Read the following passage and answer the accompanying questions - Leaving Cert Physics - Question 13 - 2022

The experiment can be illustrated with the following components:

1. Light Source: A coherent light source (like a laser)
2. Diffraction Grating: An opaque material with periodic slits.
3. Screen/Spectrometer: A surface where the resulting pattern can be observed.

The light source directs waves toward the diffraction grating, and the screen captures the resulting interference pattern.

In the experiment, a series of fringes (alternating dark and bright lines) can be observed on the screen. This pattern arises due to constructive and destructive interference of light waves, indicating the wave nature of light.

The observed interference pattern, consisting of alternating bright and dark fringes, is a hallmark of wave behavior. It demonstrates that light can interfere with itself, a characteristic feature of waves, implying that light travels in waveforms rather than as particles.

The ray diagram should depict a converging lens, where:

1. Incident Ray: A ray approaching the lens parallel to the principal axis.
2. Refracted Ray: A ray that passes through the lens and diverges.
3. Focal Point: Mark the focal point on the opposite side of the lens where the rays converge.
4. Virtual Image Location: Extend the refracted rays backward to determine where they appear to diverge from, indicating the virtual image's position.

The formula for the period of a simple pendulum is given by: $T = 2\pi\sqrt{\frac{L}{g}}$ Where:

• $T$ is the period (2 s),
• $g$ is the acceleration due to gravity (assume $g = 9.8 \, m/s^2$).

Rearranging the formula to solve for $L$: $L = \frac{g T^2}{4\pi^2}$ Substituting the values: $L = \frac{9.8 \, m/s^2 \times (2 \, s)^2}{4 \pi^2} \approx 0.993 \, m$

The mass of Saturn can be calculated using the formula: $M = \frac{4}{3}\pi R^3 \rho$ Given that:

• R for Saturn is approximately 1.16 x 10^7 m,
• Average density (\rho) is roughly 0.687 g/cm³ or 687 kg/m³. Substituting: $M = 4/3 * \pi * (1.16 \times 10^7)^3 * 687 \approx 5.7 \times 10^{26} \text{ kg}$

The acceleration due to gravity can be given by: $g = \frac{GM}{R^2}$ Where:

• $G$ is the gravitational constant ($6.674 \times 10^{-11} \, m^3/kg/s^2$),
• $M$ is the mass of Saturn,
• $R$ is the radius of Saturn (approx $6.6742 \times 10^7$ m). Substituting: $g = \frac{6.6742 \times 10^{-11} * 5.7 \times 10^{26}}{(6.6742 \times 10^7)^2} \approx 11.2 \, m/s^2$

The formula for period is: $T = 2\pi\sqrt{\frac{L}{g}}$ Using the length from earlier calculations and the calculated acceleration due to gravity on Saturn:

• $L \approx 0.993 \, m$,
• $g \approx 11.2 \, m/s^2$: $T = 2\pi\sqrt{\frac{0.993}{11.2}} \approx 1.87 \, s$