Question

36.1 single-slit diffraction
it is possible for a dark fringe for two different wavelengths to occur at the same angle. consider light of wavelength 600. nm and 500. nm. they both have a dark fringe at 1.85 mrad. for what minimum slit width is this possible (in mm)?
mm
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Explanation:

Step1: Recall Single - Slit Diffraction Formula

The formula for the dark fringes in single - slit diffraction is \(a\sin\theta = m\lambda\), where \(a\) is the slit width, \(\theta\) is the angle of the dark fringe, \(m\) is the order of the dark fringe, and \(\lambda\) is the wavelength of the light.

For the first wavelength \(\lambda_1 = 600\space nm=600\times 10^{-9}\space m\) and the second wavelength \(\lambda_2 = 500\space nm = 500\times 10^{-9}\space m\), and the same angle \(\theta=1.85\space mrad = 1.85\times 10^{-3}\space rad\).

Let \(m_1\) be the order of the dark fringe for \(\lambda_1\) and \(m_2\) be the order of the dark fringe for \(\lambda_2\). Then we have \(a\sin\theta=m_1\lambda_1\) and \(a\sin\theta = m_2\lambda_2\). So \(m_1\lambda_1=m_2\lambda_2\).

Substituting the values of \(\lambda_1\) and \(\lambda_2\), we get \(m_1\times600\times 10^{-9}=m_2\times500\times 10^{-9}\), which simplifies to \(6m_1 = 5m_2\).

To find the minimum non - zero integers \(m_1\) and \(m_2\) that satisfy this equation, we can see that \(m_1 = 5\) and \(m_2=6\) (since we need the smallest positive integers such that the ratio holds).

Step2: Calculate Slit Width \(a\)

We can use either of the two equations \(a=\frac{m_1\lambda_1}{\sin\theta}\) or \(a=\frac{m_2\lambda_2}{\sin\theta}\). Let's use \(m_1 = 5\) and \(\lambda_1=600\times 10^{-9}\space m\), \(\theta = 1.85\times 10^{-3}\space rad\).

First, calculate the numerator: \(m_1\lambda_1=5\times600\times 10^{-9}\space m=3000\times 10^{-9}\space m = 3\times 10^{-6}\space m\)

Then, since \(\sin\theta\approx\theta\) for small angles (because \(\theta\) is in radians and is very small, \(\sin\theta\approx\theta\) when \(\theta\) is in radians and \(\theta\ll1\)), we have \(\sin\theta\approx1.85\times 10^{-3}\space rad\)

Now, \(a=\frac{m_1\lambda_1}{\sin\theta}=\frac{3\times 10^{-6}\space m}{1.85\times 10^{-3}\space rad}\)

\(a=\frac{3}{1.85}\times 10^{-3}\space m\approx1.6216\times 10^{-3}\space m\)

Convert to millimeters: Since \(1\space m = 1000\space mm\), \(a = 1.6216\times 10^{-3}\times1000\space mm=1.62\space mm\) (we can also check with \(m_2 = 6\) and \(\lambda_2 = 500\times 10^{-9}\space m\): \(m_2\lambda_2=6\times500\times 10^{-9}\space m = 3000\times 10^{-9}\space m=3\times 10^{-6}\space m\), and \(a=\frac{3\times 10^{-6}}{1.85\times 10^{-3}}\approx1.62\times 10^{-3}\space m = 1.62\space mm\))

Answer:

\(1.62\) (or approximately \(1.62\) mm, the calculation may have a slight variation depending on the level of approximation, but the key is to use the relationship between the orders of the dark fringes and the wavelengths)