The experiment by the English scientist Thomas Young in 1801 demonstrated the inseparability of the wave and particle nature of light or wave – particle duality.Morever , he was able to determine the wavelength of light by this experiment . The importance of this experiment was such that , Richard Feyman , one of the greatest physicts opined that all of quantum mechanics could be gleaned from carefully thinking through the implications of this single experiment. In this post , we will look at the key theoretical aspects of YDSE and problems related to it.Young’s double slit experiment
Points to note –
- The light used for this experiment is usually ( for best results ) collimated – light rays are distinct , parallel to each other and doesn’t mix with each other , monochromatic – having a single wavelength [ to validate the results] and coherent – having a single phase.
- The single slit ensures that light from only a single direction falls on the double slits.
- The light passing through S1 and S2 will be coherent as their source is the same.
- A bright fringe is created directly opposite the mid-point between the two slits. Here , the distances between the fringe and the two slits ( l1 and l2 ) are equal and waves contain the same no of wave lenghts – thus resulting in constructive interference . In addition , constructive interference produces more bright fringes on both sides of the middle , wherever the difference between l1 and l2 is an integer number of wave lenghts – λ , 2λ, 3λ etc.
- For dark fringes, l2 is larger than l1 by exactly one half a wavelength. Additional dark fringes are created at points where the difference between l1 and l2 equals an odd integer no of half wavelengths like 1[λ/2] , 3[λ/2]..etc . Dark fringes indicate destructive interference.
- Angle θ for maximum interference ( constructive) is given by sin θ = m λ/d where m = 0,1,2,3.. and d is the distance between S1 and S2.
- Angle θ for minimum interference ( destructive ) is given by sin θ = [ m + ½ ] λ/d where m = 0,1,2,3….
Conditions for Constructive & Destructive Interference –
Consider a point on a distant screen , such that the distance D from the slits is D >> d .Now , the small arc of the circle from the chosen point is almost a straight line and the path difference , Δx = d sin θ .
- The condition for constructive interference is Δ x = ± nλ ( n = o,1,2,3..) and for maxima , d sin θ = ± nλ .
- The condition for destructive interference is Δ x = ± [ n – ½ ] λ ( n = 1,2,3..) and for minima d sin θ = ± [ n – ½]λ .
- When D >> d , then sin θ = tan θ = θ = y/D .
- Position of nth bright fringe is yn = nλ [ D/d ] .
- Position of the nth dark fringe is yn = [ n – ½ ] λ D/d .
Fringe width –
Fringe width ( β ) is defined as the distance between two sucessive maxima or minima. It is given by β = λD/d .
- β is independent of n ( fringe order) as long as d and θ are small , i.e fringes are evenly spaced.
- For red light β will be higher as β is proportional to λ .
- If this experiment is done in a medium other than air , say water with a refractive index of μ , then , β’ = β/μ .
Maximum order for Interference Fringes –
- When n << d/λ , y/D = nλ / d .
- When n ≈ d /λ , then n = d sin θ /λ .
- Therefore n max = [ d/λ ] and n min = [ d/λ + ½ ] .
1 . When the rays are not parallel to the Principal axis –
- Δ x = d sin θ – yd/D
2 . When the source is placed beyond the central line –
- Maxima : Δ x = nλ
- Minima : Δ x = ( 2n – 1 ) λ / 2
3 . When a transparent glass slab of thickness t and refractive index μ is placed in one of the incoming wave paths , due to the increase of the path by ( μ – 1 ) t , the fringe pattern undergoes a shift , s given by –
- s = D/d ( μ – 1 ) t
Now on to some problems ……..
1. In a Young’s experiment, the upper slit is covered by a thin glass pate of refractive index 1.4, while the lower slit is covered by another glass plate, having the same thickness as the first one but having refractive index 1.7. Interference pattern is observed using light of wavelength 5400 Å. It is found that the point P on the screen, where the central maximum(n = 0) fall before the glass plates were inserted, now has ¾ the original intensity. It is further observed that what used to be the fifth maximum earlier lies below the point P while the sixth minima lies above P. Calculate the thickness of glass plate. (Absorption of light by glass plate may be neglected). ( IIT JEE 1997 )
From the given data ,μ1 = 1.4 and μ2 = 1.7 . Let t be the thickness of each glass plates. Path difference at O, due to insertion of glass plates will be
Δx = ( μ2– μ1) t = (1.7 – 1.4) t = 0.3 t …(1) . The point to deduce here is that the path dfference would be lying between 5 λ and 5λ + λ/2 , since the 5th maxima and the 6th minima lies on both sides of O .
