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

[Patrickas]

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?

 

[Patrickas]

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...

 

[Redbelly98]

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.