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When light passes through a single slit, it doesn't just keep traveling in a straight line. Instead, it spreads out, creating a distinct pattern called a diffraction pattern. This pattern is typically observed on a screen positioned far away from the slit. The pattern consists of alternating bright and dark areas known as fringes. At the center, there is a bright fringe, and on both sides, there are multiple dark and bright fringes.
Different light wavlengths, like the one given in the problem, create different diffraction patterns. This means the specific arrangement of dark and bright fringes is directly related to the properties of the light wave and the slit. In single-slit diffraction, the dark fringes appear at specific angles where the light waves interfere destructively, which we'll explore in the next section.

Destructive interference is a key concept when explaining the dark fringes found in a diffraction pattern. This occurs when waves overlap in such a way that they cancel each other out, leading to areas of darkness on the screen. In a single-slit diffraction scenario, this happens when the light traveling through the slit meets itself in such a way that the crest of one wave aligns with the trough of another.
The condition for complete destructive interference (where we see a dark fringe) in single-slit diffraction can be described mathematically by the equation: \[ a \sin \theta = m \lambda \]where:

• \( a \) is the width of the slit
• \( \lambda \) is the wavelength of light
• \( m \) is the integer order marking the fringe (1, 2, 3, ...)

Using this formula, we can predict where the dark fringes will appear by determining the specific angles \( \theta \) where the light interferes destructively.

The path difference in diffraction is the extra distance that some light waves have traveled compared to others. This difference in distance is critical in determining whether we will observe constructive or destructive interference at a certain point on the screen.
In the case of single-slit diffraction, the path difference is given by the width of the slit and the angle at which the light waves meet each other after passing through the slit. For destructive interference, this path difference results in the condition:\[ a \sin \theta = m \lambda \],where the light waves cancel out and we see a dark fringe. This equation is pretty straightforward, but it's powerful because it allows us to understand and predict the formation of dark fringes in the diffraction pattern.

Calculating the diffraction angle involves determining the specific angles \( \theta \) at which destructive interference occurs. To find these, we use the diffraction condition:\[ a \sin \theta = m \lambda \],where \( m \) is the order of the dark fringe. For the problem at hand, students need to recognize that the range of \( \sin \theta \) is restricted from -1 to 1, which significantly limits \( m \).
Practically, the largest value that \( \sin \theta \) can take is 1, which simplifies the calculation of \( m \):\[ m = \frac{a}{\lambda} \]Given \( a = 0.0666 \text{ mm} \) and \( \lambda = 585 \text{ nm} \), substituting these values allows us to calculate \( m \approx 114 \). Hence, the diffraction angle for the most distant dark fringe approaches 90°, as the sine of 90° is 1. This explanation provides a clear comprehension of the calculation process used in the exercise.