Electromagnetic Waves - SS2 Physics Past Questions and Answers - page 2

The transverse nature of electromagnetic waves refers to the oscillation of electric and magnetic fields perpendicular to the direction of wave propagation. Unlike longitudinal waves, where the oscillation occurs parallel to the direction of wave propagation, electromagnetic waves exhibit perpendicular oscillations. This unique characteristic allows electromagnetic waves to exhibit various phenomena, such as polarisation and interference.

For example, consider a light wave. As it travels through space, the electric field oscillates vertically, while the magnetic field oscillates horizontally, both perpendicular to the direction of wave propagation. This transverse nature of light waves enables them to be polarised using filters that allow only certain orientations of the electric field to pass through.

The transverse nature of electromagnetic waves directly relates to their polarization. Polarisation refers to the alignment of the electric field vector in a particular direction, while the magnetic field vector remains perpendicular to it. This alignment can be achieved through various mechanisms.

One example of polarization is the use of polarising filters. These filters are designed to transmit light waves oscillating in a specific direction while blocking waves oscillating perpendicular to that direction. By selectively allowing only certain orientations of the electric field to pass through, the transmitted light becomes polarised.

Polarisation finds applications in numerous real-world scenarios. For instance, polarised sunglasses utilise the principle of polarisation to reduce glare caused by reflected light. By selectively blocking horizontally polarised light that is typically reflected off surfaces such as water or roads, the sunglasses enhance visual clarity and reduce eye strain.

Explanation: The speed of light in a vacuum is approximately 3 × 108 m/s. To find the wavelength, we use the formula wavelength = speed of light/frequency. Substituting the given values, we get wavelength = 3 × 108 m/s / (100 × 106 Hz) = 3 × 104 m.

Explanation: Using the formula frequency = speed of light/wavelength, we can calculate the frequency. The speed of light in a vacuum is approximately 3 × 108 m/s. Converting the wavelength from nm to metres (500 nm = 500 × 10-9 m), we get frequency = (3 × 108 m/s) / (500 × 10-9 m) = 6 × 1014 Hz.

Explanation: The energy of a photon can be calculated using the formula energy = Planck's constant × frequency. Planck's constant is approximately 6.63 × 10-34 J·s. Substituting the given values, we get energy = (6.63 × 10-34 J·s) × (2 × 1014 Hz) = 4.48 × 10-19 J.

Explanation: The intensity of an electromagnetic wave is related to the square of the amplitude of the electric field. Using the formula intensity = (electric field amplitude)2 / (2 × impedance of free space), we can solve for the electric field amplitude. The impedance of free space is approximately 377 Ω. Substituting the given value and solving for the amplitude, we get (electric field amplitude)2 = (5 W/m2) × (2 × 377 Ω) = 3770 V2/m2. Taking the square root, we find the amplitude to be approximately 2.2 V/m.