Superposition of Waves | Oscillations, Waves & Optics - Physics PDF Download

Superposition of Two Disturbances and Production of Polarized Wave 

Superposition of Two Waves with Parallel Electric Field

Let us consider the propagation of two linearly polarized electromagnetic waves (both

propagating along the z axis) with their electric vectors oscillating along the x axis. The

electric fields associated with the waves can be written in the form

E₁ = xˆa₁ cos (kz −ω t +θ₁)                                             (1)

E₂ = xˆa₂ cos (kz −ω t +θ₂)                                             (2)

where a₁ and a₂ represent the amplitudes of the waves, ˆx represents the unit vector

along the x axis, and θ₁ and θ₂ are phase constants. The resultant of these two waves is

given by

E = E₁ + E₂                                                                     (3)

which can always be written in the form

E = xˆa cos (kz −ω t +θ )                                                  (4)

where                              (5)

represents the amplitude of the wave. Equation (4) tells us that the resultant is also a

linearly polarized wave with its electric vector oscillating along the same axis.

Superposition of Two Waves with Mutually Perpendicular Electric field

We next consider the superposition of two linearly polarized electromagnetic waves (both

propagating along the z axis) but with their electric vectors oscillating along two

mutually perpendicular directions. Thus, we may have

E₁ = xˆa₁ cos kz −ωt                                                      (6)

E₂ = yˆa₂ cos kz −ωt +θ                                                 (7)

For θ = nπ , the resultant will also be a linearly polarized wave with its electric vector

oscillating along a direction making a certain angle with the x axis; this angle will

depend on the relative values of a₁ and a₂ .

To find the state of polarization of the resultant field, we consider the time variation of

the resultant electric field at an arbitrary plane perpendicular to the z axis which we may,

without any loss of generality, assume to be z = 0 .

If E_x and E_y represent the x and y components of the resultant field  E = (E₁ + E₂) , then

E_x = a₁ cosωt                                                                (8)

and E_y = a₂cos(ωt - θ)                                                   (9)

where we have used Equations (6) and (7) with z = 0 .

For θ = nπ the above equations simplify to

E_x = a₁ cosωt and E_y = (-1)ⁿa₂cosωt                            (10)

from which we obtain (independent of t)         (11)

where the upper and lower signs correspond to n even and n odd, respectively. In the

E_x E_y plane, Eq. (11) represents a straight line; the angle φ that this line makes with the

E_x axis depends on the ratio a₂/a₁ In fact φ = (12)

The condition θ = nπ implies that the two vibrations are either in phase (n = 0, 2, 4....) or

out of phase (n =1,3,5,...) . Thus, the superposition of two linearly polarized

electromagnetic waves with their electric fields at right angles to each other and

oscillating in phase is again a linearly polarized wave with its electric vector, in general,

oscillating in a direction which is different from the fields of either of the two waves.

Following figures shows the plot of the resultant field corresponding to Eq. (10) for various values of a₂/a₁ . The tip of the electric vector oscillates (with angular frequency ω) along the thick lines shown in the figure. The equation of the straight line is given by Eq. (11).

For θ ≠ nπ (n = 0,1, 2,...), the resultant electric vector does not, in general, oscillate along a straight line. We first consider the simple case corresponding to θ = π/2 with a₁ = a₂ Thus,

E_x = a₁cos ωt (13) and  E_y = a₁cos ωt (14)

If we plot the time variation of the resultant electric vectors whose x and y components

are given by Eqs. (13) and (14), we find that the tip of the electric vector rotates on the

circumference of a circle (of radius 1 a ) in the counterclockwise direction as shown in

Fig. (c) below, and the propagation is in the +z direction which is coming out of the page. Such a wave is known as a right circularly polarized wave (usually abbreviated as a RCP wave). That the tip of the resultant electric vector should lie on the circumference of circle is also obvious from the fact that

E_x² + E_y² = a₁² (Independent of t )

For θ = 3π/2, E_x = a₁cos ωt               (15) and

E_y = -a₁sin ωt                                    (16)

which would also represent a circularly polarized wave; however, the electric vector will

rotate in the clockwise direction [Fig. (g)]. Such a wave is known as a left circularly

polarized wave (usually abbreviated as a LCP wave). 

the tip of the electric vector rotates on the circumference of an ellipse. As can be seen from the figure, this ellipse will degenerate into a straight line or a circle when θ becomes an even or an odd multiple of π/2 . In general, when  a₁ ≠ a₂ , one obtains an elliptically polarized wave which degenerates into a straight line for θ = 0,π ,2π ,... etc.

The Phenomenon of Double Refraction

When an unpolarized light beam is incident normally on a calcite crystal, it would in general, split up into two linearly polarized beams as shown in Fig. (a). The beam which

travels undeviated is known as the ordinary ray (usually abbreviated as the o − ray) and

obeys Snell’s laws of refraction. On the other hand, the second beam, which in general

does not obey Snell’s laws, is known as the extraordinary ray (usually abbreviated as the

e − ray).

