Extension: Calculating Maxima and Minima

There is a formula we can use to determine where the peaks and minima are in the interference spectrum. There will be more than one minimum. There are the same number of minima on either side of the central peak and the distances from the first one on each side are the same to the peak. The distances to the peak from the second minimum on each side is also the same, in fact the two sides are mirror images of each other. We label the first minimum that corresponds to a positive angle from the centre as \(m=1\) and the first on the other side (a negative angle from the centre) as \(m=-1\), the second set of minima are labelled \(m=2\) and \(m=-2\) etc.

The equation for the angle at which the minima occur is given in the definition below:

Definition: Interference Minima

The angle at which the minima in the interference spectrum occur is:

\(\sin\theta =\frac{m\lambda }{w}\)

where

θ is the angle to the minimum

w is the width of the slit

λ is the wavelength of the impinging wavefronts

m is the order of the minimum, \(m=±1,±2,±3,...\)

Example: Diffraction Minimum

Question

A slit with a width of \(\text{2 511}\) \(\text{nm}\) has red light of wavelength \(\text{650}\) \(\text{nm}\) impinge on it. The diffracted light interferes on a surface. At which angle will the first minimum be?

Step 1: Check what you are given

We know that we are dealing with diffraction patterns from the diffraction of light passing through a slit. The slit has a width of \(\text{2 511}\) \(\text{nm}\) which is \(\text{2 511} \times \text{10}^{-\text{9}}\) \(\text{m}\) and we know that the wavelength of the light is \(\text{650}\) \(\text{nm}\) which is \(\text{650} \times \text{10}^{-\text{9}}\) \(\text{m}\). We are looking to determine the angle to first minimum so we know that \(m=1\).

Step 2: Applicable principles

We know that there is a relationship between the slit width, wavelength and interference minimum angles:

\(\sin\theta =\frac{m\lambda }{w}\)

We can use this relationship to find the angle to the minimum by substituting what we know and solving for the angle.

Step 3: Substitution

\begin{align*} \sin \theta & = \dfrac{\text{650} \times \text{10}^{-\text{9}}\text{ m}}{\text{2 511} \times \text{10}^{-\text{9}}\text{ m}} \\ \sin \theta & = \dfrac{\text{650}}{\text{2 511}} \\ \sin \theta & = \text{0.258861012} \\ \theta & = \sin ^{-1}\text{0.258861012} \\ \theta & = 15 ° \end{align*}

The first minimum is at \(\text{15}\)\(\text{°}\) from the centre maximum.

Example: Diffraction Minimum

Question

A slit with a width of \(\text{2 511}\) \(\text{nm}\) has green light of wavelength \(\text{532}\) \(\text{nm}\) impinge on it. The diffracted light interferes on a surface, at what angle will the first minimum be?

Step 1: Check what you are given

We know that we are dealing with diffraction patterns from the diffraction of light passing through a slit. The slit has a width of \(\text{2 511}\) \(\text{nm}\) which is \(\text{2 511} \times \text{10}^{-\text{9}}\) \(\text{m}\) and we know that the wavelength of the light is \(\text{532}\) \(\text{nm}\) which is \(\text{532} \times \text{10}^{-\text{9}}\) \(\text{m}\). We are looking to determine the angle to first minimum so we know that \(m=1\).

Step 2: Applicable principles

We know that there is a relationship between the slit width, wavelength and interference minimum angles:

\(\sin\theta =\frac{m\lambda }{w}\)

We can use this relationship to find the angle to the minimum by substituting what we know and solving for the angle.

Step 3: Substitution

\begin{align*} \sin \theta & = \dfrac{\text{532} \times \text{10}^{-\text{9}}\text{ m}}{\text{2 511} \times \text{10}^{-\text{9}}\text{ m}} \\ \sin \theta & = \dfrac{\text{532}}{\text{2 511}} \\ \sin \theta & = \text{0.211867782} \\ \theta & = \sin ^{-1}\text{0.211867782} \\ \theta & = 12.2 ° \end{align*}

The first minimum is at \(\text{12.2}\)\(\text{°}\) from the centre peak.

From the formula \(\sin\theta =\frac{m\lambda }{w}\) you can see that a smaller wavelength for the same slit results in a smaller angle to the interference minimum. This is something you just saw in the two worked examples. Do a sanity check, go back and see if the answer makes sense. Ask yourself which light had the longer wavelength, which light had the larger angle and what do you expect for longer wavelengths from the formula.

Example: Diffraction Minimum

Question

A slit has a width which is unknown and has green light of wavelength 532 nm impinge on it. The diffracted light interferes on a surface, and the first minimum is measure at an angle of \(\text{20.77}\)\(\text{°}\)?

Step 1: Check what you are given

We know that we are dealing with diffraction patterns from the diffraction of light passing through a slit. We know that the wavelength of the light is \(\text{532}\) \(\text{nm}\) which is \(\text{532} \times \text{10}^{-\text{9}}\) \(\text{m}\). We know the angle to first minimum so we know that \(m=1\) and \(\theta =20.77°\).

Step 2: Applicable principles

We know that there is a relationship between the slit width, wavelength and interference minimum angles:

\(\sin\theta =\frac{m\lambda }{w}\)

We can use this relationship to find the width by substituting what we know and solving for the width.

Step 3: Substitution

\begin{align*} \sin \theta & = \dfrac{\text{532} \times \text{10}^{-\text{9}}\text{ m}}{w} \\ \sin \text{20.77} ° & = \dfrac{\text{532} \times \text{10}^{-\text{9}}}{w} \\ w &= \dfrac{\text{532} \times \text{10}^{-\text{9}}}{\text{0.3546666667}} \\ w &= \text{1 500} \times \text{10}^{-\text{9}} \\ w &= \text{1 500}\text{ nm} \end{align*}

The slit width is \(\text{1 500}\) \(\text{nm}\).

Extension: Calculating Maxima and Minima