Question

For the interference pattern to be clearly visible on the screen, the distance (D) between the slits and the screen should be much larger than the distance (d) between the two slits (S₁ and S₂ ), i.e., D » d.

Answer 1

The condition for constructive interference at P is,

∆l = y_n\(\cfrac{d}{D}\) = nλ …. (1)

y being the position (y-coordinate) of n^th bright fringe (n = 0, ±1, ±2, …).

∴ y = nλ\(\cfrac{d}{D}\) ….. (2)

Similarly, the position of m^th (m = +1, ±2,…) dark fringe (destructive interference) is given by,

∆l = y_m\(\cfrac{d}{D}\) = (2m – 1)λ giving

y_m = (2m – 1)λ\(\cfrac{d}{D}\) …(3)

The distance between any two consecutive bright or dark fringes, i.e., the fringe width

= W = ∆y = y_n+1 – y_n = λ\(\cfrac{d}{D}\) …(4)

Conditions given by Eqs. (1) to (4) and hence the location of the fringes are derived assuming that the two sources S₁ and S₂ are in phase. If there is a non-zero phase difference between them it should be added appropriately. This will shift the entire fringe pattern but will not change the fringe width.

Several phenomena that we come across in our day to day life are caused by interference and diffraction of light. These are the vigorous colours of soap bubbles as well as those seen in a thin oil film on the surface of water, the bright colours of butterflies and peacocks etc. Most of these colours are not due to pigments which absorb specific colours but are due to interference of light waves that are reflected by different layers.

Interference due to a thin film :

The brilliant colours of soap bubbles and thin films on the surface of water are due to the interference of light waves reflected from the upper and lower surfaces of the film. The two rays have a path difference which depends on the point on the film that is being viewed. This is shown in above figure.

The incident wave gets partially reflected from upper surface as shown by ray AE. The rest of the light gets refracted and travels along AB. At B it again gets partially reflected and travels along BC. At C it refracts into air and travels along CF. The parallel rays AE and CF have a phase difference due to their different path lengths in different media. As can be seen from the figure, the phase difference depends on the angle of incidence θ₁ i.e., the angle of incidence at the top surface which is the angle of viewing, and also on the wavelength of the light as the refractive index of the material of the thin film depends on it. The two waves propagating along AE and CF interfere producing maxima and minima for different colours at different angles of viewing. One sees different colours when the film is viewed at different angles.

As the reflection is from the denser boundary, there is an additional phase difference of π radians (or an additional path difference λ). This should be taken into account for mathematical analysis.