Question

Explain YDSE

Answer 1

Young's Double Slit Experiment (YDSE)

$\bull$ For an interference pattern to be observable,

• The waves must be of the same type and must meet at a point.
• The waves are coherent, i.e. the waves from each source maintain a constant phase difference.
• The waves must have the same wavelength and roughly the same amplitude.
• The waves must be both either unpolarised or have the same plane of polarisation.

$\bull$ Two sources are said to be coherent

if waves from the sources have a constant phase difference between them.

$\bull$ Young’s Double Slit Experiment A simple experiment of the interference of light was demonstrated by Thomas Young in 1801.

$\bull$ It provides solid evidence that light is a wave.

Interference fringes consisting of alternately bright and dark fringes (or bands) which are equally spaced are observed. These fringes are actually images of the slit.

At O, a point directly opposite the mid-point between S1 and S2, the path difference between waves$S_2O – S_1O$ is zero.

Thus constructive interference occurs and the central fringe or maxima is bright.

Suppose P is the position of the nth order bright fringe (or maxima). The path difference between the two sources S1 and S2 must differ by a whole number of wavelengths.

$S_2 P –S_1 P = nλ$

As the distance D is very much larger than a, the path difference $S_2 P – S_1 P$ can be approximated by dropping a perpendicular line S1 N from S1 to S2 P such that $S_1 P ≈ NP$

and

the path difference$\boxed{ S_2 P –S_1 P ≈ S_2N = nλ}$

From geometry, S2N = a sin θ where a is the distance between the centres of the two slits.

Equating, a sin θ = nλ and re-arranging,

$\boxed {sin θ = {nλ\over a}}$

But from geometry,

$\boxed{tan θ = {x_n \over D}}$

where x_n = distance of nth order fringe from the central axis Since θ is usually very small,

tan θ ≈ sin θ

i.e. $\boxed{{ x_n \over D} ={ nλ\over a }}$ or $\boxed{x_n= {nλD\over a}}$

Thus the separation between adjacent fringes (i.e. fringe separation) is,

$\boxed{Δx = x_{n+1} – x_n ={ (n+1) λ D\over a }– {nλ D\over a} ={ λD\over a}}$

Thus,

Fringe separation $\boxed {Δx ={ λ D\over a}}$

Clearly Δx is a constant if λ, D and a are kept constant. If all factors are kept constant, the fringes are evenly spaced near the central axis.

various diagrams