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Explain Young's double slit experiment (3 marks)

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Young’s Double Slit Experiment A simple experiment of the interference of light was demonstrated by Thomas Young in 1801. 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 S2O – S1O 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.

S2 P –S1 P = nλ

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

the path difference S2 P –S1 P ≈ S2N = 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, sin θ = nλ/a

But from geometry, tan θ = xn / D where xn = distance of nth order fringe from the central axis Since θ is usually very small, tan θ ≈ sin θ i.e. xn / D = nλ/a or xn= nλD/a

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

Δx = xn+1 – xn = (n+1) λ D/a – nλ D/a = λD/a

Thus,

Fringe separation Δx = λ D/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.

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