The appearance of two beams is due to the phenomenon of double refraction, and a

crystal such as calcite is usually referred to as a double refracting crystal. If we put a

Polaroid PP′ behind the calcite crystal and rotate the Polaroid (about NN′ ), then for two

positions of the Polaroid (when the pass axis is perpendicular to the plane of the paper)

the e - ray will be completely blocked and only the o-ray will pass through.

Fig (a) When an unpolarized light beam is incident normally on a calcite

crystal, it would in general, split up into two linearly polarized beams. (b) If we rotate the

crystal about NN′ then the e-ray will rotate about NN′ .

On the other hand, when the pass axis of the Polaroid is in the plane of the paper (i.e.,

along the line PP′ ), then the o - ray will be completely blocked and only the e-ray will

pass through. Further, if we rotate the crystal about NN′ then the e -ray will rotate about

the axis [see Fig. (b)].

The velocity of the ordinary ray is the same in all directions, the velocity of the extraordinary ray is different in different directions; a substance (such as as calcite, quartz) which exhibits different properties in different directions is called an anisotropic substance. Along a particular direction (fixed in the crystal), the two velocities are equal; this direction is known as the optic axis of the crystal. In a crystal such as calcite, the two rays have the same speed only along one direction (which is the optic axis); such crystals are known as uniaxial crystals. The velocities of the ordinary and the extraordinary rays are given by the following equations:

(Ordinary ray) and (extraordinary)

where n_o and n_e are refractive index for O- ray and e - ray and θ is the angle that the ray

makes with the optic axis; we have assumed the optic axis to be parallel to the z axis.

Thus, c/n_o and c/n_e are the velocities of the extraordinary ray when it propagates parallel

and perpendicular to the optic axis.

Fig. (a) In a negative crystal, the ellipsoid of revolution (which corresponds to the extra

ordinary ray) lies outside the sphere; the sphere corresponds to the ordinary ray. (b) In a

positive crystal, the ellipsoid of revolution (which corresponds to the extraordinary ray)

lies inside the sphere.

Quarter Wave Plate and Half Wave Plate

Let electric field vector (of amplitude E₀ ) associated with the incident linearly polarized

beam makes an angle φ with optic axis which is parallel to z-axis and incident on calcite

crystal of thickness d whose optic axis is parallel to the surface

Such beam while traveling in calcite crystal splits into two components. The z-axis whose

amplitude is E₀ cosφ passes through as an extraordinary ray (e-ray) propagates with

velocity c/n_e . The y-axis whose amplitude is 0 E sinφ passes through as an ordinary ray

(o-ray) propagates with velocity c/n_o .

Since n_o ≠ n_e the two beams will propagate with different velocities, thus when they

come out of the crystal, they will not be in phase. Let on the plane x = 0 , the beam is

incident then

E_y = E₀sinφcos(kx −ωt) and E_z = E₀cosφcos(kx −ωt)

Thus at x = 0 , we have 

E_y = E₀sinφcosωt and E_z = E₀cosφcosωt

Inside crystal the two components will be

E_y = E₀sinφcos(n_okx −ωt) and E_z = E₀cosφcos(n_ekx −ωt)

If thickness of the crystal is d then on emerging surface we have

E_y = E₀sinφcos(n_okd −ωt) and E_z = E₀cosφcos(n_ekd −ωt)

Thus the phase difference between e-ray and o-ray is δ = kd(n_o - n_e)

If phase difference isδ =π / 2 , then d= λ/4(n_o - n_e)(Quarter Wave Plate)

If phase difference isδ =π , then d= λ/2(n_o - n_e)(Half Wave Plate)

Wollaston Prism

A Wollaston prism is used to produce two linearly polarized beams. It consists of two

similar prisms (calcite) with the optic axis of the first prism parallel to the surface and the

optic axis of the second prism parallel to the edge of the prism as shown below. Let us

first consider the incidence of a z polarized beam as shown in Fig. (a). The beam will

propagate as an o − ray in the first prism (because the vibrations are perpendicular to the

optic axis) and will see the refractive index n₀ . When this beam enters the second prism,

it will become an e -ray and will see the refractive index ne . For calcite no > ne and the

ray will bend away from the normal. Since the optic axis is normal to the plane of paper,

the refracted ray will obey Snell’s laws, and the angle of refraction will be given by

n_osin 20^o = n_esinr₁

where we have assumed the angle of the prism to be 20^o . Assuming n_o ≈ 1.658 and

n_e ≈ 1.486  , we readily get r₁ ≈ 22.43^o

Thus the angle of incidence at the second surface is i₁ = 22.43^o − 20^o = 2.43^o . The output angle θ₁ is given by n_esin 2.43^o =  sinθ₁⇒ θ₁ = 3.61^o

We next consider the incidence of a y - polarized beam as shown in Fig. (b). The beam

will propagate as an e -ray in the first prism and as an o -ray in the second prism. The

angle of refraction is now given by

n_esin20 = n_osinr₂ ⇒ r₂ ≈ 17.85^o

Thus the angle of incidence at the second interface is

i₂ = 20^o −17.85^o = 2.15^o

The output angle θ₂ is given by

n_osin2.15^o = sinθ₂ ⇒ θ₂ ≈ 3.57^o 

Thus, if an unpolarized beam is incident on the Wollaston prism, the angular separation

between the two orthogonally polarized beams is θ =θ₁ +θ₂ ≈ 7.18^o .

