Young's Double Slit Experiment Derivation: Explanation, Solved Examples

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Young's Double Slit Experiment Derivation

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Considering two waves interfering at point P, having different distances. Consider a monochromatic light source ‘S’ kept at a relevant distance from two slits namely S₁ and S₂. S is at equal distance from S₁ and S₂. S_O, we can assume that S₁ and S₂ are two coherent sources derived from S.

Young's Double Slit Experiment Derivation

The light passes through these slits and falls on the screen that is kept at the distance D from both the slits S₁ and S₂.

• d is the separation between both the slits.
• S₁ is opened
• S₂ is closed
• Screen opposite to the S₁ is closed
• Screen opposite to S₂ is illuminating

Thus, an interference pattern takes place when both the slits S₁ and S₂ are open. When the slit separation ‘d ‘and the screen distance D are kept unchanged , to reach point P the light waves from slits S₁ and S₂ must travel at different distances. It implies that there is a path difference in Young double slit experiment between the two slits S₁ and S₂.

The video below explains this:

Young's Double Slit Experiment Detailed Video Explanation:

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Fringe’s Role in Young's Double Slit Experiment

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Fringe’s with maximum intensity:

For N=0 , the light fringe is known as the central fringe. The fringes of higher order are asymmetrically positioned along the central fringe. The position of the bright fringe is denoted by the nth fringe.

Y(brightness) = (nπ\d)D (n=0 , n=+1 -1, +2 -2….)

Fringe’s with minimum intensity:

Y(dark) = (2n-1)πD/2d (n=0 , n=+1 -1, +2 -2….)

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Fringe’s Width in Young's Double Slit Experiment

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Fringe width is the distance between two adjacent bright and adjacent dark fringes.

Let's consider the two fringes between at position n and n+1

So, fringe width = ((n+1 π\d ) D - (nπ\d)D = πD/ d

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Intensity of Fringe in Young's Double Slit Experiment

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For two slit sources S1 and S2 the resultant intensity at point p is noted by-

I = I₁ + I₂ +2√( I₁ X IΠ)CosΠ

Let I₁ = I₂ = I₀

I = I₀ + I₀+2√( I₀ X I₀) CosΠ

I = 2 I₀ + 2(I₀) CosΠ

I = 2 I₀ (1 + CosΠ )

I = 4 I₀ cos² (Π /2)

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Position of Fringe in Double Slit Experiment

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Position of Fringe in Double Slit Experiment are elaborated below:

Position of Bright Fringes

For maximum intensity or bright fringe to be formed at P

Path difference: Δz = nλ (n = 0, ±1, ±2, . . . .)

i.e., xd/D = nλ

or

x = nλD/d

The distance of the n^th  bright fringe from the centre is

x_n = nλD/d

Similarly, the distance of the (n-1)^th bright fringe from the centre is

x _(n-1)= (n -1)λD/d

Fringe width,  β = x_n  – x _(n-1) = nλD/d – (n -1)λD/d = λD/d

(n = 0, ±1, ±2, . . . .)

Position of Dark Fringes

For minimum intensity or dark fringe to be formed at P,

Path difference: Δz = (2n + 1) (λ/2) (n = 0, ±1, ±2, . . . .)

i.e., x = (2n +1)λD/2d

The distance of the n^th  dark fringe from the centre is:

x_n = (2n+1)λD/2d

Similarly, the distance of the (n-1)^th bright fringe from the centre is:

x _(n-1)= (2(n-1) +1)λD/2d

Fringe width, β = x_n  – x _(n-1) = (2n + 1) λD/2d – (2(n -1) + 1)λD/2d = λD/d

(n = 0, ±1, ±2, . . . .)

Also check:

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Things to Remember

• The young double slit experiment shows that matter and energy can display both wave and particle characteristics.
• The young double slit experiment used diffracted light from a single source passed into two more slits, used as coherent sources.
• Laser lights are rarely used as coherent sources in present-day experiments.
• Double slit experiment uses two sources of light placed at some distance, apart from each other
• When the slit separation ‘d ‘and the screen distance D is kept unchanged, to reach point P the light waves from slits S₁ and S₂ must travel at different distances. There is a path difference in Young double slit experiment between the two slits S₁ and S₂.
• Bright fringe formula is Y(brightness) = (nπ\d)D (n=0 , n=+1 -1, +2 -2….)
• Formula for Fringe width = ((n+1 π\d ) D - (nπ\d)D = πD/ d

Young Double Slit Experiment Free PDF

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