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

An experiment to demonstrate the wave nature of light can involve a single slit diffraction setup. The diagram should include:

• A light source emitting coherent light.
• A diffraction grating or slit.
• A screen or spectrometer on which the interference pattern (series of fringes) can be observed.

The experiment would show a series of bright and dark fringes on the screen, which are characteristic of interference patterns created by the wave nature of light. These fringes demonstrate the constructive and destructive interference of light waves.

The observations show interference patterns, which can only be explained by the wave nature of light. This phenomenon indicates that light behaves like a wave, spreading out and overlapping to create regions of higher intensity (bright fringes) and lower intensity (dark fringes).

In the ray diagram, demonstrate that when parallel rays from a distant object pass through the converging lens, they converge at a focal point. If the object is placed within the focal length of the lens, the rays diverge, and their extensions indicate that a virtual image is formed on the same side as the object.

Using the formula for the period of a simple pendulum, $T = 2\pi \sqrt{\frac{L}{g}}$, where $T$ is the period, $L$ is the length of the pendulum, and $g$ is the acceleration due to gravity, we rearrange to find:

$L = \frac{gT^2}{4\pi^2}$
Given that $T = 2s$ and $g = 9.8 \, m/s^2$, we substitute to find $L$:

$L = \frac{9.8 \times (2)^2}{4\pi^2} \approx 0.99 \, m$

Using the formula for gravitational force, we have: $F = \frac{GMm}{R^2}$
Rearranging gives:
$M = \frac{FR^2}{Gm}$
Where:

• $F = mg$ (weight of Titan),
• $G = 6.674 \times 10^{-11} \, m^3 kg^{-1} s^{-2}$,
• And $R = 1.16 \times 10^7 \text{ m}$.
  Calculating gives: $M = 1.16 \times 10^7 \times 5820000 \div (6.674 \times 10^{-11} \times 1000) \approx 1.22 \times 10^{23} \, kg$

Using the formula: $g = \frac{GM}{R^2}$
Substituting known values: $g = \frac{6.674 \times 10^{-11} \times 1.22 \times 10^{23}}{(1.16 \times 10^7)^2} \approx 11.2 \, m/s^2$

Using the pendulum period formula again,
$T = 2\pi \sqrt{\frac{L}{g}}$
With $L = 0.99 \, m$ and $g = 11.2 \, m/s^2$:

$T = 2\pi \sqrt{\frac{0.99}{11.2}} \approx 1.87 \, s$