sin(theta)=p*lambda, where p=1,2,3...

Now, why can't we replace a/2 with anything with an odd denominator, such as a/3? This is because the Huygens-Fresnel principle only holds for points that are a multiple of half the wavelength apart. This means that when considering points a/3 distance away, the secondary waves from these points will not interfere destructively. This is why we cannot use a/3 in the above equation.

Moving on to the double-slit diffraction pattern observed in your lab section, the two less noticeable diffraction minima on each side are known as secondary maxima. These occur due to the interference between the secondary waves from the two slits. The general condition for destructive interference in a double-slit diffraction pattern is given by d*sin(theta)=p*lambda, where d is the distance between the two slits.

 

[astrokreng]

I can explain the logic behind the selection of points a/2 and a/4 in the equations for destructive interference in single-slit diffraction.

When a wave passes through a single slit, it diffracts and creates a pattern of bright and dark fringes on a screen placed behind the slit. The bright fringes occur at points where the waves from different parts of the slit arrive in phase and constructively interfere, while the dark fringes occur where the waves arrive out of phase and destructively interfere.

In order to determine the conditions for destructive interference, we need to consider the path difference between the waves coming from different parts of the slit. For a single slit, the path difference is directly proportional to the angle of diffraction (theta) and the width of the slit (a).

When we consider a point on the wave front and its paired point a/2 distance away, the path difference between these two points is half a wavelength, as mentioned in the first equation. This results in destructive interference at that particular angle.

Similarly, if we consider a point on the wave front and its paired point a/4 distance away, the path difference between these two points is a full wavelength. This also results in destructive interference at that angle.

Now, when we replace a/2 with any other value, such as a/3, the path difference between the two points will not be a multiple of half a wavelength. This means that the waves will not be completely out of phase and there will not be complete destructive interference.

As for the double-slit diffraction, the two less noticeable diffraction minima on each side are called secondary maxima. These occur due to the interference between the waves passing through the two slits. The positions of these minima can also be calculated using the equation asin(theta)=p*lambda, where p=1,2,3... However, their p value would be 2, as they are the second set of bright fringes on either side of the central maximum.

I hope this explanation helps to clarify the logic behind the equations and their applications in single-slit and double-slit diffraction.