Question #87758
Your teacher gives you a piece of cardboard with two pinholes on it, saying that they are separated by 205 μm ± 3%. In order to test this, you set up a double slit apparatus with the screen at a distance of 3.00 m away from the cardboard. You use a helium-neon laser (wavelength 638.2 nm) to generate an interference pattern. You measure the distance from the first to the seventh nodal line to be 56.9mm. Did your teacher give you accurate value for the separation of the pinholes?
1
Expert's answer
2019-04-16T10:14:34-0400

The angles of the nodal lines are determined by the relation dsinθ=nλd \sin \theta = n \lambda, where dd is the distance between the pinholes, λ\lambda is the wavelength of light, and n=0,1,2,n = 0, 1, 2, \ldots For expected small angles, we can use the approximation sinθθ\sin \theta \approx \theta. Then, the angular distance between the first and the seventh nodal line is θ=6λ/d\theta = 6 \lambda / d, and the corresponding distance on the screen will be L=θD=6λD/dL = \theta D = 6 \lambda D / d, where DD is the distance to the screen. The distance between the pinholes is measured with 3% precision, so the distance LL, being inversely proportional to dd, will also be determined with about 3% precision. Calculating this quantity, we have

L=6λDd=6(638.2×109m)3m205×106m=0.056m=56mm.L = \frac{6 \lambda D}{d} = \frac{6 \cdot \left( 638.2 \times 10^{-9}\, \text{m} \right) \cdot 3\, \text{m}}{205 \times 10^{-6}\, \text{m}} = 0.056\, \text{m} = 56\, \text{mm} \, .

The difference between this number and the measured value is 0.9 mm, which is 0.9/56.9 = 0.016, or 1.6% of the measured value, well within the 3% error. Thus, the teacher gave a fairly accurate value for the separation between the pinholes.


Answer: yes.


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