What is the Width of the Central Maximum in Single Slit Diffraction?

In summary, the width of the central maximum in a single slit diffraction pattern can be calculated using the equation sin(alpha) = (#)*l/a, where "#" is any half integer. For the given problem with a single slit of width 10-4m and a helium-neon laser with wavelength 6.328*10-7m, the width of the central maximum is approximately 0.19m when the observing screen is 10m away.
  • #1
Patrickas
20
0

Homework Statement


A single slit 10-4m wide(a) is illuminated by plane waves from helium-neon laser ( l = 6.328*10-7m). If the observing screen is 10m(D) away, determine the width of the central maximum(y).



Homework Equations



sin(alpha)=(m+0.5)*l/a


tan(aplha)~sin(alpha)=y/2*D =(I add 2 because its double the distance from the middle to the first minimum above and below the maximum)

y=((1+0.5)*l*2*D)/d


The Attempt at a Solution


plug in the numbers and i get it as 0.19m which is incorrect.

Help anyone?
 
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  • #2
Patrickas said:

Homework Statement


A single slit 10-4m wide(a) is illuminated by plane waves from helium-neon laser ( l = 6.328*10-7m). If the observing screen is 10m(D) away, determine the width of the central maximum(y).



Homework Equations



sin(alpha)=(m+0.5)*l/a


tan(aplha)~sin(alpha)=y/2*D =(I add 2 because its double the distance from the middle to the first minimum above and below the maximum)

y=((1+0.5)*l*2*D)/d


The Attempt at a Solution


plug in the numbers and i get it as 0.19m which is incorrect.

Help anyone?

alright i seem to have chosen wrong m... i don't understand why is there a path difference of half the wave length then? I mean I understand the double slit interference- that the additional lengh 1 ray must go must be equal to full wave to get constructive interferance. But with single slit i just get lost...
 
  • #3
Patrickas said:

Homework Equations



sin(alpha)=(m+0.5)*l/a
Another way to think of that is:

sin(alpha) = (#)*l/a​

where "#" can be any half integer, i.e. ..., -3/2, -1/2, +1/2, +3/2, ...

The two dark bands closest to the central maximum correspond to "#" = -1/2 and +1/2. So use 1/2, then double the result to get the full width.
 

Related to What is the Width of the Central Maximum in Single Slit Diffraction?

1. What is single slit diffraction and why is it important in science?

Single slit diffraction is a phenomenon in which a wave, such as light or sound, passes through a single narrow opening and spreads out into a series of concentric circles or arcs. It is important in science because it helps us understand the behavior of waves and their interactions with different objects and barriers.

2. How does the width of the slit affect the diffraction pattern?

The width of the slit directly affects the diffraction pattern. A narrower slit will produce a wider diffraction pattern, while a wider slit will produce a narrower diffraction pattern. This is because the narrower the slit, the more the wave is diffracted and spread out.

3. What is the relationship between the wavelength of the wave and the diffraction pattern?

The wavelength of the wave is inversely related to the size of the diffraction pattern. This means that as the wavelength decreases, the diffraction pattern becomes wider and vice versa. This relationship is known as the diffraction limit.

4. Can single slit diffraction be observed with all types of waves?

Yes, single slit diffraction can be observed with all types of waves, including light, sound, and water waves. However, the diffraction patterns may vary depending on the properties of the specific wave being used.

5. How is single slit diffraction different from double slit diffraction?

Single slit diffraction is the diffraction of a wave through a single narrow opening, while double slit diffraction is the diffraction of a wave through two narrow openings. In double slit diffraction, the two slits act as coherent sources, creating a more complex diffraction pattern compared to single slit diffraction.

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