Young’s Double Slit Experiment

Young’s double slit Experiment

Young’s interference experiment, also called Young’s double-slit interferometer, was the original version of the modern double-slit experiment, performed at the beginning of the nineteenth century by Thomas Young. This experiment played a major role in the general acceptance of the wave theory of light. Young’s double slit experiment uses two coherent sources of light placed at a small distance apart, usually only few orders of magnitude greater than the wavelength of light is used.

Young’s double slit Experiment

Let A and B be two fine slits, a small distance d apart. Let them be illuminated by a monochromatic light of wavelength l. MN in the screen is at a distance D from the slits AB. The waves from A and B superimpose upon each other and an interference pattern is obtained on the screen. The point C is equidistant from A and B and therefore the path difference between the waves will be zero and so the point C is of maximum intensity. It is called the central maximum.

For another point P at a distance ‘x’ from C, the path difference at P = BP – AP.

Now AB = EF = d, AE = BF = D

ΔBPF (Using Pythagoras Theorem):

BP² = [BF² + PF²]½ = \({{\left[ {{D}^{2}}\,+\,{{\left( x\,+\,\frac{d}{d} \right)}^{2}} \right]}^{1/2}}\,=\,D{{\left[ 1\,+\,\frac{{{\left( x+\frac{d}{2} \right)}^{2}}}{{{D}^{2}}} \right]}^{1/2}}\)

Similarly in ΔAPE:

\(AP\,=\,D\,{{\left[ 1\,+\,\frac{{{\left( x\,-\,\frac{d}{2} \right)}^{2}}}{{{D}^{2}}} \right]}^{1/2}}\)

∴ BP – AP = \(D\left[ 1+\frac{1}{2}\frac{{{\left( x+\frac{d}{2} \right)}^{2}}}{{{D}^{2}}} \right]\,-\,D\left[ 1\,+\,\frac{1}{2}\frac{{{\left( x\,-\,\frac{d}{2} \right)}^{2}}}{{{D}^{2}}} \right]\)

After expanding the above equation binomially,

\(BP\,-\,AP=\,\frac{1}{2D}\,\left[ 4\times \frac{d}{2} \right]\,=\,\frac{xd}{D}\)

For bright fringes (constructive wavelength) the path difference is integral multiple of wavelength i.e., path difference is nλ.

∴ \(n\lambda \,=\,\frac{xd}{D}\)

⇒ \(x\,\,=\,\,\frac{n\lambda D}{d}\)

Where, n = 0, 1, 2, 3, 4 …

(x therefore represents distance of nth bright fringe from C)


∴ n = 0           x₀ = 0

n = 1    \({{x}_{1}}\,=\,\frac{\lambda D}{d}\)

n = 2    \({{x}_{2}}\,=\,\frac{2\lambda D}{d}\)

n = 3    \({{x}_{3}}\,=\,\frac{3\lambda D}{d}\)

And so on.

\({{x}_{1}}\,=\,\frac{\lambda D}{d}\) is the distance of the 1st bright fringe and therefore

\({{x}_{n}}\,=\,\frac{n\lambda D}{d}\) will be the distance of the nth bright fringe from C.

Therefore, separation between the centres of two consecutive bright fringe is the width of a dark fringe.

∴ β₁ = xn – xn₋₁ = \(\frac{\lambda D}{d}\)

Similarly for dark fringes,

xn = (2n – 1)  \(\frac{\lambda D}{2d}\)

For n = 1, \({{x}_{1}}\,=\,\frac{\lambda D}{2d}\) → Position of 1st dark fringe.

For n = 2, \({{x}_{2}}\,=\,\frac{3\lambda D}{2d}\) → Position of 2nd dark fringe.

The separation between the centres of two consecutive dark interference fringes is the width of a bright fringe.

∴ β₂ = xn – xn₋₁ =\(\frac{\lambda D}{d}\)

The separation between the centres of two consecutive dark interference fringes is the width of a bright fringe. All bright and dark fringes are equal width as b₁ = b₂.