(a) Find the location of central maximum (bright fringe with zero path difference) on the y-axis. (b) Find the light intensity at point O relative to the maximum fringe intensity. (c) Now if 600 nm lights is replaced by white light of range 400 to 700 nm, find the wavelengths of the light that form maxima exactly at point O. [All wavelengths in the problem are for the given medium of refractive index 4/3. Ignore dispersion]

To tackle your questions, we need to delve into the principles of interference and diffraction, particularly in the context of a double-slit experiment. Let's break this down step by step.

Finding the Central Maximum Location

The central maximum occurs at the point where the path difference between the light waves from the two slits is zero. In a typical double-slit setup, this point is located directly in line with the midpoint between the two slits on the y-axis. Mathematically, this can be expressed as:

• Path Difference (Δx) = d sin(θ)

For the central maximum, we set Δx = 0, which implies:

• sin(θ) = 0

This occurs when θ = 0, meaning the central maximum is located at the origin (0, 0) on the y-axis. Thus, the location of the central maximum is:

• Point O: (0, 0)

Calculating Light Intensity at Point O

The intensity of light at any point in an interference pattern can be described using the formula:

• I = I₀ cos²(Δφ/2)

Where:

• I₀ is the maximum intensity,
• Δφ is the phase difference between the two waves.

At the central maximum (point O), the phase difference Δφ is zero, leading to:

• I = I₀ cos²(0/2) = I₀

Thus, the intensity at point O is equal to the maximum fringe intensity, which means:

• Intensity at O = I₀

Determining Wavelengths for Maxima with White Light

When we replace the monochromatic light (600 nm) with white light, we need to find the wavelengths that will also create maxima at point O. The condition for constructive interference at any point is given by:

• d sin(θ) = mλ

For the central maximum, m = 0, which simplifies to:

• λ = 0

However, we need to consider the effective wavelengths in the medium with a refractive index of 4/3. The wavelength in the medium (λ') is given by:

• λ' = λ/n

For white light, we consider the range of wavelengths from 400 nm to 700 nm. Thus, the effective wavelengths in the medium will be:

• λ' = λ/(4/3) = (3/4)λ

Calculating for the boundaries:

• For λ = 400 nm: λ' = (3/4) * 400 nm = 300 nm
• For λ = 700 nm: λ' = (3/4) * 700 nm = 525 nm

Therefore, the wavelengths that will form maxima at point O in the medium are those that satisfy:

• 300 nm ≤ λ' ≤ 525 nm

In conclusion, the wavelengths of light that create maxima at point O when using white light in a medium with a refractive index of 4/3 are those that fall within the range of 300 nm to 525 nm. This range includes some of the violet and green light from the visible spectrum.