$single slit diffraction pattern$

This apparatus allows us to vary the slit width, but the pattern is still projected on a distant screen to make the diffraction effects cl.

Here we use phasor addition to derive the intensity as a function of angle for both situations.

Lets assume that the slit is constant width and very tall compared with that width, so that we can consider the system as two-dimensional.

With light at normal incidence, the pattern is symmetrical about the axis of the slit.

On a distant screen, the light arriving on the axis from all points in the slit has travelled an equal distance from the slit, so the centre of the pattern is a maximum.

We call the slit width a, and we imagine it divided into two equal halves.

Using the Huygens construction, we consider a point at the very top of the slit, and another point a distance a 2 below it, i.e.

Consider parallel rays from both points, at angle to the axis of symmetry.

Why parallel Because the screen is distant.

Typically in diffraction experiments, the slit is 10 m wide, while the distance to the screen might be 1 m.).

$single slit diffraction pattern$

On the other side of the axis of symmetry, sin a is also a minimum.

Each half is divided into quarters, and light from a source in the first quarter cancels that from one in the second quarter.

Similarly, sources in the third quarter are nullified by those in the fourth quarter.

So this diagram represents the second order minima, where sin (a2), or sin 2 a.

To calculate I ( ), we use Huygens construction and phasor addition.

The path difference between rays from successive slices of the slit are equal, and so too are the angles between successive phasors.

The slices of the slit all have equal width and length, so the lengths of all the phasors are equal.

For a large number of slices, the phasors approximate the arc of a circle, as shown in red.

Now 2 is also the phase difference between the first and the last phasor, which is 2 a sin.

On the axis, where the phasors are all in phase, the phasor sum is the straight line shown in red at right.

This is the amplitude of the diffraction patter at 0, which we call A 0.

By the definition of angle, 2 A 0 R, which gives us R A 0 2.

As weve seen in several previous chapters, the intensity I is proportional to square of the amplitude A.

So, using.

The vertical black line scanning across the graph is in step with the phasor diagram.

To my eyes, at least, the difference in brightness seems much less than the difference in the intensity graph.

$single slit diffraction pattern$

Single Slit Diffraction Pattern