**Fringe width**is the distance between two successive bright fringes or two successive dark fringes. In the interference pattern, the fringe width is constant for all the fringes. Fringe width is independent of order of fringe. Fringe width is directly proportional to wavelength of the light used. It is given by:

$β=dλD $

**Angular fringe width**is given by:

$tanθ≈θ=Dβ =dλ $

**Example:**In Young's double slit experiment the two slits are illuminated by light of wavelength $5890_{∘}A$ and the distance between the fringes obtained on the screen is $0.2_{∘}$. If the whole apparatus is immersed in water then find the angular fringe width. (refractive index of water is $4/3$)

**Solution:**

For interference in YDSE:

$dsinθ=nλ$

For small $θ$, $sinθ$ can be approximated to $θ$

So, $dθ=nλ$

or, $θ∝λ$

Now as the set up immersed in water, $λ$ will become $μλ =4/3λ =43λ $

So, $λ_{2}λ_{1} =θ_{2}θ_{1} $

or, $3λ/4λ =θ_{2}0.2_{∘} $

or, $θ_{2}=0.2_{∘}×43 =0.15_{∘}.$