MEASUREMENT OF WAVE LENGTH BY NEWTON’S RINGS EXPERIMENT
è The effect can be used to determine the wavelength of a light source if the radius of curvature R, of the lower surface of the lens is known:-
• Where = Radius of the ring
t = PA = air film
R = Radius of curvature of the lens
è Diameter. D = 2R
Applying the theorem of intersecting chords
t Assuming the lens to have large radius of curvature R then t is very small and hence is very very small
2tR =
2t = = pass difference
• For a bright ring: Path difference = 2t = (
v
• Where n = 0, 1, 2, 3,…………
• First bright ring, n = 0
Second bright ring, n = 1 etc.
For a dark ring:
Path difference = 2t =
Where n = 0, 1, 2, 3, ………..
• Central dark spot, n = 0
• First dark ring, n = 1
• Second dark ring, n = 2 etc.
###### Problem 48
A set of Newton’s ring produced by monochromatic light is viewed through a traveling microscope. The diameter of the and (n + 5 dark rings are found to be 0.143cm and 0.287cm respectively. The radius of the curvature of Plano convex lens used is 64.5cm.
Determine the wavelength of the light used.
###### Problem 49
Calculate the radius of curvature of a Plano convex lens used to produce Newton’s rings with a flat glass plate if the diameter of the tenth dark ring is 4.48mm, viewed by normally reflected light of wavelength
5.0 m. What is the diameter of the twentieth bright ring?
WEDGE FRINGES
This is another form of interference produced by a thin wedge shaped film of air, the thickness of which gradually increases from zero along its length.
HOW A WEDGE IS FORMED
The wedge can be formed from two microscope slides clamped at one end and separated by a thin piece of paper at the other end so that the wedge angle is very small.
Monochromatic light from an extended source is partially reflected vertically downwards by the glass plate G.
When the microscope is focused on the wedge, bright and dark equally spaced fringes are seen parallel to the edge of contact of the wedge.
Some of the light falling on the wedge is reflected upwards from the bottom surface of the top slide and the rest which is transmitted through the wedge is reflected upwards from the top surface of the bottom slide.
Let ―‖ be thickness of the air wedge at P.
If a dark fringe is formed at P then:
Path
difference between the rays at
P = 2l =
• Where n = 0, 1, 2, 3, …………………
• First bright fringe, n = 0
• Second bright fringe, n = 1
DETERMINATION OF ANGLE OF THE WEDGE
In the figure above, let be the angle between the slides/plates (=wedge angle).
For a dark fringe at P we have:
è
If is in radians then we have:
è ———————- (3)
• For the (n + K dark fringe at Q, we have:
è here = Distance moved by the microscope
= wavelength of the light used.
##### Special case
If K = 1, then fringe width.
###### Problem 50
An air wedge is formed between two glass plates which are in contact at one end and separated by a piece of thin metal foil at the other end. Calculate the thickness of the foil if 30 dark fringes are observed between the ends when light of wavelength 6 m incident normally on the wedge.
###### Problem 51
A wedge air film is formed by placing aluminum foil between the two glass slides at a distance of 75mm from the line of constant of the slides. When the air wedge is illuminated normally by a light of wavelength 5.6 m interference fringe are produce parallel to the line of contact which have a separation of 1.2mm. Calculate the angle of the wedge and the thickness of the foil.
###### Problem 52
A narrow wedge of air is enclosed between two glass plates. When the wedge is illuminated by a beam of monochromatic light of wavelength 6.4 m parallel fringes are produced which are 0.3mm apart. Find the angle of the wedge.
###### Problem 53
A wedge shaped film of air is formed between two parallel sided glass plates by means of a straight piece of wire. The two plates are in contact along one edge of the film and the wire is parallel to this wedge. In such experiment, using light of wavelength 589nm, the distance between the 7th and 169th dark fringes was 26.3mm and the distance between the junction of the glass plates and the wire was 35.6mm. Calculate the angle of the wedge and the diameter of the wire.
###### Problem 54
An air wedge is formed between two flat glass plates of length 45mm by using a spacer at one end. Wedge fringes spaced 0.29mm apart are observed by reflection when light of wavelength 430nm is directed normally at the wedge. Calculate the angle of the wedge and the thickness of the spacer.
APPLICATION OF INTERFERENCE
(1 (1) QUALITY TESTING OF OPTICAL SURFACES
Fringes of equal thickness are useful for testing the quality of optical components.
##### Example
In making of optical flats the plate under test is made to form an air wedge with a standard plane glass surface.
Any uneven part of the surface which require more grinding will show up irregularities. The grinding of a lens surface may be checked if it is placed on an optical flat and Newton‘s rings observed in monochromatic light.
