How does the fringe width, in Young's double-slit experiment change when the distance of separation between the slits and screen is doubled?

Hint: The fringe width in young’s double-slit experiment is proportional to the distance of separation between the slit and the screen and inversely proportional to the separation of the slits. We can use this relation to determine the fringe width when the screen distance is changed.

Formula used: In this solution, we will use the following formula
 $ \Delta y = \dfrac{{\lambda D}}{d} $ where $ \Delta y $ is the fringe width of the interference pattern, $ \lambda $ is the wavelength of the light, $ D $ is the distance between the sources of light, and $ d $ is the distance between the two coherent slits.

Complete step by step answer
In Young’s double-slit experiment, when a coherent source of light passes through the two slits, it forms an interference pattern on the screen.
The distance $ (y) $ of bright fringe/maxima from the centre of the screen can be calculated as:
 $ {y_n} = \dfrac{{n\lambda D}}{d} $ where $ n $ is the order of the maxima.
The fringe width of two fringes can be determined as the distance between two bright fringes that is
 $ {y_{n + 1}} - {y_n} = \dfrac{{(n + 1)\lambda D}}{d} - \dfrac{{n\lambda D}}{d} $
This can be simplified to
 $ {y_{n + 1}} - {y_n} = \dfrac{{\lambda D}}{d} $
Hence the fringe width $ \Delta y = \dfrac{{\lambda D}}{d} $ is proportional to the distance of the screen of from the slits i.e. $ \Delta y \propto D $
Hence, if the separation of the screen from the screen is doubled, the fringe width will also be doubled.

Note
Here we have assumed that when we double the distance of the screen from the slits, the medium between the slits and the screen is still air otherwise the fringe widths would be affected. The fringe width is also independent of the order of the maxima and will stay constant regardless of which fringes we are taking into account. This also implies that all the fringes will have the same width.