Two slits \(4.0 \times 10^{-6} \mathrm{m}\) apart are illuminated by light of wavelength \(600 \mathrm{nm} .\) What is the highest order fringe in the interference pattern?

The double-slit experiment is one of the most compelling demonstrations of the wave-like nature of light. In this classic setup, a light source illuminates two closely spaced slits, and the light passing through them interferes with itself on the other side. This creates an interference pattern consisting of a series of bright and dark fringes on a screen positioned behind the slits.

The pattern results from the constructive interference, where waves combine to increase amplitude (bright fringes), and destructive interference, where waves cancel each other out (dark fringes). The appearance of the fringes depends on a variety of factors including the wavelength of the light, the distance between the slits, and the distance from the slits to the screen.

Understanding the double-slit experiment is fundamental for students as it leads to insights into the behavior of waves and the fundamental principles of quantum mechanics, revealing the dual particle-wave nature of light and matter.

Wavelength conversion is a basic yet essential skill in physics, particularly when working with light and optics. The exercise provided requires the conversion of a light wavelength from nanometers (nm) to meters (m), which is central for further calculations in optics.

To convert the wavelength given in nanometers to meters, you multiply by the factor of \(10^{-9}\), since there are \(10^9\) nanometers in a meter. In the exercise, a wavelength of \(600\) nm is converted to \(600 \times 10^{-9}\) m. This conversion allows us to use the wavelength in equations that require SI units, making it possible to calculate other optical phenomena accurately.

Optics calculations involve equations that describe how light behaves as it encounters various objects and materials. One such fundamental equation is used for the double-slit experiment, which describes the relationship between the order of the fringe (m), the wavelength of light (\(\lambda\)), and the distance between the slits (d).

The equation \(m\lambda = d \sin \theta\) allows us to solve for various properties of the interference pattern. Here, \(\sin \theta\) represents the sine of the angle that light is diffracted, and 'm' is the order of the fringe, which corresponds to the number of wavelengths that fit into the path difference between the two slits. Simple algebraic manipulation of these formulas can provide a wealth of information about the properties of the interference pattern observed in the double-slit experiment.

The 'order of fringe' in an interference pattern refers to the number assigned to each bright or dark band, which corresponds to the level of constructive or destructive interference. A first-order fringe (\(m=1\)) is the first bright band from the central maximum, the second-order (\(m=2\)) is the next, and so on.

During calculations, especially in the double-slit experiment context, we might seek the highest order fringe that can be observed. This is determined by the physical constraints of the setup, like the wavelength of light and distance between the slits. With \(\sin \theta\) at its maximum value of 1 (at the angle that causes the most deflection before the wavefronts cannot meet), one can solve the fringe order equation for 'm' to find the highest order that fits within the physical limits of the setup, which is an integer value resulting from simplifying the wavelength and slit separation ratio. In the provided exercise, the highest order fringe came out to be 6 after rounding down the calculated value.