How are interference fringes formed in a wedge-shaped thin film and what is the fringe width?

In one line

A wedge-shaped film has thickness increasing steadily from the edge, so each fringe follows a line of constant thickness, giving straight, equally spaced fringes parallel to the edge with fringe width \(\beta = \lambda/(2\mu\theta)\).

Explained simply

Place two microscope slides on each other and slip a thin paper strip under one end: the air gap between them is a wedge. Under sodium light you see straight dark and bright bands running parallel to the edge where the slides touch. Each band is a "contour line" of equal film thickness.

How it works

1. Two surfaces inclined at a small angle \(\theta\) enclose a film of index \(\mu\). At distance \(x\) from the edge, thickness \(t = x\tan\theta \approx x\theta\).
2. For near-normal incidence, path difference in reflected light is \(2\mu t - \lambda/2\).
3. Dark fringe: \(2\mu t = n\lambda\). Bright fringe: \(2\mu t = (2n-1)\lambda/2\).
4. For the \(n\)th dark fringe at \(x_n\): \(2\mu x_n\theta = n\lambda\); for the next, \(2\mu x_{n+1}\theta = (n+1)\lambda\).
5. Subtracting: fringe width \(\beta = x_{n+1} - x_n = \lambda/(2\mu\theta)\).
6. At the edge \(t = 0\), so the edge appears dark in reflected light.
7. Since \(\beta\) does not depend on \(n\), fringes are equally spaced. Smaller \(\theta\) gives wider fringes.

Key formula

\(\beta = \dfrac{\lambda}{2\mu\theta}\). \(\beta\): fringe width (m), \(\lambda\): wavelength (m), \(\mu\): film index (1 for air), \(\theta\): wedge angle in radians.

Quick example

An air wedge (\(\mu = 1\)) shows fringes of width 0.3 mm with \(\lambda = 600\) nm. Then \(\theta = \lambda/(2\beta) = 600\times10^{-9}/(2\times0.3\times10^{-3}) = 10^{-3}\) rad. If the wedge is 5 cm long, the spacer thickness is \(t = L\theta = 0.05\times10^{-3} = 50\) μm. This is how the thickness of a thin wire or paper is measured.

Common mistakes

• Using \(\theta\) in degrees instead of radians.
• Saying the fringes are circular; for a wedge they are straight and parallel to the edge.
• Forgetting that the edge (zero thickness) is dark in reflected light.

How MU asks it

MU keeps the wedge qualitative: "Explain the formation of fringes in a wedge-shaped film" or "State conditions of maxima and minima for a wedge-shaped film" (3 to 5 marks). Draw the wedge with \(\theta\), \(t\) and \(x\) marked, state the conditions and the fringe width, and mention that fringes are straight and equidistant. Engineering use: testing optical flatness and measuring thin wire diameter.