A CD or DVD displays a rainbow-like pattern due to the diffraction of light as it passes through the microscopic grooves etched onto the surface. The grooves act as a series of narrow slits, causing the light to diffract. The spacing of these grooves is very small, typically on the order of 1-2 micrometers (µm). This close spacing is crucial for creating a noticeable diffraction effect.

When white light (containing all colours) shines on the CD/DVD, each colour experiences a different angle of diffraction. This is because the wavelength of each colour is different. Shorter wavelengths (e.g., violet) diffract at a smaller angle than longer wavelengths (e.g., red). Therefore, each colour of light is diffracted to a different extent, resulting in the separation of colours that we observe as a rainbow-like pattern.

The spacing of the grooves determines the angular spacing between the different colours. The diffraction angle (θ) for a given wavelength (λ) can be approximated by a sin θ = mλ, where 'a' is the width of the groove (effectively the gap size) and 'm' is an integer. Because the groove spacing is relatively constant, the angular separation between the colours is also relatively constant. This is why the pattern is consistent across the disc.

In summary, the diffraction of light by the grooves on a CD/DVD, combined with the varying wavelengths of colours, creates a rainbow-like pattern. The spacing of the grooves is essential for producing a distinct and observable diffraction effect, leading to the separation of colours.

Transverse Waves: In transverse waves, the displacement of the medium is perpendicular to the direction of wave propagation. Imagine shaking a rope up and down; the wave travels horizontally along the rope, but the individual segments of the rope move vertically. The crests and troughs of a transverse wave are clearly defined.

Example: Light waves are transverse waves. Electromagnetic waves, such as light and radio waves, are transverse. They propagate through space by the oscillation of electric and magnetic fields, which are perpendicular to the direction of propagation. No physical medium is required for the propagation of light.

Longitudinal Waves: In longitudinal waves, the displacement of the medium is parallel to the direction of wave propagation. Imagine pushing and pulling a slinky; the compressions and rarefactions (regions of high and low density) travel along the slinky in the same direction as the pushing and pulling motion.

Example: Sound waves are longitudinal waves. Sound waves propagate through a medium (like air, water, or solids) by the compression and rarefaction of the medium. The particles of the medium vibrate back and forth in the same direction as the wave is traveling.

Propagation: Transverse waves propagate by the oscillation of the medium, transferring energy through the disturbance. Longitudinal waves propagate by the vibration of the medium particles, transferring energy through the compression and rarefaction of the medium. The type of wave determines how the energy is transferred and the characteristics of the wave itself.

Given:

• Wavelength of red light (λ) = 600 nm = 600 x 10^-9 m
• Width of the slit (a) = 0.1 mm = 0.1 x 10^-3 m
• Distance to the screen (D) = 1.5 m

To find: The distance (y) between the first dark fringe and the first bright fringe.

Solution:

1. The spacing of the bright fringes (Δy) is given by: Δy ≈ aθ = mλ/a. For the first bright fringe (m=1), Δy ≈ λ/a.
2. The angle of diffraction (θ) is related to the distance to the screen (D) and the distance from the slit to the screen (y) by: tan θ = y/D. Since θ is small, y ≈ D.
3. Therefore, Δy ≈ λ/a. We can also express the distance from the slit to the screen as y = D. The distance from the slit to the point where the bright fringe is formed is y. The distance from the slit to the point where the dark fringe is formed is y + a/2. The distance between the first bright and dark fringe is y + a/2 - (y + a/2) = a. This is incorrect.
4. Let's use the formula for the distance between the first dark and bright fringes: y = (mλD)/a. For the first bright fringe (m=1), y = λD/a.
5. Substituting the given values: y = (600 x 10^-9 m * 1.5 m) / (0.1 x 10^-3 m) = (900 x 10^-9 m²) / (0.1 x 10^-3 m) = 9000 x 10^-6 m = 0.009 m = 9 mm

Answer: The distance between the first dark fringe and the first bright fringe is 9 mm.