# Interference of light through two narrow holes

##### Demonstration

This is a simple demonstration of the interference of the light. From this, students can calculate the wavelength of the light source.

#### Apparatus and materials

diode laser (Class 2)

wooden stand with clamp, 2

aluminium foil, 0.1 mm thicknewss, 5 cm x 5 cm

metre scale

sharp needle

travelling microscope

screen - A4 graph paper pasted on cardboard

#### Health & Safety and Technical notes

Cheap laser pointers or toys cannot be recommended for use in UK schools. Only class 2 types are suitable.

Make two narrow, very close circular holes in the aluminium foil by pressing the needle tip gently into the foil on a smooth plane surface.

Set up the apparatus, as shown in the diagram. The laser source, aluminium foil and screen should all lie in a straight line.

#### Procedure

a Switch on the source and make fine adjustments so that two holes in the foil are illuminated equally.

b Move the screen about 1 metre from the aluminium foil, so that the interference pattern falls on the graph paper. You should get a pattern of alternate dark and bright lines, of equal width and equal intensity, within a circular spot of light (see diagram above).

c Measure and record the average fringe width (β), with the help of graph paper markings.

d Measure and record the distance between the foil and the screen (D).

e Using a travelling microscope, measure the distance between the centres of the holes, x (see diagram).

f Calculate the wavelength of the light, λ, using the formula λ = Xβ/D

#### Teaching notes

1 Derivation of the formula:

Let S1 and S2 be the centres of the holes acting like coherent sources.
Secondary wave fronts coming from these points undergo superposition (interference), leading to bright and dark fringes on the screen.

The ray diagram is shown above.
S2N is the path difference between S1P and S2P, where P is the position of a bright fringe.
The condition for constructive interferences gives S2N = nλ, where n is an integer.

In triangle S1NS2, sinθ = S2N/S1S2 = nλ/X

From triangle PRO, since θ is a small angle, sinθ = PR /OP ≈ Y/OR = Y/D

From the above relations, nλ/X = Y/D or Y = nλD/X

Let Y1 and Y2 be the distances of two consecutive bright fringes, 1st and 2nd fringes.
i.e. Y1 = λD/s , Y2 = 2λD/s

Therefore fringe width = Y2-Y1= λD/X
Or λ = BX /D

2 Experimental results conducted at physics laboratory CMRIT BANGALORE, using a Class 2 diode laser of wavelength 630 – 680 nm, 1 mW power:

For X = 0.046 cm,
D = 140 cm
0.2 cm = fringe separation
the calculated λ is 657 nm.

For comparison, the result from a diffraction grating experiment was λ = 653 nm

The photo shows an interference pattern recorded with a 3.1 Megapixel digital camera.

This template with two sets of semi-circular lines can be used with an OHT to simulate the interference of coherent waves from two point sources.

This experiment was submitted by Tukaram Shet, Senior Lecturer in Physics at CMRIT Bangalore.

This experiment was safety-checked in January 2007