Diffraction is one of the wave properties of light - Leaving Cert Physics - Question 13 - 2022

A typical experiment is the double-slit experiment, which involves:

1. A light source illuminating two closely spaced slits.
2. A screen positioned behind the slits to observe the light pattern. The diagram should label:

• Light source
• Slits
• Screen
• Interference pattern (series of bright and dark fringes).

In this experiment, a series of alternating bright and dark fringes are observed on the screen, known as an interference pattern. This pattern is the result of constructive and destructive interference of the light waves passing through the slits.

The presence of the interference pattern indicates that light behaves as a wave, as the bright fringes correspond to points of constructive interference and the dark fringes correspond to destructive interference. This phenomenon cannot be explained by a particle model and implies that light has wave properties.

A ray diagram showing a converging lens should include:

1. A converging lens positioned vertically.
2. An object placed closer to the lens than its focal point.
3. Rays emanating from the object:
   • One ray parallel to the principal axis and refracted through the focal point.
   • Another ray passing through the optical center of the lens.
4. Indicate the point where the rays diverge to form a virtual image on the same side of the lens as the object.

To find the length (L) of the pendulum, we can use the formula for the period of a simple pendulum: $T = 2\pi\sqrt{\frac{L}{g}}$ where:

• T = 2 s,
• g = 9.81 , m/s^2

Rearranging this formula gives: $L = \frac{g T^2}{4\pi^2}$ Substituting the known values: $L = \frac{9.81 \, m/s^2 \times (2 \, s)^2}{4\pi^2} \approx 0.993 \, m$.

The mass of Saturn can be calculated using: $M = \frac{g R^2}{G}$ where:

• $R = 1.16 \times 10^6 \, m$ (radius above Saturn's surface),
• $g = 9.81 \, m/s^2$ (acceleration due to gravity),
• $G = 6.674 \times 10^{-11} \, m^3 kg^{-1} s^{-2}$ (gravitational constant). After calculations, we find that: $M \approx 5.68 \times 10^{26} \, kg$.

Using the formula to find the acceleration due to gravity: $g = \frac{GM}{R^2}$ Substituting the known values: $g = \frac{(6.674 \times 10^{-11} \, m^3 kg^{-1} s^{-2}) \times (5.68 \times 10^{26} \, kg)}{(5820000 \, m)^2} \approx 11.2 \, m/s^2$.

The period can be recalculated using the adjusted gravitational force: $T = 2\pi\sqrt{\frac{L}{g}}$ Where:

• $g\text{(on Saturn)} \approx 11.2 \, m/s^2$ Substituting values provides: $T \approx 2\pi \sqrt{\frac{0.993 \, m}{11.2 \, m/s^2}} \approx 1.87 \, s$.