(2) MAKING NON-REFLECTING GLASS / BLOOMING LENSES
In optical instruments containing lenses or prisms light is lost by reflection at each refracting surface which results in reduced brightness of the final image.
The amount of light reflected at the surface can be appreciably reduced by coating it with a film of transparent material example magnesium fluoride. The process is called Blooming.
Blooming is the process of depositing a transparent film of a substance such as magnesium fluoride on a lens to reduce (or eliminate) the reflection of the light at the surface.
è The film is about one-quarter of a wavelength thick and has a lower refractive index than the lens.
Light reflected from the top (ray 1) and from the bottom (ray 2) surfaces of the film interfere destructively.
##### Condition
Destructive interference occurs when:
Path difference =
This is because the phase change which occur by reflection is 180 which is equivalent to v
• Where = thickness of the film
When light passes from one medium to another its speed and wavelength change but frequency (f) remain unchanged.
If are the speeds of light in air and in the film respectively then:
• Let n be refraction index of the material of the film.
è
è
• Substitute equation (2) in equation (1)
v
##### Problem 56
Expensive cameras have non-reflecting coated lens. Normally the lens surface is covered with a thin transparent film of fluoride of refractive index 1.43. suppose that no light of wave length 6000 is reflected when the incidence is norm
ally. Calculate the thickness of the film.
COLOUR IN THIN FILMS OF OIL / SOAP BUBBLE
When a drop of oil spreads on water a thin film of oil is formed on its surface.
In broad day light, the film appears to be made up of beautiful rainbow colors.
The formation of these colors in a thin film is due to interference phenomenon.
The interference phenomenon is the effect which happens when light wave gate reflected from the two opposite surfaces of a thin film.
When the path difference gives constructive interference for light of wavelength, the corresponding color is seen in the film.
The path difference varies with the thickness of the film and the angle of viewing both of which affect the color produced at any one part.
The colours seen in soup bubbles are also produced in this way.
DIFFRACTION OF LIGHT WAVES
Diffraction of light is the phenomenon of spreading or bending of light waves as they pass through a narrow opening (aperture) or round the edge of a barrier.
The diffracted waves subsequently interfere with each other producing regions of reinforcement and weakening
• Behind the obstacles / apertures at which diffraction occurs, a diffraction pattern of dark and bright fringes is formed.
• Diffraction is regarded as being due to the superposition of secondary wavelets from coherent sources on the unrestricted part of a wave front that has been obstructed by an obstacle aperture.
DIFFRACTION EFFECTS
• Diffraction of light is quite pronounced when the width of the opening is comparable with the wavelength of the light.
• This can be illustrated below
• In figure (i) and (ii) above the apertures are of the same size but the wave length of the incident waves are different.
• The waves are diffracted more when the wavelength is longer
In figure (iii) and (iv) the two waves have the same wavelength but the sizes of the apertures are different.
The diffraction is more in case of narrow slit.
TYPES OF DIFFRACTION
Diffraction of light is of two types (i) Fraunhofer diffraction
(ii) Fresnel diffraction
FRAUNHOFER DIFFRACTION
This is the type of diffraction which takes place at a narrow slit when parallel rays of light (plane wave front) are incident on it.
Both the light source and the receiving screen should be at infinite distance from the narrow slit.
Here a lens is used to obtain the diffraction pattern on a screen placed at a distance equals focal length () of the lens.
FRENSEL DIFFRACTION
This is a type of diffraction which takes place at a narrow slit when non-parallel rays of light are incident on it.
In this type of diffraction either the light source or the receiving screen or both are at finite distances from the narrow slit.
DIFFRACTION OF LIGHT AT A SINGLE SLIT
If a plane wave front is incident on a slit AB, then there will be the formation of a bright band on a screen at the centre followed by dark and bright bands of decreasing intensity on either sides of the central bright band.
A graph of fringe brightness against position is as shown in the figure above.
• Consider a plane wave front of monochromatic light of wavelength to
be incident on a slit AB
• By Huygen‘s principle, every point on the portion AB of the wavefront acts as a source of secondary wavelets spreading out in all directions.
• Those secondary wavelets which go straight across the slits arrive on the screen at point O in the same phase and hence the intensity at point O is maximum.
• Consider the 1st dark band (1st minimum) to be formed at any angle to the direction of the incident beam.
• 1st dark band (1st minimum) is obtained on the screen at P where:
Path difference,
From triangle ABN
Sin
But , BN =
Sin
For 1st dark band (1st minimum)
• Therefore, the general condition for a minimum for a single slit diffraction is
Where n = 3, ………….
The sign indicate that there are two nth order minima, one on each side of the original direction beam.
The variation of intensity with angle of direction for a single slit diffraction pattern is a shown below.
