PHYSICS As LEVEL(FORM FIVE) NOTES – WAVE MOTION-2(2)

Measurement of Wavelength by Newton’s Rings Experiment

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 r = Radius of the ring

t = PA = air film thickness

R = Radius of curvature of the lens

Diameter, D = 2r

Applying the theorem of intersecting chords:

Assuming the lens has a large radius of curvature R, then t is very small and hence t² is negligible.

Therefore,

2tR = r²

2t = path difference

• For a bright ring: Path difference = 2t = 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 rings produced by monochromatic light is viewed through a traveling microscope. The diameter of the and (n + 5)th dark rings are found to be 0.143 cm and 0.287 cm respectively. The radius of curvature of the plano-convex lens used is 64.5 cm.

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 4.48 mm, 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 the thickness of the air wedge at P.

If a dark fringe is formed at P then:

Path difference between the rays at P = 2t =

• 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)th 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 two glass slides at a distance of 75 mm from the line of contact of the slides. When the air wedge is illuminated normally by light of wavelength 5.6 , interference fringes are produced parallel to the line of contact which have a separation of 1.2 mm. 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 , parallel fringes are produced which are 0.3 mm 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 589 nm, the distance between the 7th and 169th dark fringes was 26.3 mm and the distance between the junction of the glass plates and the wire was 35.6 mm. 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 45 mm by using a spacer at one end. Wedge fringes spaced 0.29 mm apart are observed by reflection when light of wavelength 430 nm is directed normally at the wedge. Calculate the angle of the wedge and the thickness of the spacer.

Application of Interference

1. Quality Testing of Optical Surfaces

Fringes of equal thickness are useful for testing the quality of optical components.

In making 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 requires more grinding will show up as 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, for 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 light at the surface.

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 occurs by reflection is 180° which is equivalent to λ/2.

• Where = thickness of the film

When light passes from one medium to another, its speed and wavelength change but frequency (f) remains unchanged.

If and are the speeds of light in air and in the film respectively, then:

• Let n be the refractive index of the material of the film.

Substitute equation (2) in equation (1):

Problem 56

Expensive cameras have non-reflecting coated lenses. Normally the lens surface is covered with a thin transparent film of fluoride of refractive index 1.43. Suppose that no light of wavelength 6000 is reflected when the incidence is normal. 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 daylight, 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 waves are 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 soap 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 around 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 or 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 wavelength 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 equal to the focal length () of the lens.

Fresnel 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 side 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 λ 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 slit 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 θ 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 θ =

But BN =

Sin θ =

For 1st dark band (1st minimum):

• Therefore, the general condition for a minimum for a single slit diffraction is:

Where n = 1, 2, 3, …

The ± sign indicates 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 shown below:

Expression of Width of Central Maximum

The width of the central maximum is the distance between the first minimum on either side of O as shown in the figure above.

Thus, the width of the central maximum is:

The 1st minimum occurs at:

Sin θ =

If θ is small then sin θ ≈ θ in radians:

—————– (1)

From the figure above:

Tan θ =

Equation (1) = Equation (2)

——————— (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 650 nm is directed normally at a single slit of width 0.14 mm in a darkened room. A screen is placed 1.50 m 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

Plane Transmission Diffraction Grating

• This is a device consisting of a large number of equidistant closely spaced parallel lines of equal width ruled on glass.
• The ruled widths are opaque to light while the space between any two successive lines is transparent and acts as parallel slits.

Application

Diffraction gratings are used for producing spectra and for measuring wavelengths accurately.

Theory

Suppose plane waves of monochromatic light of wavelength λ fall on a transmission grating in which the slit separation (called grating spacing) is d.

Let N be the number of lines per meter of the grating (= Grating Constant).

• Consider wavelets coming from corresponding points A and B on two successive slits and travelling at an angle θ to the direction of the incident beam.

Path difference, BC = d sin θ

Where n is an integer giving the order of the spectrum.

Here reinforcement of the diffracted wavelets occurs in the direction θ 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 sin θ cannot be greater than 1 (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 θ = 1, i.e.,

Problem 59

Monochromatic light of wavelength 600 nm is incident normally on an optical transmission grating of spacing 2.00 .

1. The angular position of the maxima.
2. The number of diffracted beams which can be observed.
3. The maximum order possible.

Problem 60

The limits of the visible spectrum are approximately 400 nm to 700 nm. 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 lines per cm when a parallel beam of white 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 the 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:

1. The wavelength of the two lights.
2. The angle at which nth order of red light will coincide with the (n+1)th order of violet light.

Problem 65

A diffraction grating has 500 lines per mm when used with monochromatic light of wavelength m at normal incidence. At what angles will bright diffraction images be observed?

Problem 66

A slit 0.1 mm wide is illuminated with a monochromatic light of wavelength 5000 . How wide is the central maximum on a screen 1 m from the slits?

Problem 67

A single narrow slit is illuminated with red light of wavelength 6328 . A screen placed 1.60 m from the slits shows a typical single-slit diffraction pattern. The separation between the first two minima is 4.0 mm. What is the width of the slit?

Problem 68

Light of wavelength 7500 passes through a slit of 1.0 mm wide. How wide is the central maximum:

1. In degrees?
2. In cm on a screen 20 cm away?