Explain the terms diffraction and interference - Leaving Cert Physics - Question 10 - 2019

Interference occurs when two or more waves overlap and combine. In the context of Young's experiment, the overlapping light waves from the slits interfere with each other, resulting in areas of constructive interference (bright fringes) and destructive interference (dark fringes).

In Young's double-slit experiment, monochromatic light passes through two closely spaced slits. The light waves emanating from each slit spread out (diffraction) and overlap. When the waves from the two slits meet, they can either amplify (constructive interference) or diminish (destructive interference) each other, resulting in a pattern of alternating bright (maxima) and dark (minima) fringes on the screen. A labelled diagram would show the two slits, the wavefronts, and the resulting interference pattern.

The experiment demonstrates that light is a wave because the creation of an interference pattern is a characteristic behavior of waves. The observable bright and dark fringes can only be explained by the wave nature of light, as particles would not exhibit this type of behavior.

Using the formula:

$d \sin \theta = n \lambda$

Where,

• (d) = distance between the slits = 0.5 m
• (n) = number of bright fringes = 13
• (\theta) can be calculated using (tan \theta = \frac{opposite}{adjacent} = \frac{0.825}{1.25} \rightarrow \theta = 0.378 rad)

Thus:

• (n \lambda = d \sin(\theta) \implies \lambda = \frac{0.5 \times 10^{-3} \times \sin(0.378)}{13} \approx 5.5 \times 10^{-7} m)

• Increase the distance between the screen and the slits.
• Decrease the distance between the slits.

When the ciliary muscles contract, the lens in the eye becomes thinner, which decreases the power of the lens. This means that it will be less effective in converging light rays, thereby affecting the eye's ability to focus on nearby objects.

The first harmonic will have one antinode at the closed end and one at the open end, while the second harmonic will have two antinodes with a node in between. The diagrams should depict the standing waves formed in the cylindrical pipe.

The wavelength can be calculated using:

$\lambda = 2 \times (0.498 - 0.167) = 0.66 m$

Using the formula:

$c = f \lambda$

Where:

• (f = 512 Hz)
• (\lambda = 0.66 m )

Thus:

$c = 512 \times 0.66 = 339 m/s$