EXPRESSION OF WIDTH OF CENTRAL MAXIMUM
• The width of the central maximum is the distance between the first minimum on the either side of O as shown in the figure above.
• Thus, the width of the central maximum is
• The 1st minimum occurs at
Sin
• If is small then sin in radian
—————– (1)
• From the figure above:
Tan
——————— (2) è equation (1) = equation (2)
v ——————— (3)
– Now P = 2K
———— (4)
##### Problem 57
Monochromatic light of wavelength 6.0 m is incident at a single slit of width 5.0 m. How many orders of diffraction minimum are visible?
##### Problem 58
A parallel beam of light of wavelength 650nm is directed normally at a single slit of width 0,14mm in a darkened room. A screen is placed 1.50m from the single slit.
1. Sketch the graph to show how the intensity of light falling on the screen varies with position across the screen.
2. Calculate the width of the central fringe.
DIFFERENCE BETWEEN INTERFERENCE AND DIFFRACTION FRINGES
 INTERFERENCE DIFFRACTION (1)Interference fringe are obtained due to the superposition of light coming from two different wave fronts originating from two coherent source (1)Diffraction fringe are obtained due to the superposition of light coming from different parts of the same wave front. (2)The width of interference fringe is generally the same. (2)The width of diffraction fringes is not the same. (3)The intensity of all bright fringes is the same. (3)The intensity of all bright fringes is not the same. It is not the same. It is maximum for central fringes and decreases sharply for first, second etc. bright fringes.
PLANE TRANSMISSION DIFFRACTION GRATING
• -This is a device consisting of a large number of equidistant closely spaced parallel lines of equal width rule on glass.
• -The ruled widths are opaque to light while the space between any two successive lines is transparent and act as parallel slits.
Application
Diffraction gratings are used for producing spectra and for measuring wavelengths accurately.
THEORY
Suppose plane waves of monochromatic light of wave length fall on a transmission grating in which the slit separation (called grating spacing) is d.
Let N be number of lines per meter of the grating (= Grating Constant)
• Consider wavelets coming from corresponding point A and B on two successive slits and travelling at an angle to the direction of the incident beam.
è Path difference, BC = d
v
Where ―n‖ is an integer giving the order of the spectrum.
Here reinforcement of the diffracted wavelets occurs in the direction and a maximum (bright band) will be obtained.
The angular distribution of diffracted beams is as shown below:
NOTE
The actual number of orders possible for a given grating will depend on the width of the grating spaced and the wavelength of the light used.
Since cannot be greater than (90 diffraction), then the maximum number of orders possible for any given wavelength cannot be greater than the whole number value of n given by
―n‖ is maximum when sin = 1 è
Maximum number of orders possible
##### Problem 59
Monochromatic light of wavelength 600nm is incident normally on an optical transmission grating of spacing 2.00 Calculate:
1. The angular position of the maxima
2. The number of diffracted beams which can be observed
3. lower-alpha;margin-left: -0.75pt;vertical-align: baseline”>
The maximum order possible
##### Problem 60
The limits of the visible spectrum are approximately 400nm to 700nm. Find the angular spread of the first -order visible spectrum produced by a plane grating having 6000 lines per centimeter when light is incident normally on the grating.
##### Problem 61
A grating has 500 lines per millimeter and is illuminated normally with monochromatic light of wavelength 5.89 m.
1. How many diffraction maxima may be observed?
2. Calculate the angular separation.
##### Problem 62
What is the angular width of the first order spectrum produced by a diffraction grating of 5000 per cm when a parallel beam of while light is incident normally on it?
##### Problem 63
A monochromatic light of wavelength 2 falls normally on a grating which has 4 lines per cm.
1. What is largest order of spectrum that can be visible?
2. Find the angular separation between the third and fourth order image.
##### Problem 64
A diffraction grating with grating of 2 m is used to examine the light from the glaring gas. It is found that the first order of violet light emerges at an angle of 11.8 and the first order of red light emerges at angle of 15.8. Calculate:
(a) The wavelength of the two lights.
(b)The angle at which nth order of red light will coincide with the (n+1 order of violet light. Problem 65
A diffraction grating has 500 lines per mm when used with monochromatic light of m at normal incidence. At what angles will bright diffraction images be observed?
##### Problem 66
A slit 0.1mm wide is illuminated with a monochromatic light of wavelength 5000. How wide is the central maximum on a screen 1m from the slits?
##### Problem 67
A single narrow slits is illuminated with red light of wavelength 6328. A screen placed 1.60m from the slits shows a typical single –slit diffraction pattern. The separation between the first two minima is 4.0mm. what is the width of the slits?
##### Problem 68
Light of wavelength 7500 passes through a slit of 1.0 mm wide. How wide is the central maximum
1. In degrees and
2. In cm on a screen 20cm away?

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