In Young’s double-slit experiment, if a monochromatic source is replaced by white light, which contains multiple wavelengths, the patterns observed on the screen will differ significantly.

First, remember that the pattern formed in the double-slit experiment consists of both bright and dark fringes due to constructive and destructive interference, respectively. The position and intensity of these fringes depend on the wavelength of the light used. For monochromatic light (light of a single wavelength), the interference pattern is stable, with bright and dark fringes evenly spaced. Each fringe is uniformly bright or dark.

When using white light, which is a combination of various wavelengths of light, each color, or each wavelength, forms its own interference pattern with slightly different fringe spacing. This occurs because the separation between fringes $ \Delta y $ is given by the formula:

$$ \Delta y = \frac{\lambda L}{d} $$

Where:

• $ \lambda $ is the wavelength of light
• $ L $ is the distance from the slits to the screen
• $ d $ is the separation between the slits

Since different wavelengths have different values for $ \lambda $, each color’s fringes will be at slightly different positions. The result is that near the center of the pattern, where there is the least path difference, all wavelengths constructively interfere to form a bright white central fringe. However, moving away from the center, the fringes start to show different colors as the path difference between the light from the two slits increases. This causes a dispersion of colors with different orders of fringes dominated by different colors. The constructive and destructive interference patterns of different wavelengths slightly offset one another.

Thus:

• Option A - "Interference pattern will disappear" is incorrect because the interference pattern does not disappear but changes due to the dispersion of colors.
• Option B - "There will be a central dark fringe surrounded by a few coloured fringes" is incorrect as the central fringe in the presence of white light is bright, not dark.
• Option C - "There will be a central bright white fringe surrounded by a few coloured fringes" is correct. This is because all the wavelengths interfere constructively at the center, creating a bright white fringe, succeeded by colored fringes due to the varying interference conditions for each wavelength.
• Option D - "All bright fringes will be of equal width" is incorrect. The width and spacing of the fringes vary by wavelength, so the interference pattern will not have uniform fringe widths.

Therefore, the correct answer is Option C: "There will be a central bright white fringe surrounded by a few coloured fringes."