Question

part a
if the angle between the first dark fringes on either side of the central maximum is 29.0 ° (dark fringe to dark fringe), what is the wavelength of the light used?
express your answer to three significant figures and include the appropriate units.
λ = value units

Explanation:

Step1: Identify the problem type

This is a diffraction problem, likely involving a single - slit diffraction (since we are dealing with dark fringes around a central maximum). The formula for the position of the \(n\)th dark fringe in single - slit diffraction is \(a\sin\theta=n\lambda\), where \(a\) is the width of the slit, \(\theta\) is the angle of the dark fringe from the central maximum, \(n\) is the order of the dark fringe, and \(\lambda\) is the wavelength of the light.

Wait, but the problem statement seems to be incomplete. We need the width of the slit \(a\) to calculate the wavelength. Since the problem is not fully provided (maybe there was a previous part where the slit width \(a\) was given, for example, if this is a follow - up question and the slit width \(a\) was, say, \(a = 0.01\space mm=1\times10^{- 5}\space m\) (just an example, but we need the actual value of \(a\) from the original problem context)). Let's assume that we missed the slit width information. But if we assume that this is a standard problem where maybe the slit width \(a\) is given (for example, if in the original problem, the slit width \(a = 1.00\times10^{-5}\space m\) and \(n = 1\) (first dark fringe), and the angle between the two first dark fringes is \(29.0^{\circ}\), so the angle for one dark fringe \(\theta=\frac{29.0^{\circ}}{2}=14.5^{\circ}\)

Step2: Apply the single - slit diffraction formula

Using \(a\sin\theta=n\lambda\), and solving for \(\lambda\), we get \(\lambda=\frac{a\sin\theta}{n}\)

Let's assume \(a = 1.00\times10^{-5}\space m\), \(n = 1\), and \(\theta = 14.5^{\circ}\)

First, calculate \(\sin(14.5^{\circ})\approx\sin(14.5)=\sin(14 + 0.5)=\sin14^{\circ}\cos0.5^{\circ}+\cos14^{\circ}\sin0.5^{\circ}\approx0.2419\times0.99997 + 0.9703\times0.008727\approx0.2419+0.00847\approx0.2504\)

Then \(\lambda=\frac{1\times10^{-5}\space m\times\sin(14.5^{\circ})}{1}\)

\(\lambda=1\times10^{-5}\space m\times0.2504 = 2.504\times10^{-6}\space m=2504\space nm\)? Wait, that can't be right. Maybe the slit width is different. Let's assume that the slit width \(a = 0.001\space mm = 1\times10^{-6}\space m\)

Then \(\lambda=\frac{1\times10^{-6}\space m\times\sin(14.5^{\circ})}{1}\approx1\times10^{-6}\times0.2504 = 2.504\times10^{-7}\space m = 250\space nm\) (but this is just a demonstration as the problem is incomplete)

Since the problem is incomplete (missing the slit width \(a\)), we can't give a correct numerical answer without the value of \(a\).

Answer:

The problem is incomplete. We need the width of the slit \(a\) to calculate the wavelength using the single - slit diffraction formula \(\lambda=\frac{a\sin\theta}{n}\) (where \(\theta=\frac{29.0^{\circ}}{2}\), \(n = 1\) for the first dark fringe). Please provide the slit width \(a\) to get the numerical value of the wavelength.