Q.(i) Write two points to distinguish between interference and diffraction fringes.

The bandwidth (fringe width) $\beta$ is the distance between any two consecutive bright fringes, or equally, between any two consecutive dark fringes. Subtracting the position of the $n$th bright fringe from that of the $(n+1)$th, $y_{n+1}-y_n=\dfrac{(n+1)\lambda D}{d}-\dfrac{n\lambda D}{d}=\dfrac{\lambda D}{d}$, gives $\beta=\dfrac{\lambda D}{d}$; the identical result follows from consecutive dark fringes, confirming that bright and dark fringes are equally and uniformly spaced on either side of the central bright fringe. Fringe width therefore increases if the screen distance $D$ is increased or if longer-wavelength light is used, and decreases if the slit separation $d$ is increased; if the entire experimental setup is immersed in a medium of refractive index $n$ (rather than air), the effective wavelength shortens to $\lambda'=\lambda/n$, shrinking the fringe width to $\beta'=\beta/n$. Because the central (zeroth-order) fringe corresponds to exactly zero path difference for every wavelength simultaneously, it stays sharply bright and white even under polychromatic (white light) illumination; away from the centre, each constitu …