Contents
Some of the most important Physics Topics include energy, motion, and force.
What do you Mean by Non-localised Fringes?
It has already been stated that for sustained interference, a pair of coherent sources is necessary. When sources are not coherent, intensity at any point changes so rapidly that no Interference fringe is observed in practice, and all points appear equally bright.
Two similar but separate sources do not form coherent sources. Even two very close points on the same source are not coherent. Hence there are special methods of creating coherent sources. Some are described below:
i) In Young’s double slit experiment, two slits S1 and S2, kept at fixed distance from light source S, act as coherent sources [Fig.].
ii) In Lloyd’s single mirror experiment, a thin illuminated slit S and its virtual image, S’, formed due to reflection from a plane mirror, serve as a pair of coherent sources [Fig.].
iii) In experiment with Fresnel’s biprism, two virtual images S1 and S2 of a source S, produced by refraction through the biprism, act as coherent sources [Fig.(c)].
Explanation of Formation of Interference Fringes
The formation of interference fringes can be explained from the principle of superposition of waves. In Young’s experiment, slits A and B are equidistant from source M. In that, case, the secondary sources will also be coherent and hence in phase. Wavefronts from these two sources, propagate one after another,
through the space between the obstacle and the screen. Hence they superpose. In Fig., solid line arcs show wavefronts in the same phase. Broken line arcs, in between represent wavefronts in the opposite phase. Clearly, waves coming from two sources proceeding towards the points a, c and e, superpose in the same phase. Thus amplitude of the resultant wave becomes maximum and constructive interference takes place along the lines leading to the points a, c and e on the screen.
On the other hand, waves directed towards points b or d, superpose in opposite phase. Thus amplitude of the resultant wave is zero and destructive interference takes place. Therefore, interference fringes consisting of bright and dark lines are displayed on screen S. If wavelength of monochromatic light be λ, the path difference between two waves along the dark lines happens to be odd multiples of \(\frac{\lambda}{2}\) and that along bright lines, even multiples of \(\frac{\lambda}{2}\).
1. If white light is taken instead of a monochromatic light in Young’s double silt experiment, the central bright fringe will be white and bright coloured fringes will be observed on either side of the central fringe. This is because wave lengths of the colours, forming white are different and therefore each one produces its characteristic interference fringe pattern. Thus fringes of different colours are produced. As the central line of the interference pattern is equidistant from each of the two coherent sources, the light of all component colours reach in phase and a bright white light is formed at that point.
2. The fringes disappear when one of the slits A or B is covered by an opaque plate and the screen gets illuminated with a uniform intensity.
3. If one of the slits is covered with a translucent paper, fringes of the same width will be formed. But the bright band will look less bright and dark band will look less dark.
Width of interference fringes: In Fig., A and B are two coherent sources of monochromatic light, S is a screen, 2d = distance between A and B, O is the mid-point of AB, D = perpendicular distance of AB from screen. As AC = BC, waves starting from two the sources in phase reach point C in phase.
Hence, resultant amplitude as well as intensity at C is maximum. All points on the line perpendicular to the plane of the paper and passing through C will be equidistant from the sources A and B . Thus a central bright fringe is formed which is a straight line.
Now a point P is taken at a distance x from C i.e., CP = x; AP and BP are joined. As the lengths of two paths are not equal there will be a phase difference between the waves reaching P from A and B. Thus whether constructive or destructive interference will take place at the point P depends on the path difference between the two incoming waves.
From the figure,
BP2 = D2 + (x + d)2 and AP2 = D2 + (x – d)2
∴ BP2 – AP2 = (x + d)2 – (x – d)2 = 4xd
or, BP – AP = \(\frac{4 x d}{B P+A P}\) – \(\frac{4 x d}{2 D}\) = \(\frac{2 x d}{2 D}\) …. (1)
[Since, D \(\gg\)d hence BP = AP ≈ D]
Thus path difference between the two waves in reaching P,
δ = BP – AP = \(\frac{2 x d}{D}\)
By the condition of interference, for n -th bright fringe at P, path difference
δ = \(\frac{2 x_n d}{D}\) = 2n ᐧ \(\frac{\lambda}{2}\) [n = 0, 1, 2, …]
or, xn = \(\frac{D n \lambda}{2 d}\) …. (2)
[xn is the distance of n-th bright fringe on either side of C]
Taking n = 1, 2, 3…, the positions of 1st, 2nd, 3rd etc. bright fringe, on either side of the central bright fringe, can be found out.
