Question

derivation for diffraction plz class 12

Answer 1

The light ray at the bottom of the slit has a phase lag, ϕϕ, compared with the ray at the top of the slit because it has to travel farther. Let's assume that the angle happens to be the one where the phase lag is 2π2π.

Now let's ask what the phase lag is for a light ray coming from somewhere in the slit between the top and bottom ray:

The light ray comes from a distance xxmeasured from the top of the slit so 0≤x≤a0≤x≤a. It is (hopefully) obvious from the diagram that the phase lag of this light ray is:

ϕ(x)=2πxa(1)(1)ϕ(x)=2πxa

Now consider two light rays, one coming from the position xx and one coming from x+a/2x+a/2. For example this could be the two rays you describe in the question, one from the top of the slit (x=0x=0) and one from the middle (x=a/2x=a/2), but we'll stick with the general case of any value of xx.

The phase lag of the ray from xx is given by equation (1) above, and the phase lag of the ray from x=a/2x=a/2 is given by:

ϕ(x+a/2)=2πx+a/2a=2π(xa+a/2a)=2π(xa+12)=2πxa+π=ϕ(x)+πϕ(x+a/2)=2πx+a/2a=2π(xa+a/2a)=2π(xa+12)=2πxa+π=ϕ(x)+π

So what we've found is that any two rays separated by a distance a/2a/2 have a phase difference of ππ. This answers your question as to why a ray from the top of the slit and the centre of the slit have a phase difference of ππ, but it's a more general result.

The importance of this result is that two waves with a phase difference of ππinterfere destructively and cancel each other out. So for every ray in the range 0≤x≤a/20≤x≤a/2, i.e. from the top half of the slit, there is a corresponding ray at x+a/2x+a/2, i.e. from the bottom half of the slit, that has a phase difference of ππ and therefore interferes destructively.

This means that all the light being emitted at the angle I've drawn interferes destructively and the total intensity is zero. This angle, i.e. the angle where the phase lag across the whole slit is 2π2π, corresponds to a dark region in the diffraction pattern.