PHYSICS As LEVEL(FORM FIVE) NOTES – WAVE MOTION-2(1)

WAVE MOTION-2

Derivation of Snell’s Law on the Basis of Huygen’s Principle

Consider a plane wave front AA’ which is incident on a transparent medium MM’.

The position of a new wave front after time t may be found by applying Huygen’s principle with points on AA’ as centers.

Those wavelets originating near the upper end of AA’ travel with a speed in the first medium and those wavelets originating near the lower end of AA’ travel with a speed in the second medium.

Hence, after time t:

• Distance travelled in the first medium = = t
• Distance travelled in the second medium = = t

Relationship Between i and r

Where i = angle of incidence, r = angle of refraction.

From triangle AOQ:

sin i = (1)

From triangle AOR:

sin r = (2)

Dividing (1) by (2):

For a given pair of media, this ratio is constant which expresses Snell‘s law. From V = :

Equation (3) becomes:

Where is the refractive index of the second medium with respect to the first.

Definition

The refractive index (η) of a material is the ratio of the sine of the angle of incidence to the sine of the angle of refraction.

Problem 36

Monochromatic light of wavelength 800 nm enters a glass plate of refractive index 1.5.

Calculate:

1. The speed of light in glass
2. The frequency of the light
3. The wavelength of light in glass

Given that C = 3 × 10⁸ m/s and 1 nm = 1 × 10^-9 m.

Interference of Light Waves

Interference is a situation in which two or more waves overlap in space. In a region where two or more light waves cross, superposition occurs giving reinforcement (addition) of the waves at some points and cancellation (subtraction) at others.

The resulting effect is called an interference pattern or system of fringes.

Definition

Interference pattern/system of fringes is a set of light bands (fringes) and dark bands (fringes) formed on a screen when interference of light waves occurs.

Interference Pattern

Where B = Bright band (fringe)

D = Dark band (fringe)

In order to bring about the interference of light waves, the first task is to produce two coherent sources of light.

Coherent sources of light are sources producing light of the same wavelength, frequency, and amplitude.

• The sources have constant phase difference.
• Two independent sources of light cannot be coherent.
• This is because the emission of light from any source is from a very large number of atoms and the emission from each atom is random.
• Therefore, there is no stable phase relationship between radiations from two independent sources.

How to Produce Coherent Sources of Light

• Coherent sources of light are obtained by splitting up the light from a single source (primary source) into two parts (secondary sources).

Methods of Producing Interference Pattern

• Interference pattern can be produced by either of the following methods:

1. Young‘s double slit
2. Lloyd‘s mirror
3. Fresnel‘s biprism
4. Newton‘s rings
5. Wedge fringes

In this experiment, monochromatic light is passed through a slit “s” and the light emerging from this slit is used to illuminate two adjacent slits.

By allowing light from these two slits to fall on the screen, a series of alternate bright and dark bands/fringes are formed on the screen.

The bright bands/fringes represent constructive interference while the dark bands/fringes represent destructive interference.

Constructive Interference

Constructive interference occurs when the interfering waves are in phase.

Waves are said to be in phase if their maximum and minimum values occur at the same instant.

Example:

When constructive interference occurs, the two interfering waves combine together to give a wave of larger amplitude and hence a bright fringe is formed.

Destructive Interference

Destructive interference occurs when the two interfering waves are out of phase.

Waves are said to be out of phase if the maximum of one wave and minimum of the other wave occur at the same instant.

Example:

When destructive interference occurs, the two interfering waves cancel each other to produce nothing and hence a dark fringe is formed.

Theory of Young’s Double Slit Experiment

Consider the following:

Let A and B be two close slits separated by a distance a.

Let O be the central bright fringe.

Suppose the bright fringe is formed at point P a distance y from O.

