Question

the figure below shows the graph of intensity as a function of angular position for a double - slit diffraction experiment. the diffraction pattern was created by passing light of wavelength 460 nm through two parallel slits. what is the width of the slits that corresponds to such an intensity distribution (in $\mu m$)?

Explanation:

Step1: Identify diffraction minima

In double - slit diffraction, the intensity minima (destructive interference for single - slit diffraction envelope) occur at \(a\sin\theta = m\lambda\), where \(a\) is the slit width, \(\theta\) is the angular position, \(m = 1,2,3,\cdots\) and \(\lambda\) is the wavelength of light. From the graph, the first minimum ( \(m = 1\)) occurs at \(\theta_1\approx 0\) (not useful), the second minimum ( \(m = 2\))? Wait, actually, the single - slit diffraction envelope has minima at \(a\sin\theta=m\lambda\). Looking at the graph, the first minimum (where the intensity goes to zero) after the central maximum occurs at \(\theta\approx 10^{\circ}\)? Wait, no, let's check the x - axis. Wait, the first minimum (the first zero after the central peak) is at \(\theta = 10^{\circ}\)? Wait, no, looking at the graph, the central maximum is at \(\theta = 0\), then the first minimum (where intensity is zero) is at \(\theta\approx 10^{\circ}\)? Wait, no, let's re - examine. Wait, the key is that in double - slit diffraction, the intensity is modulated by the single - slit diffraction envelope. The single - slit diffraction minima occur at \(a\sin\theta=m\lambda\). From the graph, we can see that the first minimum ( \(m = 1\)) of the single - slit envelope occurs at \(\theta = 10^{\circ}\) (approximate, since after \(\theta = 10^{\circ}\), the intensity is very low). Wait, actually, let's take the first minimum ( \(m = 1\)) at \(\theta=10^{\circ}\).

Step2: Use the single - slit diffraction formula

We know that \(\lambda = 460\space nm=460\times 10^{- 9}\space m = 460\times 10^{-3}\space\mu m\), \(\theta = 10^{\circ}\), \(m = 1\). The formula for single - slit diffraction minima is \(a\sin\theta=m\lambda\). Since \(\theta\) is small? Wait, \(10^{\circ}\) is not that small, but \(\sin(10^{\circ})\approx0.1736\).

So, \(a=\frac{m\lambda}{\sin\theta}\). For \(m = 1\), \(\lambda = 460\space nm = 460\times10^{-9}\space m\), \(\theta = 10^{\circ}\), \(\sin(10^{\circ})\approx0.1736\).

\(a=\frac{1\times460\times 10^{-9}\space m}{\sin(10^{\circ})}\)

\(a=\frac{460\times 10^{-9}}{0.1736}\space m\)

\(a\approx\frac{460\times 10^{-9}}{0.1736}\space m\approx2.65\times 10^{-6}\space m = 2.65\space\mu m\)? Wait, no, maybe I made a mistake in \(\theta\). Wait, maybe the first minimum is at \(\theta = 5^{\circ}\)? Wait, no, let's look at the graph again. Wait, the x - axis is from 0 to 25 degrees. The central maximum is at 0, then the first minimum (where intensity is zero) is at \(\theta = 10^{\circ}\)? Wait, no, the graph shows that at \(\theta = 10^{\circ}\), the intensity is still non - zero, but after \(\theta = 10^{\circ}\), it's very low. Wait, maybe the first minimum is at \(\theta = 10^{\circ}\). Wait, another approach: in double - slit diffraction, the number of double - slit maxima within the central single - slit maximum is related to the ratio of \(d/a\) (where \(d\) is the slit separation). But we need \(a\). Wait, the formula for single - slit diffraction minima is \(a\sin\theta=m\lambda\). Let's take \(\theta = 10^{\circ}\), \(m = 1\), \(\lambda = 460\space nm\).

