Question

A Young's double-slit experiment is performed with monochromatic green light \((\lambda=540 \mathrm{nm})\). The separation between the slits is \(0.100 \mathrm{~mm},\) and the interference pattern on a screen has the first side maximum \(5.40 \mathrm{~mm}\) from the center of the pattern. How far away from the slits is the screen?

Short answer

Answer: The distance between the slits and the screen in the double-slit experiment is approximately 1.00 meter.

Step-by-step solution

Identify the given data

We are given the following information: - Wavelength of monochromatic green light, \(\lambda = 540 \, \mathrm{nm}\) - Slit separation, \(d = 0.100 \, \mathrm{mm}\) - Position of the first side maximum, \(y = 5.40 \, \mathrm{mm}\) The goal is to find the distance between the slits and the screen, \(L\).

Convert units

In order to work with consistent units, let's convert the given values to the SI unit system (meters). - \(\lambda = 540 \times 10^{-9} \, \mathrm{m}\) - \(d = 0.100 \times 10^{-3} \, \mathrm{m}\) - \(y = 5.40 \times 10^{-3} \, \mathrm{m}\)

Apply the formula for destructive interference condition

The condition for destructive interference (dark fringes) in Young's double-slit experiment is: \(m \lambda = (y - y_0)\frac{d}{L}\) Here, \(m\) is an integer called the order of the fringe (in this case \(m = 1\) for the first side maximum), \(y\) is the position of the dark fringe from the central axis (we will use \(y=y_0+5.40\times 10^{-3}\,\mathrm{m}\)), \(y_0\) is the position of the central bright fringe (usually called the central maximum), \(d\) is the slit separation, and \(L\) is the distance between the slits and the screen. We want to solve for \(L\), so we can rewrite the formula as: \(L = \frac{d (y - y_0)}{m \lambda}\)

Substituting the known values and solving for L

Substituting the values of \(\lambda\), \(d\), and \(y\), and the fact that \(m = 1\), into the formula, we get: \(L = \frac{(0.100 \times 10^{-3}(5.40 \times 10^{-3} - y_0))}{1 \times 540 \times 10^{-9}}\) We assume that \(y_0\) is negligible compared to \(5.40 \times 10^{-3}\,\mathrm{m}\), so the formula becomes: \(L = \frac{(0.100 \times 10^{-3})(5.40 \times 10^{-3})}{1 \times 540 \times 10^{-9}}\) Solving for \(L\), we get: \(L \approx 1.00 \, \mathrm{m}\) So, the distance between the slits and the screen in the double-slit experiment is approximately 1.00 meter.