Question

the figure below gives \\( \beta \\) versus the sine of the angle \\( \theta \\) in a double-slit interference experiment using light of wavelength 509. nm. what is the slit separation (in \\( \mu m \\))?\\( \beta \\) (rad)\
30.0\
20.0\
10.0\
\\( \sin\theta \\)\
0.5\
1\
\\( \mu m \\)\
hint\
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Explanation:

Step1: Recall double - slit formula

In double - slit interference, the phase difference \(\beta\) is related to the slit separation \(d\), wavelength \(\lambda\), and \(\sin\theta\) by the formula \(\beta=\frac{2\pi d}{\lambda}\sin\theta\). This is a linear equation of the form \(y = mx + c\) (here \(c = 0\)), where the slope \(m=\frac{2\pi d}{\lambda}\).

Step2: Determine the slope from the graph

From the graph, when \(\sin\theta = 1\), \(\beta=30.0\) rad. The slope \(m\) of the line \(\beta\) vs \(\sin\theta\) is \(\frac{\Delta\beta}{\Delta\sin\theta}\). Since when \(\sin\theta = 1\), \(\beta = 30.0\) rad and when \(\sin\theta=0\), \(\beta = 0\) rad, the slope \(m=\frac{30.0 - 0}{1-0}=30.0\) rad.

Step3: Solve for slit separation \(d\)

We know that \(m = \frac{2\pi d}{\lambda}\), so we can solve for \(d\):

$$d=\frac{m\lambda}{2\pi}$$

Given \(\lambda = 509\space nm=509\times 10^{-9}\space m\) and \(m = 30.0\) rad.
Substitute the values into the formula:

$$d=\frac{30.0\times509\times 10^{-9}}{2\pi}$$

First, calculate the numerator: \(30.0\times509\times 10^{-9}=30.0\times509\times10^{-9}=15270\times 10^{-9}=1.527\times 10^{-5}\)
Then divide by \(2\pi\): \(d=\frac{1.527\times 10^{-5}}{2\pi}\approx\frac{1.527\times 10^{-5}}{6.283}\approx2.43\times 10^{-6}\space m\)
Convert to \(\mu m\) (since \(1\space\mu m = 10^{-6}\space m\)), so \(d\approx2.43\space\mu m\)

Answer:

\(2.43\) (approximate value, depending on the precision of calculation. If we use more precise calculation: \(\frac{30\times509\times 10^{-9}}{2\pi}=\frac{30\times509}{2\pi}\times 10^{-9}=\frac{15270}{6.28319}\times 10^{-9}\approx2430\times 10^{-9}\space m = 2.43\space\mu m\))