**Given Data:**
- Diameter of the wire (d) = 0.04 mm = 0.004 cm
- Distance of the wire from the edge of the plates (x) = 10 cm
- Wavelength (λ) = 589.3 nm = 589.3 × 10^(-7) cm

**To Find:**
- Fringe width (w)

**Assumptions:**
- The wire does not touch the plates or disturb the air film between them.
- The air film is wedge-shaped and thin.

**Explanation:**
When a thin film is formed between two surfaces, interference of light occurs due to the phase difference between the light waves reflected from the upper and lower surfaces of the film.

**1. Deriving the Formula for Fringe Width:**
The formula for the fringe width in terms of the wavelength of light (λ), refractive index of the medium (μ), and the angle of incidence (θ) is given by:
w = (λ / 2μθ)

In the given problem, the air film is assumed to be very thin, so we can consider it to be a wedge-shaped film. The angle of the wedge is very small, and we can approximate the angle of incidence (θ) as the angle between the normal to the plates and the light ray incident on the film.

**2. Calculating the Angle of Incidence (θ):**
Using trigonometry, we can calculate the angle of incidence (θ) as:
θ = tan^(-1)(d / x)

Substituting the given values:
θ = tan^(-1)(0.004 / 10)
θ ≈ 0.023°

**3. Calculating the Refractive Index (μ):**
The refractive index of air (μ) is approximately 1. The refractive index of the medium is not given explicitly, so we assume it to be air.

**4. Calculating the Fringe Width (w):**
Using the formula for fringe width:
w = (λ / 2μθ)
w = (589.3 × 10^(-7) / 2 * 1 * 0.023)
w ≈ 0.0126 cm

**5. Conclusion:**
The fringe width is approximately 0.0126 cm. Since the wire is very thin (diameter = 0.004 cm), its effect on the fringe width is negligible, and it can be considered as zero (tare r to be zero almost).