λ = xs where x = fringe separation
d s = double slit
d = distance between slits and screen
λ = 0.005625 x 0.00025
2.075
λ = 0.000000677m
= 6.7 x 10 m
= 670 x 10 m
= 670 nm
Conclusion
When the laser was turned on it diffracted at the two slits producing two waves. Two source superposition took place as the waves overlapped travelling in the same direction.
The bright fringes on the screen called maximas were due to constructive superposition where the waves have arrived in phase. For two waves to arrive in phase the path difference or ‘extra distance travelled’ must be equal to a whole number of wavelengths i.e. 1, 2, 3 etc.
The dark fringes are called minimas. They separate the bright fringes and are due to deconstructive superposition. The waves have arrived antiphase and
therefore the path difference between any two waves is not equal to a whole number of wavelengths, i.e. , 1, 2, etc.
The path differences for minimas and maximas on the screen increase from the central fringe. The positioning is shown below.
The bright and dark fringes on the screen stay fixed in position because light from a laser is monochromatic meaning it has only one frequency and it has a constant phase relationship, therefore it is coherent.
The two waves produced from the double slit made a good superposition paten because the waves had the same amplitude.
Task3
The wavelength of red light is about 700nm. The value I obtained, 670nm, is a good value as it is quite close. To allow my value to have a range, the errors must be assessed and then the percentage uncertainties can be calculated.
The most obvious probable errors were in the measurements, especially when measuring small lengths such as the double slit spacing. To work out the percentage uncertainties of measurements the following equation must be used
absolute uncertainty x100
actual value
The errors have been estimated as follows:
d = 2.075m ± 0.002
Percentage uncertainty = 0.002 x 100 = 0.1%
2.075
s = 0.25mm ± 0.1
Percentage uncertainty = 0.1 x 100 = 4%
0.25
x = 0.5625cm ± 0.05
percentage uncertainty = 0.02 x100 = 3.5%
0.5625
Percentage uncertainty for final result can be found from the sum of all
percentage uncertainties.
0.1 + 4 + 3.5 = 7.6 %
670 x 7.6 = 50.92
100
Range = 619nm – 721nm
Task 4
If one slit had been covered up the pattern seen on the screen would be slightly different to the pattern produced from the double slit experiment. I say slightly different because the pattern for the double slit experiment exists in the envelope of the single slit pattern.
A single slit does not produce as many fringes and would not follow the equally spaced pattern produced from a double slit.
The central fringe is the brightest, and also twice the width of the other fringes.
Task 5
As the microwaves diffract at the metal plates, two sets of coherent waves are produced. Microwaves travel from the point of diffraction to O in phase producing constructive superposition; this is the central maxima. The path difference is zero. The path difference increases from the central maxima by one whole wavelength for each subsidiary maxima.
Second maxima = 2λ
Second minima = 1λ
First maxima = 1λ
First minima = λ
Central maxima = 0λ
First minima = λ
First maxima = 1λ
Second minima = 1λ
Second maxima = 2λ
If X is the first subsidiary maxima, the path difference must be equal to one whole wavelength.
Path difference to X needs to be calculated to determine the wavelength of the microwaves.
S to X = 7.4 x 3 = 22.2 S to X = 6.5 x 3 = 19.5
P.d = 1λ = 22.2 – 19.5 = 2.7cm
Microwaves wavelength are about 0.1m –1mm.
Increasing the width of the slits is not a good idea for increasing the intensity of the wave at the maxima, because the diffraction would be less.
The narrower the slit, the greater the diffraction.