Concept

If the two coherent monochromatic light sources are A and B, then a bright fringe can be seen at P only if the path difference (BP – AP) is a whole number of wavelengths:

BP – AP = nλ, where n = 0, 1, 2, 3, …

If the difference is an odd number of half wavelengths, then darkness is obtained at the point considered:

BP – AP = (n + ½)λ, where n = 0, 1, 2, 3, …

In the figure above, for a bright fringe to be seen at P, we have:

• Path difference = BP – AP = BP – NP = BN

Let θ be the angular displacement of the bright fringe from the central bright fringe.

Consider triangles ABN and PMO, angle m(BN) = m(PMO) = θ.

From triangle BAN:

sin θ = BN / AB = y / a

Since θ is very small, sin θ ≈ tan θ ≈ θ (in radians).

From triangle PMO:

tan θ = y / D

Where D = distance from the slits to the screen.

Equating the two expressions:

y / a = y / D

Therefore:

y = nλD / a

This gives the displacement of the nth bright fringe from the central bright fringe at O.

Note

• Central bright fringe, n = 0
• First bright fringe, n = 1
• Second bright fringe, n = 2
• etc.

Similarly, the displacement from O for a dark fringe is given by:

y = (n + ½)λD / a

• First dark fringe, n = 0
• Second dark fringe, n = 1
• Third dark fringe, n = 2
• etc.

All fringes are of the same width.

Some Points About Young’s Interference Fringes

1. If the source slit “S” is put very close to the double slits, the fringe separation does not change but the fringe intensity increases.
2. If the double slits are very close so that their separation distance approaches zero, then the fringe separation increases.
3. If one of the double slits is covered up, then no fringes can be seen on the screen.
4. If the source slit S is made wider, then the fringes on the screen disappear.
5. If white light is used instead of monochromatic light source, then the central fringe at O is white but the other bright fringes on either side of the central fringe are colored with violet near O and red far away from O.

Angular Width of a Fringe

If P is the position of the first bright fringe, then OP = y.

Since θ is very small, tan θ ≈ θ (in radians).

From the expression of fringe width:

Fringe width β = λD / a

Problem 37

Two slits are at a distance of 0.2 mm apart and the screen is at a distance of 1 m. The third bright fringe is found to be displaced 7.5 mm from the central fringe. Find the wavelength of the light used.

Problem 38

A yellow light from a sodium vapour lamp of wavelength 5893 Å is directed upon two narrow slits 0.1 cm apart. Find the position on the screen 100 cm away from the slits.

Problem 39

In Young’s experiment, the distance of the screen from the two slits is 1.0 m. When light of wavelength 6000 Å is allowed to fall on the slits, the width of the fringes obtained on a screen is 2.0 mm. Determine:

1. The distance between the two slits
2. The width of the fringes if the wavelength of the incident light is 4800 Å.

Problem 40

In double slit experiment, light has a frequency of 6 × 10¹⁴ Hz. The distance between the centres of adjacent bright fringes is 0.75 mm. What is the distance between the slits if the screen is 1.5 m away?

Given that the speed of light in free space is 3 × 10⁸ m/s.

Problem 41

In a Young’s double slit experiment, two narrow slits 0.8 mm apart are illuminated by the same source of yellow light (wavelength 5893 Å). How far apart are the adjacent bright bands in the interference pattern observed on a screen 2 m away?

Problem 42

In Young’s double slit experiment, the angular width of a fringe formed on a distant screen is 0.1°. The wavelength of light used is 6000 Å. What is the spacing between the slits?

Problem 43

A beam of light consisting of two wavelengths 6500 Å and 5200 Å is used to obtain interference fringes in a Young’s double slit experiment.

1. Find the distance of the third bright fringe on the screen from the central maximum for the wavelength 6500 Å.
2. What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?

The distance between the slits is 2 mm and the distance between the plane of the slits and the screen is 120 cm.

Problem 45

A Young’s double slit arrangement produces interference fringes for sodium light (wavelength 5893 Å) that are 0.20 mm apart. What is the angular fringe separation if the entire arrangement is immersed in water? Given that the refractive index of water = 4/3.

