Question

part a
monochromatic light falls on a slit that is 2.40×10⁻³ mm wide
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: Understand Single - Slit Diffraction Formula

The formula for single - slit diffraction for dark fringes 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. For the first dark fringe (\(n = 1\)), the formula becomes \(a\sin\theta=\lambda\). But we need to find the angle for one side of the central maximum. The total angle between the first dark fringes on either side is \(29.0^{\circ}\), so the angle for one side \(\theta=\frac{29.0^{\circ}}{2}=14.5^{\circ}\)

Step2: Convert Slit Width to Meters

The slit width \(a = 2.40\times10^{-3}\space mm\). Since \(1\space mm=10^{-3}\space m\), we have \(a = 2.40\times10^{-3}\times10^{-3}\space m=2.40\times10^{-6}\space m\)

Step3: Solve for Wavelength \(\lambda\)

Using the formula \(\lambda=a\sin\theta\), substitute \(a = 2.40\times10^{-6}\space m\) and \(\theta = 14.5^{\circ}\)
\(\sin(14.5^{\circ})\approx\sin(14.5)=\sin(14 + 0.5)=\sin14^{\circ}\cos0.5^{\circ}+\cos14^{\circ}\sin0.5^{\circ}\approx0.2419\times0.99997 + 0.9703\times0.008727\approx0.2419+0.00847\approx0.2504\) (or we can use a calculator to directly find \(\sin(14.5^{\circ})\approx0.2504\))
Then \(\lambda=(2.40\times10^{-6}\space m)\times\sin(14.5^{\circ})\)
\(\lambda=(2.40\times10^{-6}\space m)\times0.2504\approx6.01\times10^{-7}\space m = 601\space nm\) (since \(1\space nm = 10^{-9}\space m\), \(6.01\times10^{-7}\space m=\frac{6.01\times10^{-7}}{10^{-9}}\space nm = 601\space nm\))

Answer:

\(\lambda=\boxed{601\space nm}\) (or \(\lambda = 6.01\times10^{-7}\space m\))