For (n + 1) -th bright fringe,
xn + 1 = \(\frac{D(n+1) \lambda}{2 d}\) …. (3)
So, the distance between two consecutive bright fringes
= xn + 1 – xn = \(\frac{D(n+1) \lambda}{2 d}\) – \(\frac{D n \lambda}{2 d}\) = \(\frac{D}{2 d}\) ᐧ λ ….. (4)
By the condition for formation of the n -th dark fringe at P is,
δ = \(\frac{2 x_n d}{d}\) = (2n + 1)\(\frac{\lambda}{2}\) or, xn = \(\frac{D}{2 d}\)(2n + 1)\(\frac{\lambda}{2}\) ……. (5)
Taking n = 0, 1, 2 etc, positions of 1st, 2nd, 3rd, etc., dark fringes on either side of the central bright fringe can be found out.
Hence for (n + 1) -th dark fringe,
xn+1 = \(\frac{D}{2 d}\)[2(n + 1) + 1]\(\frac{\lambda}{2}\) ….. (6)
Hence the distance between two consecutive dark fringes
= xn + 1 – xn = \(\frac{D}{2 d}\)[2(n + 1) + 1]\(\frac{\lambda}{2}\) – \(\frac{D}{2 d}\)(2n + 1)\(\frac{\lambda}{2}\)
= \(\frac{D}{2 d}\) ᐧ λ ……. (7)
Thus the distance between two consecutive bright or dark fringes is the same. This distance is called fringe width.
Therefore if y is the fringe width then,
y = \(\frac{D \lambda}{2 d}\) ……. (8)
It is clear from equation (8) that—
- Since there is no ‘n’ in the expression of y, it can be said that fringe width does not depend on order of the fringe. All fringes are of the same width.
- Fringe width is directly proportional to wavelength of light used. For a greater wavelength, fringe width increases, i.e., fringes will be wider. Similarly, for shorter wavelengths, fringes will be thinner.
- If value of D is large, fringe width will increase.
- If value of d is small, fringe width will increase.
- Also if the total experiment is conducted in any other medium, wavelength decreases (λ’ = \(\frac{\lambda}{\mu}\)) . Hence, fringe width decreases.
Further, in whichever position the screen is placed in front of sources A and B, the interference pattern is always observed. As interference patterns are not restricted to a fixed place, these are called non-localised fringes.
Angular fringe width: If the angular position of n-th fringe on the screen be θn then,
θn = \(\frac{x_n}{D}\) = \(\frac{\frac{D n \lambda}{2 d}}{D}\) = \(\frac{n \lambda}{2 d}\)
For (n + 1) -th fringe, θn + 1 = \(\frac{(n+1) \lambda}{2 d}\)
So, angular separation between two successive fringes i.e., angular width of fringe,
θ = θn + 1 – θn = \(\frac{(n+1) \lambda}{2 d}\) – \(\frac{n \lambda}{2 d}\) = \(\frac{\lambda}{2 d}\) ……. (9)
- The angular width of the fringe does not depend on the position of the screen.
- Angular width decreases with the increase of the separation between the coherent sources and vice versa. (∵ θ ∝ \(\frac{1}{2 d}\))
- If the entire experiment is performed inside any liquid (refractive index = µ say), the angular width decreases
(∵ λliquid < λair).
Optical path: The equivalent optical path of the distance covered by a certain monochromatic light through a certain medium in a certain interval of time is the path covered by light in the same interval of time through vacuum.
Suppose, refractive index of a medium for a certain monochromatic ray is µ. Velocity of the ray in that medium is v.
∴ µ = \(\frac{c}{v}\) or c = µv …… (10)
[where, e = velocity of light in vacuum]
Now if the said monochromatic light takes time t to travel through the said medium then,
v = \(\frac{x}{t}\)
And if light can travel distance l in the same interval of time through vacuum then,
c = \(\frac{l}{t}\)
Hence from equation (10) we get,
\(\frac{l}{t}\) = µ ᐧ \(\frac{x}{t}\) or, l = µ ᐧ x
Here l, according to the definition of optical path, is the equivalent optical path of x.
For more than one medium, we can write optical path,
l = Σµixi
Displacement of fringes due to introduction of a thin plate: As shown in Fig., A and B are two monochromatic, coherent sources of light, S is the screen, 2d = distance between A and B, O is the mid-point of AB, D = perpendicular distance of AB from the screen. Waves coming from the coherent sources produce interference pattern on the screen. Generally point C is the position of the central bright fringe as AC = BC. A point P is considered on the screen. Now a glass plate is introduced perpendicularly in the path AP. Let its thickness be t and the refractive index of glass be µ. It results in the shifting of the entire fringe along with its central fringe.
