INTERFERENCEINTERFERENCE

DIVISION OF WAVEFRONTDIVISION OF WAVEFRONT

Division of wavefrontWhen light from a single point source is incident on two small slits, two coherent beams of light can be produced. Each slit acts as a secondary source due todiffraction.

If an extended source is used, each part of the wavefront will be incident on the slit at a different angle. Each part of the source will then produce a fringe pattern, but slightly displaced. When the intensity of all the patterns is summed the overall interference pattern may be lost. However a line source parallel to the slits is anexception.

Young's Slits ExperimentThe diagram below shows light from a single source of monochromatic light incident on a double slit. The light diffracts at each slit and the overlapping diffraction patterns produce interference.

The first bright fringe is observed at P. Angle PMO is θ

N is a point on BP such that NP = AP

Since P is the first bright fringe BN = λ

For small values of θ, AN cuts MP at almost 900 giving angle MAQ = θ and hence angle ΒΑΝ = θ

Again providing θ is very small, sin θ = tan θ = θ in radians

From triangle BAN: θ = λ / dand from triangle PMO: θ = Δx / D

So λ / d = Δx / D

Therefored

λDΔx

Two points to note:1. This formula only applies if x<<D, which gives θ

small. This is likely to be true for light waves but not for microwaves.

2. The position of the fringes is dependent on the wavelength. If white light is used we can expect overlapping colours either side of a central white maximum. The red, with the longer wavelength, will be the furthest from this white maximum (Δxred > Δxviolet since λred > λviolet).

Example

A laser beam is incident on two narrow slits of separation (0.250±0.002)mm. An interference pattern is produced on a screen (2.455±0.005)m from the slits. The spacing across ten fringes is measured as (62.0±0.5)mm.

(a) Calculate the wavelength of the light.

Δx = 62.0x10-3/10 and Δx= λD/d

62.0x10-3/10 = λx2.455/0.250x10-3

λ=631nm

(b) State the uncertainty in the wavelength and express the wavelength as (value±absolute uncertainty).

Using % uncertainties:Uncertainty in d =(0.002/0.250)x100=0.8%

Uncertainty in D=(0.005/ 2.455)x100=0.2%

Uncertainty in Δx=(0.5/62.0)x100=0.8%

The uncertainty in D can be ignored since it is less then one third of the others.

Total % uncertainty is = =1.1%

1.1% of 631nm = 7nm so wavelength = (631 ± 7)nm

2 20.8 0.8

(c)State one way in which you might be able to improve the accuracy of the experiment, give a reason.

Measure across more fringes, for example twenty fringes. This would reduce the percentage uncertainty in Δx.

(d)State, with a reason, if the fringe spacing increases, decreases or remains the same if red laser light is replaced by a Helium-Cadmium laser with a wavelength of 422nm.

Since the wavelength has decreased the fringe spacing decreases, since Δx λ