Problem 46

In Young’s double slit experiment, the distance between the slits is 1 mm and the distance of the screen from the slits is 1 m. If light of wavelength 6000 Å is used, then find the distance between the second dark fringe and the fourth bright fringe.

Fringe Shift

When a thin transparent plate of thickness t and refractive index μ is introduced in the path of one of the interfering waves (say in the path P), then the effective path in air is increased by (μ – 1) t due to the introduction of the plate.

• Effective path difference in air = (μ – 1) t

For a bright fringe, path difference = nλ.

For the next bright fringe:

Fringe width β = λD / a

Substitute the expression for velocity and wavelength as needed.

Thus, with the introduction of the transparent plate in the path of one of the slits, the entire fringe pattern is displaced through a distance Δx given by either equation (5) or equation (7) towards the side on which the plate is placed.

Problem 47

In the figure below, S₁ and S₂ are two coherent light sources in a Young’s two slit experiment separated by a distance 0.5 mm and O is a point equidistant at a distance 0.8 m from the slits. When a thin parallel-sided piece of glass (G) of thickness 3.6 μm is placed near S₁ as shown, the central fringe system moves from O to point P. Calculate OP. (The wavelength of light used = 6.0 × 10^-7 m).

Conditions Necessary for Sustained Interference of Light Waves

• Sustained interference is an interference pattern in which the positions of maxima and minima remain fixed.
• In order for light waves to produce an interference pattern which can be observed on a screen, the following conditions must be satisfied:
• The light should be monochromatic. If this is not so, fringes of different colors will overlap.
• The two sources producing interference must be coherent.
• The two interfering waves must have the same plane of polarization.
• To observe interference fringes clearly, it is necessary that the fringe width is sufficiently large. This is possible if:

1. The two coherent sources are parallel and close to each other.
2. The distance between slits and screen is reasonably large.

Bi Prism

This is a glass prism with an obtuse angle that functions as two acute angle prisms placed together.

A double image of a single object is always formed by means of this prism.

The device was used by Fresnel to produce two coherent beams for interference experiments.

Monochromatic light from a narrow slit S falls on a bi prism as shown in the figure above.

Two virtual images S₁ and S₂ of S are formed by refraction at each half of the bi prism and these act as coherent sources which are close together.

An interference pattern, similar to that given by the double slit but brighter, is obtained in the shaded region where the two refracted beams overlap.

The theory and the expression for the fringe spacing are the same as for Young’s method.

Newton’s Rings

This is an interference effect discovered by Newton.

Interference fringes are formed by placing a slightly plano-convex lens on a flat glass plate H.

A glass plate G reflects the light that comes from the monochromatic light source towards the lens L downwards.

Both reflected rays are analyzed by the traveling microscope M, and at the point of contact of the lens, a dark spot is seen surrounded by a series of alternate bright and dark rings. These are called Newton’s rings.

Definition

Newton‘s ring is a circular interference fringe formed between a lens and a glass plate with which the lens is in contact.

Consider the air film PA of thickness t.

Some of the incident light is reflected at P towards the microscope M.

Some of the light passes to point A where it is reflected by the glass plate H towards the microscope M.

For the nth dark ring, path difference = 2t = nλ, where n = 0, 1, 2, 3, …

Central dark spot corresponds to n = 0.

Mean Free Path of a Gas Molecule

At constant pressure, the mean free path of a gas molecule λ is directly proportional to its absolute temperature T:

λ = K T

Where K is a constant of proportionality.

If λ₁ and λ₂ are mean free paths of a gas molecule at temperatures T₁ and T₂ respectively:

λ₁ = K T₁, λ₂ = K T₂

Pressure P of the Gas

At constant temperature, the mean free path of a gas molecule is inversely proportional to pressure:

P λ = K

Where K is a constant of proportionality.

Generally, we may write:

P₁ λ₁ = P₂ λ₂