Hence , Δx is taken as 5λ + Δ …(2) . Here , Δ < λ/2 . Now , the phase difference at O will be –
Φ = 2π /λ . Δx = 2π /λ ( 5λ + Δ ) = ( 10 π + 2π/λ Δ ) …… (3)
Now , we know that I (Φ ) = Imax cos² [ Φ / 2 ] and the intensity at O is given as 3/4 I max . Therefore ,
3/4 I max = I (Φ ) = Imax cos² [ Φ / 2 ] or 3/4 = cos² [ Φ / 2 ] ….. (4)
Substituting the value of eq (3 ) in eq. (4) and solving , we get Δ = λ /6 .
Hence , Δx = 5λ + λ /6 = 31 λ / 6 . Now from eq. (1) Δx = 0.3 t . Therefore ,
t = 31 λ/6 x 0.3 = 9.3 x 10-6 m = 9.3 μm .
2.A coherent parallel beam of microwaves of wavelength λ = 0.5 mm falls on a Young’s double slit apparatus. The separation between the slits is1.0 mm. The intensity of microwaves is measured on a screen placed parallel to the plane of the slits at a distance of 1.0 m from it as shown in the figure. a ) If the incident beam falls normally on the double slit apparatus, find the y-coordinates of all the interference minima on the screen. b ) If the incident beam makes an angle of 30o with the x-axis (as in the dotted arrow shown in figure), find the y-coordinates of the first minima on either side of the central maximum. ( IIT JEE 1998 )
Solution –normal path of incident rays
Refer [variations.2] above . For minimum intensity , d sin θ = ( 2n – 1 ) λ /2 so , sin θ = ( 2n – 1 ) λ /2d . Subtituting the values from the given data,
sin θ = 2n – 1 /4 . The point to note – as sin θ ≤ 1 , ( 2n – 1 ) λ /2 ≤ 1 , means that n ≤ 2.5 . Hence n can either be 1 or 2 .
If n = 1 , sin θ1 = 1/4 and tan θ1 = 1 /√15 . Conversely, if n = 2 , then sin θ2 = 3/4 and tan θ2 = 3 /√7 .
When D = 1m , y = tan θ . Hence , y1 = 1/√15 m = 0.26 m .Similarily , y2 = 3 / √7 m = 1.13 m .
As there can be minima on either side of O , there ideally should be 4 minima at positions ± 0.26 m and ± 1.13 m .
Part b ) If α = 30° , then Δx1 = d sin 30 = 0.5 mm which is equal to the given λ . Hence , the path difference betwwen the rays is λ .
At central maxima , the net path difference is zero . Hence , d sin θ = d sin α and θ = α = 30° or tan θ = 1/√3 = y0/D . Therefore , y0 = 0.58m.
Let X1 and X2 be the minimas flanking the central maxima. Now , considering X2 –
Δx2 – Δx1 = λ /2 therefore , Δx2 = 3λ /2 as Δx1 = λ . Thus , d sin θ2 = 3λ /2 = ¾ .
tan θ2 = y2/D = 3/√7 , hence , y2 = 1.13m . Working out in the same manner for X1 , tan θ1 = y1/D = 1/√15 , hence , y1 = 0.26m.
3. In a Young’s double slit experiment, two wavelengths of 500 nm and 700 nm were used. What is the minimum distance from the central maximum where their maximas coincide again? Take D/d = 103. Symbols have their usual meanings. ( IIT JEE 2004 )
Taking n1 as the bright fringe of wavelength λ1 which is given as 500 nm and n2 as that of the bright fringe λ2 given as 700nm , then –
n1 . λ1D/d = n2 . λ2 D/d . Therefore , n1/n2 = λ2/λ1 = 7/5 . The inference here is that the 7th maxima of λ1 will coincide with the 5th maxima of λ2 . Hence , the minimum distance = n1λ1D/d = 7 x 5 x 10 -7 x 10³ = 3.5 mm