We now understand the diffraction effect of a single slit on light passing through. But what about two slits?
When light passes through two or more slits, it interferes. A pattern can be observed if the waves are coherent. The effect continues with multiple slits or a diffraction grating.
Key Concepts
The Young's double slit experiment consisted of a beam of coherent light passing through two identical slits before striking a screen.
The result is a pattern of evenly spaced bright fringes separated by dark fringes. It is caused by the interference of the light from the two slits as the two paths to each point on the screen vary in distance. The central maximum is most intense.
The fringe spacing can be calculated as follows:
\(s={D\lambda\over d}\)
- \(s\) is the distance between consecutive maxima (m)
- \(D\) is the distance from the slits to the screen (m)
- \(\lambda\) is the wavelength of light (m)
- \(d\) is the distance separating the centres of the two slits (m)
The fringe spacing is increaed by increasing the distance between the slits and the screen or by reducing the slit separation. It is also increased by increasing the wavelength of the light.
When we look beyond the interference pattern shown above, we notice that the intensity of the maxima appears not to follow a straightforward pattern of reducing outward from the centre.
The two-slit interference pattern is modulated by the one-slit diffraction effect.
This affects the distribution of intensity against distance on the screen from the centre by capping the double slit pattern within the intensity of the diffraction pattern.
When light shines through multiple slits, an interference pattern forms similar to that through a double slit.
\(n\lambda = d \sin \theta\)
- \(n\) is the number (order) of the maximum, where the central maximum is order 0
- \(\lambda\) is the wavelength of the light (m)
- \(d\) is the distance separating the slits (m)
- \(\theta\) is the angle subtended by the maximum and the centre
Subsidiary maxima form in between the overall interference maxima.The number of subsidiary maxima produced is equal to the number of slits minus 2.
The ratio of \(\text{slit width}\over \text{separation}\) can be obtained using a combination of the single slit diffraction equation and the multiple slit interference equation. At the first diffraction minimum:
\(\theta = {\lambda\over b}={n\lambda\over d}\Rightarrow {b\over d}={1\over n}\)
Interference patterns emerge from a variety of types of coherent sources such as electromagnetic waves and sound.
Interference can be modelled using a simulation.
Use quizzes to practise application of theory.
START QUIZ!
The image represents the intensity distribution for the interference of light passing through N narrow slits.
What is N?
The number of subsidiary maxima = N - 2
The image represents the intensity distribution for the interference of light passing through N narrow slits.
The first diffraction minima (\(\theta ={\lambda\over a}\)) occur at the same angle as the fifth interference maxima (\(\theta={5\lambda\over d}\)).
\({\lambda\over a}={5\lambda\over d}\)
The ratio \({a\over d}={1\over 5}\)
The image represents the intensity distribution for the interference of light passing through N narrow slits.
What is the order of the fringe marked with an arrow?
Fringe spacing is constant and the order is obtained by counting outward from the centre. The 4th order fringe is not visible because it falls in the diffraction minima.
The image represents the intensity distribution for the interference of light passing through N narrow slits.
The ratio \(\text{slit separation}\over \text{slit width} \) is:
The first diffraction minima (\(\theta ={\lambda\over a}\)) occur at the same angle as the fourth interference maxima (\(\theta={4\lambda\over d}\)).
\({\lambda\over a}={4\lambda\over d}\)
The ratio \({d\over a}={4\over 1}\)
The diagram represents the intensity distribution for the same wavelength light passing through two different sets of slits.
The slits have...
The position of the first diffraction minima are the same, which means that the slit width is unchanged. The positions of the interference maxima are different, so the spacing has changed.
The diagram represents the intensity distribution for the same wavelength light passing through two different sets of slits.
The ratio \(\text{slit separation on the left} \over\text {slit separation on the right}\) is:
The 5th maximum on the left has the same position as the 4th maximum on the right. The ratio of the fringe spacing is \(4\over 5\).
Since \(s={D\lambda\over d}\), fringe spacing is inversely proportional to slit separation.
