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

This experiment demonstrates that light behaves as a wave through the phenomenon of interference. The formation of alternating bright and dark fringes clearly indicates the wave nature of light as this pattern occurs only when light waves overlap in a coherent manner. If light were a particle, we would not expect to observe such a fringe pattern. Consequently, the result can only be explained using the wave theory of light.

To calculate the wavelength, we start with the formula for the position of bright fringes:

$d \sin(\theta) = n\lambda$

Where:

• $d$ = distance between the slits = 0.5 mm = 0.5 x $10^{-3}$ m
• $\theta = \tan^{-1}(\frac{y}{L})$ where $y$ is the distance from the center to the nth bright fringe, and $L$ is the distance to the screen (1.25 m).
• For the 13th bright fringe, $y = 0.0165 m$ (1.65 cm per 13 fringes)

Calculating $\theta$ gives us:

$\tan \theta = \frac{y}{L} = \frac{0.0165}{1.25} = 0.0132$

From this, find $\theta$:

$\theta \approx 0.0129 rad$

Now, inserting values into the formula to find $\lambda$:

$n = 6, \text{ using the equation: } \lambda = \frac{d \sin(\theta)}{n}$

$\lambda = \frac{0.5 \times 10^{-3} m \cdot 0.378}{6} = 5.5 \times 10^{-7} m$

Thus, the wavelength is approximately 550 nm.

• Move the screen further away from the slits, which increases the distance over which the light travels, resulting in more spacing between bright fringes.
• Decrease the distance between the slits, which will cause wider spreading of the wavefronts due to increased diffraction.

When the ciliary muscles contract, the lens becomes thinner, which increases the focal length of the lens. This results in a decrease in the power of the lens because power is defined as the inverse of the focal length ($P = \frac{1}{f}$). Thus, a thinner lens has a lower power, allowing for focusing on distant objects.

Using the formula for the fundamental frequency of a pipe closed at one end:

$f = \frac{c}{\lambda}$

Where:

• $c = 339 m/s$ (speed of sound in air),
• For the first harmonic, the wavelength is:
• $L = \frac{\lambda}{4} = 0.167 m$, thus:

$\lambda = 4L = 4 \times 0.167 = 0.668 m$

So, the wavelength of the sound wave is approximately 0.67 m.

We already calculated the wavelength to be approximately 0.67 m and the frequency given is 512 Hz. Using the speed formula:

$c = f \lambda$

So,

$c = 512 \times 0.67 \approx 343.44 m/s$

Thus, the speed of sound in air is approximately 343 m/s.