91Ó°ÊÓ

In two-slit interference, the intensity pattern forms due to the constructive and destructive interactions of light waves from both slits. To find the intensity at a specific point in the pattern, we use the formula:

• \( I = I_{0} \cos^{2}(\phi / 2) \)

Here,

• \(I\) is the intensity at the point,
• \(I_{0}\) is the peak intensity at the central maximum,
• \(\phi\) is the phase difference in radians.

To solve the problem, first convert the phase difference from degrees to radians since trigonometric functions require radians. For \(60^{\circ}\), the conversion is \(\phi = \pi / 3\). Then, substitute

• \(\phi = \pi / 3\) (radians)

into the equation to calculate

the intensity \(I\).

Calculate \(I = I_{0} \cos^{2}(\pi / 6)\). Since

• \(\cos(\pi / 6) = \sqrt{3}/2\),

we find \(I = I_{0} (3/4)\).
Thus, at \(60^{\circ}\), the intensity is three-fourths (or 75%) of the peak intensity \(I_{0}\).

Phase difference (\(\phi\)) measures how "out of step" or "in step" waves from the two slits are with each other. It’s crucial in determining where peaks and troughs in the interference pattern occur. The formula used here helps calculate the intensity by

• relating \(\phi\) to the position within the fringe pattern.

phase difference \(\phi\) is determined by factors like
path difference and wavelength. It is typically expressed in radians for calculations involving trigonometric functions. In this problem, the given phase difference is initially \(60^{\circ}\), which means that the path of one wave is ahead of the other by that angle. Transforming this into radians yields

• \(\phi = \pi / 3\)

for further use in calculations. This conversion is essential since trigonometric functions, like cosine, require radian measures to calculate associated intensities or displacements.

Path difference refers to the physical length difference between the paths traveled by two waves from their respective slits to a common point. This difference dictates the phase difference (\(\phi\)) and thus the interference pattern formed.
The formula for path difference is:

• \(d = \lambda \phi / (2\pi)\),

where

• \(d\) is the path difference,
• \(\lambda\) is the wavelength of the light,
• \(\phi\) is the phase difference in radians.

In the context of this problem, substituting the given values into the formula

• (\(\phi = \pi / 3\), and \(\lambda = 480 \text{ nm}\))

enables you to find the path difference. This computes to

• \(d = 480 \text{ nm} \times \pi / 3 / (2\pi) = 80 \text{ nm}\).

Thus, the path difference where the phase difference is \(60^{\circ}\) in the interference pattern is

• 80 nm.

The wavelength of light (\(\lambda\)) is a fundamental property dictating how light waves propagate and interfere. It is defined as the distance between successive crests (or troughs) of a wave and is key to understanding interference phenomena.
In calculations involving interference, the wavelength helps determine both path and phase differences. It plays a part in calculating path differences since changes in wavelength will affect the distance the waves must travel to recombine constructively or destructively.
In this exercise, the wavelength

• \(\lambda = 480 \text{ nm}\)

is critical in computing the path difference required for a given phase difference. The relationship between wavelength and both path and phase differences allows us to describe entire interference phenomena and predict where constructive or destructive interference will occur in the pattern.