27.4 AIR WEDGE
27.4.1 Interference Pattern Produced by Air Wedge
 |
Figure 27. 10 A thin glass plate P is placed on another plate Q, forming a thin air wedge between them. One end is in contact while the other is slightly separated by a thin material, creating a varying thickness of air. |
🔬 Apparatus
- A thin glass plate P, similar to a microscope slide, is placed on top of another glass plate Q. One end of plate P touches plate Q, while the other end rests on a very thin material (such as paper or a fine wire), forming a thin air wedge.
- Light from a monochromatic source is directed towards a tilted glass plate, where it is partially reflected.
- A microscope is positioned above the setup, directly over plates P and Q, to observe the interference fringes formed.
 |
Figure 27.11 Light is incident nearly normal onto plate P. Part of the light is reflected at point A from the lower surface of P, forming ray 1, which contributes to the interference pattern.
✨ Principle
- Light is incident almost normally onto plate P. A portion is reflected at point A from the lower surface of P, forming Ray 1.
- The remaining light travels through the air wedge and is partially reflected at point R from plate Q, forming Ray 2.
- There exists an optical path difference (p) between Ray 1 and Ray 2, which depends on the thickness AR of the air wedge.
- When p = (m + 1/2)λ, constructive interference occurs, producing a bright fringe.
- When p = mλ, destructive interference occurs, producing a dark fringe.
- At the contact edge where P touches Q, a dark fringe is observed even though the path difference is zero. This happens because the reflected light from Q undergoes a phase reversal of λ/2.
- As we move away from this edge, the air wedge thickness increases. An increase of λ/4 produces a bright fringe, and another λ/4 produces a dark fringe.
- Thus, the thickness increases progressively, resulting in parallel and equally spaced fringes.
- The wedge thickness varies along its length: certain points satisfy the condition for dark fringes, while others produce bright fringes. The observed pattern consists of alternating bright and dark bands.
 |
| Figure 27.12 (a) The thickness of the air wedge increases gradually along its length. Certain thicknesses (XX’, BB’, DD’) produce dark fringes, while others (AA’, CC’) produce bright fringes due to interference. (b) The resulting pattern consists of alternating bright and dark fringes observed on the surface of plate P. |
27.4.2 Formula for Fringe Separation
 |
Figure 27.13 The diagram shows the side view of an air wedge formed between two glass plates. The wedge angle θ is related to its thickness H and length L, where tan θ = H / L. |
Figure 27.13 shows the side view of an air wedge. We have
$tan \ \theta = \frac{H}{L}$
where H is the thickness of the material placed at the edge of the air wedge L is the length of the air wedge
Suppose that a bright fringe of the m-th order is formed at distance xm from one end of the wedge and at air thickness tm. Then the optical path difference p there must be
$p = 2t_m = \left(m + \frac{1}{2}\right) \lambda$
$t_m = \frac{1}{2} \left(m + \frac{1}{2}\right)\lambda$
and $tan \ \theta = \frac{t_m}{x_m}$
Hence, $\frac{t_m}{x_m}=\frac{H}{L}$
$x_m = \left(\frac{L}{H}\right) t_m$
$= \left(\frac{L}{H}\right) \left[\frac{1}{2} \left(m + \frac{1}{2}\right)\lambda \right]$
Consider the adjacent bright fringe of the (m + 1)-th order which is at distance xm+1 and at air thickness tm+1.
$t_{m+1} = \frac{1}{2} \left(m + 1 + \frac{1}{2}\right)\lambda$
and $tan \ \theta = \frac{t_{m+1}}{x_{m+1}}$
Hence, $\frac{t_{m+1}}{x_{m+1}}=\frac{H}{L}$
$x_{m+1} = \left(\frac{L}{H}\right)t_{m+1}$
$= \left(\frac{L}{H}\right) \left[\frac{1}{2} \left(m + 1 + \frac{1}{2}\right)\lambda \right]$
The separation y between the two adjacent bright fringes is
$y = x_{m+1} −x_m$
$= \left(\frac{L}{H}\right) \left[\frac{1}{2} \left(m + 1 + \frac{1}{2}\right)\lambda \right] − \left(\frac{L}{H}\right)\left[\frac{1}{2} \left(m + \frac{1}{2}\right)\lambda \right]$
$y = \frac{1}{2} \left(\frac{L}{H}\right) \lambda$
Angular size of air wedge:
$tan \ \theta = \frac{H}{L}$
$= \frac{1}{2} \left(\frac{\lambda}{y}\right)$
EXAMPLE 27.5
An air wedge is formed by placing a piece of thin paper at the edges of a pair of glass plates. Light of wavelength 600 nm is incident normally onto the plates. Fringes are observed, with fringe separation of 0.25 mm. The length of the air wedge is 5.0 cm.
(a) Determine the thickness of the piece of paper.
(b) Estimate the number of bright fringes that are formed on the plate.
Answer
(a)
$y = \frac{1}{2} \left(\frac{L}{H}\right) \lambda$
The thickness of the piece of paper is
$H = \frac{1}{2}\left(\frac{Lλ}{y}\right)$
$= \frac{(5.0 cm)(600 \times 10^{-9} m)}{2(0.025 cm)} = 6.0 \times 10^{-5} \ m$
(b)
$L ≈ ny$
where n is the number of bright fringes.
$n ≈ \frac{L}{y}$
$n= \frac{50 \ mm}{0.25 \ mm} = 200$
EXAMPLE 27.6
 |
| Figure 27.14 A thin material of thickness 5.0 μm creates an air wedge between glass plates A and B. When light of wavelength 500 nm is incident normally, an interference pattern of bright fringes is formed along the wedge. |
Figure 27.14 shows how a piece of material of thickness 5.0 μm forms an air wedge between a pair of glass plates A and B. Light of wavelength 500 nm is incident normally onto the plates. Estimate the number of bright fringes produced.
Answer
$y = \frac{1}{2}\left (\frac{L}{H}\right)\lambda$
But, $L ≈ ny$
where n is the number of bright fringes.
$n ≈ \frac{L}{y}$
$= \frac{2H}{\lambda}$
$= \frac{2(5.0 \times 10^{-6}) }{(500 \times 10^{-9})} = 20$
Post a Comment for "Air Wedge Interference Explained (Fringe Formation, Formula & Solved Examples)"