[Physics Class Notes] on Young's Double Slit Experiment Derivation Pdf for Exam

In 1801, an English physician and physicist established the principle of light interference, where he made a pinhole camera in cardboard and allowed sunlight to pass through it.

This white light was then allowed to fall upon another cardboard having two pinholes placed together symmetrically. 

The emerging light was received on a plane screen placed at some distance. 

At a given point on the screen, the waves emerging from two holes had different phases, interfering to give a pattern of bright and dark areas.

Such a variation of intensity on the plane screen demonstrated the light waves emerging from the two holes.

Therefore, this pattern of bright (constructive fringe) and dark (destructive fringe) areas can be sharply defined only if the light of a single wavelength is used.

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Young Double Slits Experiment Derivation

The two waves interfering at P have covered different distances.

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Let the waves from two coherent sources of light be represented as 

                  y1 = a sinωt…(1)

                  y2 = b Sin(ωt + Φ)…(2)

Here, a and b are amplitudes of the two waves resp. 

Φ is the constant phase angle by which the second wave leads the first wave.

Applying the superposition principle, the displacement(y) of the resultant wave at time (t) would be:

 y  = y1 +  y2 =   a sinωt +  b sin(ωt + Φ)

                        Expanding sin(ωt + Φ) = sin ωt cosΦ + cosωt . sinΦ

                          =    sinωt  (a+ b cosΦ ) + cosωt . b sinΦ

Put = a + b cosΦ = R cosӨ ….(3)

and b sinΦ = R sinӨ ….(4)

y  = sinωt. R cosӨ + cosωt . R sinӨ 

We get,

             y  = R sin(ωt + Ө)……(4)

The wave equation (4) represents the harmonic wave of amplitude R.

Now, squaring (3) and (4) and adding, we get

R2 (cos2Ө + sin2Ө) = (a + b cosΦ)2+ (b sinΦ)2

R2.1 = a2+ b2 Cos2Φ + 2ab cosΦ + b2Sin2 Φ

a2+ b2(Sin2 Φ+Cos2Φ)+2abcosΦ

R = √a2+ b2 + 2ab cosΦ

Since, intensity I α  amplitude R2.

I  α a2+ b2 + 2ab cosΦ

For constructive fringe: 

I should be maximum for which cosΦ = max or +1; Φ = 0, 2π, 4π…

I.e.,

 Φ = 2nπ 

(n = integer such that n = 0,1,2…)

If x is the path difference between the two waves reaching point P (in Fig.2) corresponding to phase difference Φ, then,

              x  = λ/2π Φ = [frac{lambda }{2pi }(2npi )]

 x = nλ….(p)                  

       

For destructive interference:

I should be minimum i.e., CosΦ= minimum when Φ = -1 or π, 3π, 5π….

         So,    

Φ = (2n – 1)π 

The path difference will be:

x  =  λ/2π Φ =  λ/2π (2n – 1)π

 x = λ/2(2n – 1)…(q)

Young’s Double Slits Formula Derivation

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Let S1 and S2 be two slits  separated by a distance d, and the center O equidistant from S1 and S2.

Let the slits be illuminated by a monochromatic source S of light of wavelength λ. 

Consider a point P at a distance y from C.

Here, O is the midpoint of  S1 and S2, and

                     D = OP₀ and D >> d

As S1S2  are perpendicular to OP₀ and S1A nearly perpendicular to O., we have

                                   ∠ S₁ S₂A = ∠ POP₀ = Ө

The path difference between two waves approaching at P is,

                     Δ x   = S₂P – S₁P = S₂P – PA (Since D>>d)

                              = S₂A = d SinӨ

                              =  d tanӨ = dy/D 

                            Δ x =  dy/D = nλ (where n is an integer)                       

So,  

                             y = nDλ/d…..(a)

At n = 0, y = 0

n = 1, y = Dλ/d

n = 2, y = 2Dλ/d

n  =3, y = Dλ/d,…etc

The centers of the dark fringes will be obtained when 

                      Δ x = dy/D = (2n – 1)λ

                                 y  = (2n -1)Dλ/d….(b)

At n =1, y =  Dλ/2d

n =2, y =  3Dλ/2d,

n  = 3, y = 5Dλ/2d,…etc

Now, to find the fringe width, subtracting equation (b)  from (a), we get

Fringe width,  w  = (2n -1)Dλ/d – nDλ/d = Dλ/d    

YDSE Derivation

If a glass slab of refractive index μ and thickness t is introduced on one of the paths of interfering waves, the optical length of this path will become μ instead of t, increasing by (t-1)μ.

This generates a path difference, given by,

 Δx =  μt – t = (t-1)μ                    

This path difference comes due to the glass slab.

If current position of fringe is y =D/d (Δx ), the new position will be 

                   y’  = D/d(Δx  + (t-1)μ) 

Lateral shift of fringe:

                              y₀  = y’ – y

                                   = D/d(t-1)μ

As, fringe width, w = D/dλ 

 We get,

   y₀   = β/λ  (t-1)μ