The diagram represents the intensity distribution for the same wavelength light passing through two different sets of slits.
The slits have...
Both fringe spacing (a result of slit separation) and the distance to the diffraction minima (a result of slit width) are different.
The diagram represents the intensity distribution for the same wavelength light passing through two different sets of slits.
The ratio of \(\text{slit width}\over \text{separation}\) and the number of slits in each set is:
The ratio is \(1\over 5\) on the left and \(1\over 4 \) on the right.
The right set must have more slits to have narrowed the interference pattern.
Light of wavelength 500 nm passes through a diffraction grating with 100 lines per mm.
The angle of the 1st order maxima is:
\(d\sinθ = nλ\Rightarrow \sin\theta={n\lambda\over d}\)
\(d = {1\over 100\times 10^3} = 10^{-5}\text{ m}\) and \(λ \)= 5 x 10-7
\(\sinθ = 0.05\)
This is equal to the angle in radians as the small angle approximation applies.
Light of two wavelengths 500 nm and 400 nm passes through a diffraction grating with 100 lines per mm.
Calculate the angular separation of the 2 different coloured 1st order lines
\(d\sinθ = nλ\Rightarrow \sin\theta={n\lambda\over d}\)
\(d = {1\over 100\times 10^3} = 10^{-5}\text{ m}\)
For \(λ\) = 5 x 10-7, \(θ = 0.05\text{ rad}\)
For \(λ\) = 4 x 10-7, \(θ = 0.04 \text{ rad}\)
A 2 mm wide laser beam is incident on a 300 line per mm grating.
What is the effective number of lines on the grating?
The laser will pass through 2 x 300 lines.
NB: A longer way of answering this question would have been to calculate the slit separation and then to divide this into the width of the beam.
A 2 mm wide laser beam is incident on a 300 line per mm grating and used to view a 600 nm line.
What is the wavelength of the resolvable line in the first order spectrum?
\(Δλ = {λ\over mN} = {600\over 600} = 1 \text{ nm}\)
The image represents light of two wavelengths passing through a diffraction grating.
There are 3 orders of line. Which orders can be resolved?
Order 1 is not resolvable as the maxima are not separated sufficiently to reach the other's minimum.
Order 2 is just resolved and 3 is more easily resolved (albeit rather faint).
Magenta light made up of red (600 nm) and blue (400 nm) is passed through a diffraction grating producing the spectrum shown.
Which lines are magenta?
Line 5 is the principle maximum - red and blue combine to make magenta.
Lines 4, 2 and 1, and 6, 8 and 9 are equally spaced blue light maxima.
Lines 3 and 1, and 7 and 9 are equally spaced red light maxima (more spread than blue).
Since 1 and 9 combine as both the 3rd order blue and 2nd order red, these are magenta.
NB: 600 nm is actually orange rather than red (but Chris is colour blind!)
Magenta light made up of red (600 nm) and blue (400 nm) is passed through a diffraction grating producing the spectrum shown.
Which lines are red?
Line 5 is the principle maximum - red and blue combine to make magenta.
Lines 4, 2 and 1, and 6, 8 and 9 are equally spaced blue light maxima.
Lines 3 and 1, and 7 and 9 are equally spaced red light maxima (more spread than blue).
BUT since 1 and 9 combine as both the 3rd order blue and 2nd order red, these are magenta.
NB: 600 nm is actually orange rather than red (but Chris is colour blind!)
Magenta light made up of red (600 nm) and blue (400 nm) is passed through a diffraction grating producing the spectrum shown.
Which lines are blue?
Line 5 is the principle maximum - red and blue combine to make magenta.
Lines 4, 2 and 1, and 6, 8 and 9 are equally spaced blue light maxima.
Lines 3 and 1, and 7 and 9 are equally spaced red light maxima (more spread than blue).
BUT since 1 and 9 combine as both the 3rd order blue and 2nd order red, these are magenta.
NB: 600 nm is actually orange rather than red (but Chris is colour blind!)
How much of Multiple slit interference have you understood?
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