This is caused due to the difference in velocity of light in air and glass.
While moving from A to P, light travels a path (AP – t) through air (µ = 1) and a path t through glass.
Hence optical path from A to P = (AP – t) ᐧ 1 + µ ᐧ t
= AP + t(µ – 1)
Difference in optical paths from the points A and B,
δ = BP – AP – t(µ – 1) = (BP – AP) – t(µ – 1)
= \(\frac{2 x d}{D}\) – t(µ – 1) [Following the equation (1)]
If P is at the centre of n -th bright fringe,
δ = 2n ᐧ \(\frac{\lambda}{2}\) = nλ, where n = 0, 1, 2,…..
∴ \(\frac{2 x d}{D}\) – t(µ – 1) = nλ or, x = \(\frac{D}{2 d}\) {nλ + t(µ – 1)}
Hence, because of the presence of glass plate, distance of n -th bright fringe from C, x = \(\frac{D}{2 d}\){nλ + t(µ – 1)}. Putting n = 0 in this equation we get the displacement of the central bright fringe due to insertion of plate.
Let the displacement be x0.
∴ x0 = \(\frac{D}{2 d}\)(µ – 1)t …….. (11)
Refractive index of glass µ is greater than 1, hence x0 is positive, i.e., central bright fringe will move further from C towards the glass plate.
Putting n = 1, displacement of the 1st bright fringe i.e., x1 is obtained.
∴ x1 = \(\frac{D}{2 d}\){λ + (µ – 1)t}
∴ Fringe width,
y = x1 – x0 = \(\frac{D}{2 d}\){λ + (µ – 1)t} – \(\frac{D}{2 d}\)(µ – 1)t = \(\frac{D}{2 d}\) ᐧ λ
It proves that though a displacement occurs in fringe pattern due to introduction of a glass plate fringe width remains unaltered.
∴ \(\frac{y}{\lambda}\) = \(\frac{2 D}{d}\) …….. (12)
From equations (11) and (12),
x0 = \(\frac{y}{\lambda}\)(μ – 1)t ………. (13)
If x0, y, t and λ are known, refractive index p of the material
of the plate can be found out from equation (13).
Also if μ is known, t can be calculated out.
Due to introduction of the glass plate, if the central bright fringe shifts a distance qual to the length of m bright fringes, then
x0 = my or, \(\frac{y}{\lambda}\)(μ – 1)t = my
or, (μ – 1)t = μλ or, m = \(\frac{(\mu-1) t}{\lambda}\)
This equation gives the equivalent number of fringe width by which a displacement of total fringe pattern occurs.
Numerical Examples
Example 1.
A screen is placed a a distance of 5 cm from a point source. A 5 mm thick piece of glass of refractive index 1.5 is placed between them. What is the length of optical path between the source and the screen?
Solution:
Length of air path between the source and screen
= 5 – 0.5 = 4.5 cm
Air path, equivalent to path through glass plate
= refractive index × thickness = 1.5 × 0.5 = 0.75 cm
∴ Length of optical path = 4.5 + 0.75 = 5.25 cm.
Example 2.
Two straight and parallel slits, 0.4 mm apart, are illuminated by a source of monochromatic light. Interference pattern of fringe width 0.5 mm is produced 40 cm away from the silts. Find the wavelength of the light used.
Solution:
In this case, 2d = 0.4 mm = 0.04 cm, y = 0.5 mm = 0.05 cm, D = 40 cm
We know, the fringe width,
y = \(\frac{\lambda D}{2 d}\)
∴ λ = \(\frac{2 d \cdot y}{D}\) = \(\frac{0.04 \times 0.05}{40}\) = 5 × 10-5 cm
= 5000 × 10-8 cm = 5000 Å
Example 3.
In a Young’s double slit experiment on interference, distance between two vertical slits was 0.5 mm and distance of the screen from the plane of slits was 100 cm. It was observed that the 4th bright band was 2.945 mm away from the second dark band. Find the wavelength of light used.
