How narrow is a slit that produces a diffraction pattem on a screen \(1.8 \mathrm{m}\) away whose central peak is \(1.0 \mathrm{m}\) wide? Assume \(\lambda=589 \mathrm{nm}\)

Understanding the angular width of a diffraction peak is fundamental to interpreting diffraction patterns. In simple terms, diffraction occurs when waves, such as light, encounter an obstacle or pass through a narrow opening, known as a slit. The waves spread out, or diffract, creating a pattern of bright and dark regions on a screen. The central bright region, or the central peak, has a certain width that can be measured in angular terms.

The formula to calculate the angular width (\theta) of the central peak is given by:
\[\theta = \frac{2 \lambda}{w}\]
where \(\theta\) represents the angular width of the central peak, \(\theta\) the wavelength of the light, and \(w\) the slit width. To measure this angle accurately, we can use trigonometric functions based on the geometry of the setup, considering the distance from the slit to the screen and the physical width of the peak observed. After finding the angle, you can rearrange this formula to solve for unknown variables such as the slit width, which is exactly what you'd need in exercises like determining the narrowness of a slit based on the diffraction pattern produced.

The wavelength of light, denoted as \(\theta\), plays a critical role in diffraction and optics. It is the distance over which the light wave's shape repeats and is a key parameter in understanding how light interacts with objects. For instance, in the case of diffraction through a slit, different wavelengths will produce diffraction patterns with varying angular widths, which affects the visibility and clarity of the fringes on the screen.

Light's wavelength is often measured in nanometers (nm), where 1 nm equals 10^-9 meters. Visible light—the light we can see—varies in wavelength from around 400 nm (violet) to 700 nm (red). In many diffraction problems, including the textbook example, a specific wavelength is given, which corresponds to yellow light in this case. This information is essential for calculating the diffraction pattern because as the wavelength changes, so does the pattern.

Calculating the width of a slit, given the diffraction pattern, is a typical application of the principles of wave optics. In the scenario provided, the process requires initially determining the angular width of the central diffraction peak. This angle can be found using:
\[\theta = \frac{1.0}{2 \cdot 1.8}\]
, where the width of the peak on the screen and the distance between the screen and the slit are known. From that angle, we calculate the slit width (w) with the formula:
\[w = \frac{2\theta}{\tan^{-1}(\theta)}\]

By substituting the angle and known wavelength into this equation, we can solve for the slit's width. It's important to convert this final measurement into practical units like micrometers (\textmu m) for easier comprehension. The ability to calculate slit width from diffraction patterns is essential for understanding optical phenomena and designing experiments involving light diffraction.