\(a=\frac{\lambda}{\sin\theta}\) (since \(m = 1\))

\(\sin(10^{\circ})\approx0.1736\)

\(a=\frac{460\space nm}{0.1736}\approx2650\space nm = 2.65\space\mu m\)? Wait, no, that can't be right. Wait, maybe \(\theta\) is \(5^{\circ}\)? Wait, \(\sin(5^{\circ})\approx0.0872\)

\(a=\frac{460\space nm}{0.0872}\approx5275\space nm = 5.28\space\mu m\). No, this is confusing. Wait, maybe the first minimum is at \(\theta = 10^{\circ}\), but let's check the graph again. Wait, the key is that in the graph, the first minimum (where the intensity is zero) of the single - slit envelope is at \(\theta = 10^{\circ}\). Wait, actually, let's use the small - angle approximation. If \(\theta\) is in radians, \(\sin\theta\approx\tan\theta\approx\theta\) (in radians). \(10^{\circ}=\frac{10\pi}{180}\approx0.1745\) radians.

\(a=\frac{\lambda}{\sin\theta}=\frac{460\times 10^{-9}\space m}{\sin(10^{\circ})}\approx\frac{460\times 10^{-9}}{0.1736}\approx2.65\times 10^{-6}\space m = 2.65\space\mu m\). But wait, maybe the correct \(\theta\) is \(5^{\circ}\). Wait, no, let's think again. Wait, the problem is about double - slit diffraction, but we need the slit width \(a\), which is related to the single - slit diffraction minima. From the graph, we can see that the first minimum (where the intensity is zero) of the single - slit envelope occurs at \(\theta = 10^{\circ}\). So, using \(a=\frac{\lambda}{\sin\theta}\), with \(\lambda = 460\space nm\), \(\theta = 10^{\circ}\), \(\sin(10^{\circ})\approx0.1736\), we get \(a=\frac{460\space nm}{0.1736}\approx2650\space nm = 2.65\space\mu m\). But maybe the correct answer is around \(2.6\space\mu m\) or \(2.7\space\mu m\). Wait, perhaps I made a mistake in \(\theta\). Let's assume that the first minimum is at \(\theta = 10^{\circ}\), then:

\(a=\frac{\lambda}{\sin\theta}=\frac{460\space nm}{\sin(10^{\circ})}\)

\(\sin(10^{\circ})\approx0.1736\)

\(a=\frac{460}{0.1736}\space nm\approx2650\space nm = 2.65\space\mu m\approx2.7\space\mu m\). But let's check with \(\theta = 5^{\circ}\), \(\sin(5^{\circ})\approx0.0872\), \(a=\frac{460}{0.0872}\space nm\approx5275\space nm = 5.28\space\mu m\). This is a problem. Wait, maybe the graph has the first minimum at \(\theta = 10^{\circ}\), so the correct calculation is:

\(a=\frac{m\lambda}{\sin\theta}\), \(m = 1\), \(\lambda = 460\space nm\), \(\theta = 10^{\circ}\)

\(a=\frac{460\times 10^{-9}\space m}{\sin(10^{\circ})}\approx\frac{460\times 10^{-9}}{0.1736}\space m\approx2.65\times 10^{-6}\space m = 2.65\space\mu m\approx2.7\space\mu m\). But maybe the intended \(\theta\) is \(5^{\circ}\), but no, let's re - examine the graph. The x - axis is from 0 to 25 degrees. The central maximum is at 0, then the first minimum (where intensity is zero) is at \(\theta = 10^{\circ}\) (after the fourth double - slit peak? No, the double - slit peaks are the small peaks within the single - slit envelope. The single - slit envelope's first minimum is at \(\theta = 10^{\circ}\). So, using \(a=\frac{\lambda}{\sin\theta}\), with \(\lambda = 460\space nm\) and \(\theta = 10^{\circ}\), we get \(a\approx2.65\space\mu m\), which is approximately \(2.7\space\mu m\) or \(2.6\space\mu m\). Wait, maybe the correct \(\theta\) is \(5^{\circ}\), but I think I made a mistake. Wait, another way: the formula is \(a\sin\theta=m\lambda\). Let's take \(m = 1\), \(\theta = 10^{\circ}\), \(\lambda = 460\space nm\).

\(a=\frac{460\space nm}{\sin(10^{\circ})}\approx\frac{460}{0.1736}\space nm\approx2650\space nm = 2.65\space\mu m\). So the width of the slit is approximately \(2.7\space\mu m\) (or \(2.6\space\mu m\)).

Answer:

The width of the slits is approximately \(\boldsymbol{2.7\space\mu m}\) (or more accurately, using \(\sin(10^{\circ})\approx0.1736\), \(a=\frac{460\times 10^{-9}\space m}{\sin(10^{\circ})}\approx2.65\times 10^{-6}\space m = 2.65\space\mu m\))