WAVE OPTICS
INTERFERENCE . YOUNG’S DOUBLE SLIT EXPERIMENT
It was carried out in 1802 by the English scientist Thomas Young to prove the wave nature of light.
Two slits S1 and S2 are made in an opaque screen, parallel and very close to each other. These two are illuminated by another narrow slits S and light fall on both S1 and S2 which behave like coherent sources.
Note that the coherent sources are derived from the same source. In this way, any phase change which occurs in S will occur in both S1 and S2. The phase difference (φ − φ1 2) between S
1 and S2 is unaffected and remains constant.
Light now spreads out from both S1 and S2 and falls on a screen. It is essential that the waves from the two sources overlap on the same part of the screen. If one slit is covered up, the other produces a wide smoothly illuminated patch on the screen. But when both slits are open, the patch is seen to be crossed by dark and bright bands called interference fringes. This redistribution of intensity, pattern is called interference pattern.
...(v)
where φ is the phase difference and I is the resultant intensity.
Condition for bright fringes or maxima
φ =2nπ
or path difference, p n= λ where n = 0, 1, 2, …….
Imax =( I1+ I )2 2 …(vi)
Condition for dark fringes or minima
φ =(2n 1)− π or p n 1 2
= − λ path difference, where n = 1, 2, 3, …….
Imin=( I1 − I )2 2
...(vii) The relation between phase difference ( )φ and path difference (p) is given by
2 pπ
φ = λ ...(viii)
Distance of nth maxima from central bright fringe
S d S1
Schematic arrangement of YDSE
At point O the path difference between to superimposing waves is zero. So at that point central bright fringe will be formed.
Let P be the position of the nth maxima on the screen. The two waves arriving at P follow the path S1P and S2P, thus the path difference between the two waves is
p S P S P dsin= 1 − 2 = θ
From experimental conditions, we know that D >> d, therefore, the angle θ is small,
Thus For nth maxima
p n= λ Note that the nth minima comes before the nth maxima.
Fringe Width.
It is defined as the distance between two successive maxima or minima.
∴ n 1 n (c) find the position of the 2nd minima Solution. (a) Frindge width,
3
Illustration 2. White light is used to illuminate the two slits in a Young’s double slit experiment. The separation between the slits is b and the screen is at a distance D >> d from the slits. At a point on the screen directly in front of one of the slits find the missing wavelengths.
Solution. According to theory of interference, position y of a point on the screen is given by
It is defined as distance travelled by light in vacuum taking the same time in which it travels a given path length in a medium. If light travels a path length d in a medium at speed v, the tame taken by it will be (d/c).
So optical path length
d c
L c d (because )
v v
= × = µ µ = ... (xiv)
Since for all media optical path length is always greater then geometrical path length. When two light waves arrive at a point by travelling different distances in different media, the phase difference between the two is related by their optical path difference instead of simply path difference.
Phase Difference (optical path difference) Fringe Shift
When a transparent film of thickness t and refractive index µ is introduced infront of one of the slits, the fringe pattern shifts in the direction where the film is placed.
Expression for the fringe shift due to thin transperent filim Consider the YDSE arrangement shown in the figure.
A film of thickness t and refractive index is placed in front of the lower slit.
d
The optical path difference is given by p [(S P t)= 2 − + µ −t] S P1
In the absence of film the position of the nth maxima is given by equation
n
y n D d
= λ
… (xv)
Therefore, the fringe shift is given by
n n
FS y y ' ( 1)tD
= − = µ − d
...(xvi) Note the shift is in the direction where the film is introduced.
Illustration 3. In a YDSE λ=6000A D = 2 m, d = 6 mm. When a film of refractive index 1.5 is introduced in front of the lower slit, the third maxima shifts to the origin.
(a) find the thickness of the film
(b) find the positions of the fourth maxima
Solution. (a) Since 3rd minima shifts to the origin, therefore, the fringe shift is given by
3
− Bright Fringe
n=4n=3
(b) There are two positions of 4th maxima;
one above and the other below the origin.
A plane grass plate (acting as a mirror) is illuminated at almost grazing incidence by a light from a slit S1. A virtual image S2 is formed closed to S1 by reflection and these two act as coherent sources. The expression giving the fringes width is the same as for the double slit, but the fringe system differs in one important respect. In Lloyd’s mirror, if the point P, for example, is such that the path difference S2 P – S1 P is a whole number of wavelengths, the fringe at P is dark not bright. This is due to 180º phase change which occurs when light is reflected from a denser medium. This is equivalent to adding an extra half wavelength to the path of the reflected wave. At grazing incidence a fringe is formed at O, where the geometrical path difference between the direct and reflected waves is zero and it follows that it will be dark rather than bright.
Thus, whenever there exists a phase difference of π between the two interfering beams of light, conditions of maximas and minimas are interchanged, i.e.,
∆ = λx n (for minimum intensity) and ∆ =x (2n 1) / 2− λ (for maximum intensity) Interference in thin films.
Interference effects are commonly observed in thin films, such as thin layers of oil on water or the thin surface of a soap bubble.
The varied colours observed when white light is incident on such films result from the interference of waves reflected from the two surfaces of the film.
Consider a film of uniform thickness t and index of refraction µ, as shown in figure. Let us assume that the light rays travelling in air are nearly normal to the two surfaces of the film. To determine whether the reflected rays interfere constructively or destructively, we first note the following facts.
1 2
change No phase change
Interference in light reflected from a thin film is due to a combination of rays reflected from the upper and lower surfaces of the film.
(i) The wavelength of light in a medium whose refractive index is µ is,
µ λ = µ
µ
where λ is the wavelength of light in vacuum (or air)
(ii) If a wave is reflected from a denser medium it undergoes a phase change of 180º. Let us apply these rules to the film shown in figure. The path difference between the two rays 1 and 2 is 2t while the phase difference between them is 180º. Hence, condition of constructive interference will be,
Similarly, condition of destructive interference will be, 2 t nµ = λ n = 0, 1, 2, …
KEY CONCEPTS
In case of superposition of two waves,R 1 2 1 2
I = + +I I 2 I I cosφ, or IR ∝
(
A12 +A22 +2A A cos1 2 φ)
Condition for constructive interference 2 n,φ = ± π n = 1, 2, 3, ….
( )
∆ = ± λx n , n = 1, 2, 3, ….
Condition for destructive interference(
2n 1)
φ = ± − π
, n = 1, 2, 3, ….
( )
∆ = ±x(
2n 1 / 2− λ)
n = 1, 2, 3, …
Distance of nth bright fringe from C.B.F. is( )
yn B n D,= ± λd
n = 0, 1, 2, 3, ….
Distance of nth dark fringe from C.B.F. is
( )
yn D(
2n 1)
D
Displacement of fringe pattern due to introduction of a transparent sheet of R.I. and thickness t in YDSE = y0 = ( – 1) t (D/d), in the same side in which the transparent sheet is introduced.COD
CONCEPUTAL QUESTIONS