Solution:
From Fig. it is observed that the distance between second dark band to 4th bright band = 2.5 × band width. If band width or fringe width is y then,
2.5 × y = 2.945
∴ y = \(\frac{2.945}{2.5}\) = 1.178 mm
= 0.1178 cm
As, y = \(\frac{D \lambda}{2 d}\)
So, λ = \(\frac{2 d \cdot y}{D}\) = \(\frac{0.05 \times 0.1178}{100}\)
= 5890 × 10-8cm = 5890 Å
Example 4.
Monochromatic light of wavelength 6000A was used to set up interference fringes. On the path of one of the interfering waves a mica sheet 12 × 10-5 cm thick was placed when the central bright fringe was found to be displaced by a distance equal to the width of a bright fringe. What is the refractive index of mica?
Solution:
On placing the mica sheet, if the central bright fringe shifts by the distance of m number of bright-fringes then,
(µ – 1 )t = µλ
Here, m = 1, λ = 6000Å = 6000 × 10-8 cm, t = 12 × 10-5cm
∴ (µ – 1) × 12 × 10-5 = 1 × 6000 × 10-8
or, (µ – 1) = \(\frac{1}{2}\)
or, µ = 1.5
Example 5.
A ray of light of wavelength 6 × 10-5 cm, after passing through two narrow slits, 1 mm apart, forms interfer-ence fringes on a screen placed 1 m away. Find the dis-tance between two successive bright bands of the fringes.
Solution:
Distance between two successive bright bands is the fringe width.
Width of the fringe, y = \(\frac{D}{2 d}\) × λ
Here, D = 1m, λ = 6 × 10-5 cm, 2d = 1 mm = 0.1cm
∴ y = \(\frac{100 \times 6 \times 10^{-5}}{0.1}\) = 0.006 cm
Example 6.
In a double slit experiment using a monochromatic light, interference fringes are formed on a screen at a particular distance from the slits. If the screen is moved towards the slits by 5 × 10-2 m, then the fringe width changes by 3 × 10-5 m. If the distance between the two slits is 10-3 m , then determine the wavelength of the light used.
Solution:
Fringe width, y = \(\frac{\lambda D}{2 d}\)
Here, λ is the wavelength of light used, D is the distance of the screen from the slits, 2d is the separation between the slits. Here screen distance changes by ΔD and fringe width changes by Δy.
∴ Δy = \(\frac{\lambda \Delta D}{2 d}\) or λ = \(\frac{\Delta y 2 d}{\Delta D}\)
Now, Δy = 3 × 10-5m, 2d = 10-3m, ΔD = 5 × 10-2m
∴ λ = \(\frac{3 \times 10^{-5} \times 10^{-3}}{5 \times 10^{-2}}\) = 6 × 10-7m = 6000 Å
Example 7.
In Young’s double slit experiment, the width of the fringe is 2.0 mm. Find the distance between 9th bright band and 2nd dark band. [HS’ 13]
Solution:
It is seen from the Fig., the distance of the 9th bright band from second dark band
= 7.5 × width of fringe = 7.5 × 0.2 cm = 1.5 cm
Example 8.
Using light of wavelength 600 nm in Young’s double slit experiment 12 bands are found on one part of the screen. If the wavelength of light is changed to 400 nm, then what will be the number of bands on that part of the screen? [IIT 01]
Solution:
In first case, wavelength, λ1 = 600 × 10-9 m; number of fringes, n1 = 12, fringe width y1. In second case, wavelength, λ2 = 400 nm = 400 × 10-9 m, number of fringes n2 and fringe width y2.
Let the distance between the slits and the screen be D, separation between the slits be 2d and length of the entire fringe pattern be x.
Example 9.
Green light of wavelength 5100 A from a narrow slit is incident on a double slit. If the overall separation of 10 fringes on a screen 200 cm away is 2 cm, find the sep-aration between the slits. [HS ’15]
Solution:
Width of 10 fringes, x = \(\frac{D}{2 d} \lambda\) = 2 cm
Here, D = distance of this screen =200 cm, 2d = separation between the two slits, λ = wavelength of light = 5100A = 5100 × 10-8 cm
∴ 2d = 10\(\frac{D}{x}\)λ = \(\frac{10 \times 200 \times 5100 \times 10^{-8}}{2}\)
= 51 × 10-3 = 0.051 cm
Example 10.
The ratio of the intensities between two coherent light sources used in Young’s double slit experiment, is n. Find the ratio of the intensities of principal maximum and minima of the band. [HS 13]
Solution:
Let the intensities of the sources be I1 and I2.
According to question, I1 = nI2
Now, if Imax and Imin be the intensities of central maximum