Intensity of Interference Maxima & Minima

Using different sources of monochromatic light demonstrates that:
- Increasing the wavelength increases the width of the fringes
The angle of diffraction of the first minima can be found using the equation:

$θ = \frac{λ}{b}$

Where:
- θ = the angle of diffraction (radians)
- λ = wavelength (m)
- b = slit width (m)

This equation explains why red light produces wider maxima
- It is because the longer the wavelength, λ, the larger the angle of diffraction, θ
It also explains the coloured fringes seen when white light is diffracted
- It is because red light (longer λ) will diffract more than blue light (shorter λ)
- This creates fringes which are blue nearer the centre and red further out

It also explains why wider slits cause the maxima to be narrower
- It is because the wider the slit, b, the smaller the angle of diffraction, θ

9-3-2-slit-width

Slit width and angle of diffraction are inversely proportional. Increasing the slit width leads to a decrease in angle of diffraction, hence the maxima appear narrower

Single Slit Geometry

The diffraction pattern made by waves passing through a slit of width b can be observed on a screen placed a large distance away

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The geometry of single-slit diffraction

If the distance, D, between the slit and the screen is considerably larger than the slit width, $D > > b$ :
- The light rays can be considered as a set of plane wavefronts that are parallel to each other

9-2-2-single-slit-geometry

Determining the path difference using two parallel waves

For two paths, r₁ and r₂, travelling parallel to each other at an angle, θ, between the normal and the slit, the path difference will be:

path difference = r₁ − r₂ = $\frac{b}{2} \sin θ$

For a minima, or area of destructive interference:

The path difference must be a half-integral multiple of the wavelength

path difference = $\frac{n λ}{2}$

Equating these two equations for path difference:

$\frac{n λ}{2} = \frac{b}{2} \sin θ$

$n λ = b \sin θ$

Where n is a non-zero integer number, n = 1, 2, 3...

Since the angle θ is small, the small-angle approximation may be used: $\sin θ \approx θ$

$n λ = b θ$

Therefore, the first minima, n = 1, occurs at:

$λ = b θ$

This leads to the equation for angle of diffraction of the first minima:

$θ = \frac{λ}{b}$

Worked example

A group of students are performing a diffraction investigation where a beam of coherent light is incident on a single slit with width, b.

The light is then incident on a screen which has been set up a distance, D, away.

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A pattern of light and dark fringes is seen.

The teacher asks the students to change their set-up so that the width of the first bright maximum increases.

Suggest three changes the students could make to the set-up of their investigation which would achieve this.

Step 1: Write down the equation for the angle of diffraction

$θ = \frac{λ}{b}$

- The width of the fringe is related to the size of the angle of diffraction, θ

Step 2: Use the equation to determine the factors that could increase the width of each fringe

Change 1

- The angle of diffraction, θ, is inversely proportional to the slit width, b

$θ \propto \frac{1}{b}$

- Therefore, reducing the slit width would increase the fringe width

Change 2

- The angle of diffraction, θ, is directly proportional to the wavelength, λ

$θ \propto λ$

- Therefore, increasing the wavelength of the light would increase the fringe width

Change 3

- The distance between the slit and the screen will also affect the width of the central fringe
- A larger distance means the waves must travel further hence, will spread out more
- Therefore, moving the screen further away would increase the fringe width

Intensity of Interference Maxima & Minima (DP IB Physics: HL)

Revision Note