# 10.5.1 Single-Slit Interference Pattern

We have had so much fun increasing the number of slits from 2 to N. Now let’s go the other way. Let’s have only one single slit. When we shine a laser beam through a slit, we get an interference pattern shown below.

Wait. If there is only one slit, there is only one wave, right? How can a wave interfere with itself? Good that you asked. Now, there is something I have been hiding from you (because you were not ready for the truth): a slit is not exactly a point source. Because a slit has a width, so it is actually a continuous line of point sources. Rays from each point in the slit interferes with rays from every other point in the slit.

It is quite easy to understand what’s happening at $\theta =0$. All the rays travel the same distance (remember the parallel ray approximation) to the centre point on the screen. The path difference is zero. All the rays arrive in-phase and interfere constructively to produce the maximum possible intensity.

What happens on either side of $\theta =0$? Since the rays must now travel different distances to arrive at the destination P, the rays no longer arrive in-phase. As q increases, more and more cancellation take place among the rays.

Eventually, the first minimum angle q1 is reached at which the first complete destructive interference occurs – all the rays from the slit superpose to zero.

The first minimum angle is given by the formula $\displaystyle \sin {{\theta }_{1}}=\frac{\lambda }{b}$

where b is the slit width.

Beyond $\theta ={{\theta }_{1}}$, there are higher order minima occurring at angles θn given by the formula $\displaystyle \sin {{\theta }_{n}}=n\frac{\lambda }{b}$. In between every two consecutive minima is a mini maximum. However, the vast majority of the total light energy is contained in the central maximum so the higher order maxima are usually too dim to be visible. For this reason, perhaps, the H2 syllabus also ends at the first minimum.

In other words, the H2 syllabus only requires a qualitative understanding of the central maximum – the intensity peaks at the centre (where C.I. of all rays occurs), and tapers off towards both sides, hitting zero at the so-called first minimum (where D.I. of all rays occurs). And the only calculation you are required to do is the first minimum angle only, namely $\displaystyle \sin {{\theta }_{1}}=\frac{\lambda }{b}$.

For example, the formula predicts that narrowing the slit causes the emergent light beam to spread wider. This may be counter-intuitive at first, but makes total sense once you remind yourself that light is just doing its wave thingy here.

Example

Parallel light of wavelength 590 nm passing through a rectangular slit is incident on a screen. Calculate the width of the central fringe w as observed on the screen.

Solution

First, we calculate the first minimum angle q1. The formula is $\displaystyle \sin {{\theta }_{1}}=\frac{\lambda }{b}$. But since q1 is expected to be a very small angle, we can save some time by using the small angle approximation $\sin {{\theta }_{1}}\approx {{\theta }_{1}}$. \displaystyle \begin{aligned}(\sin {{\theta }_{1}}&=\frac{\lambda }{b})\\{{\theta }_{1}}&=\frac{{590\times {{{10}}^{{-9}}}}}{{0.60\times {{{10}}^{{-3}}}}}\\&=9.833\times {{10}^{{-4}}}\text{ rad}\end{aligned}

Since q1 is very small, w can be approximated by the arc length subtended by 2q1. \begin{aligned}(s&=r\theta )\\w&=L(2{{\theta }_{1}})\\&=2.4(2\times 9.833\times {{10}^{{-4}}})\\&=4.72\text{ mm}\end{aligned}

Demonstration

Laser Beam through a Single-Slit

Applet

Single-Slit (Walter Fendt)

Concept Test

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