Rochon Prism

We next consider the Rochon prism which consists of two similar prisms of (say) calcite;

the optic axis of the first prism is normal to the face of the prism while the optic axis of

the second prism is parallel to the edge as shown in figure. Now, in the first prism both

beams will see the same refractive index n_o ; this follows from the fact that the ordinary

and extraordinary waves travel with the same velocity  along the optic axis of the

crystal. When the beam enters the second crystal, the ordinary ray (whose D is normal to

the optic axis) will see the same refractive index and go undeviated as shown in figure. On the other hand, the extraordinary ray (whose D is along the optic axis) will see the refractive index n_e and will bend away from the normal.

We assume the angle of the prism to be 25^o . The angle of refraction will be determined from

n_osin25^o = n_esinr

This sinr =  = 1.658/1.486 x 0.423 ≈ 0.472 ⇒r = 28.2^o

Therefore the angle of incidence at the second surface will be 28.2^o − 25^o = 3.2^o . The

emerging angle will be given by sinθ = n_esin (3.2^o) ⇒θ ≈ 4.8^o

Example 1: Light strikes a water surface at the polarizing angle. The part of the beam refracted into the water strikes a submerged glass slab (index of refraction,1.50 ), as shown in Figure. The light reflected from the upper surface of the slab is completely polarized. Find the angle between the water surface and the glass slab.

For the air-to-water interface,

 (θ_p= 53.1^o)

and (1.00)sinθ_p = (1.33)sinθ₂

For the water to glass interface,

⇒ θ₂ = 48.4^o

The angle between surfaces is θ = θ₃ - θ₂ = 11.5^o

Example 2: Plane- polarized light is incident on a single polarizing disk with the direction of E₀ parallel to the direction of the transmission axis. Through what angle should the disk be rotated so that the intensity in the transmitted beam is reduced by a factor of
(a) 3.00, (b) 5.00, (c) 10,0?

I = I_maxcos²θ ⇒ θ =

(a)  ⇒ θ =  = 54.7^o

(b) ⇒ θ =  = 63.4^o
(c) ⇒ θ = = 71.6^o

Example 3: In figure below, suppose that the transmission axes of the left and right

polarizing disks are perpendicular to each other, also, let center disk be rotated on the common axis with an angular speedω . Show that if unpolarized light is incident on the left disk with intensity I_max of the beam emerging from right disk is

For incident unpolarized light of intensity I_max :

After transmitting 1^st disk:

After transmitting 2^nd disk:

After transmitting 3^rd disk:

Where the angle between the first and second disk is θ =ω t

Using trigonometric identities cos²θ = 1/2(1 + cos2θ)

Ans cos²(90^o - θ) = sin²θ = 1/2(1-cos2θ)

We have

since θ =ω t , the intensity of the emerging beam is given by

Example 4: A half-wave plate and a quarter-wave plate are placed between a polarizer P₁ and an analyzer P₂ . All of these are parallel to each other and perpendicular to the direction of propagation of unpolarized incident light (see the figure). The optic-axis of the half-wave plate makes an angle of 30^o with respect to the pass-axis of P₁ and that of the quarter-wave plate is parallel to the pass-axis of P₁ .

(a) Determine the state of polarization for the light after passing through (i) the half- wave plate and (ii) the quarter-wave plate.

(b) What should be the orientation of the pass-axis of P₂ with respect to that of P₁ such that the intensity of the light emerging from P₂ is maximum?

(a) After passing through HWP the incident ray will split into an o -ray of

amplitude E₀sin30^o = E₀/2 and of e-ray of amplitude E₀cos30^o = E₀√3/2. After emerging out of HWP they will have phase difference ofπ . Superposition of these o -

and e -ray will produce linearly polarized light at the output of HWP. The electric field components of o -ray and e -ray at the output of HWP is

 =

=

The electric field components of linearly polarized light at the output of HWP is

The optic axis of QWP is parallel to the 1 P i.e along x -axis. The electric field component of incident linearly polarized light makes zero angle with the optic axis as a result the incident light will simply pass through QWP as a e -ray. Therefore at the output of the QWP light will be linearly polarized with the equation  

(b) The orientation of the pass axis of P₂ should be parallel to the pass axis of P₁ to allow maximum intensity of light to pass through P₂.

Example 5: Two orthogonally polarized beams (each of wavelength 0.5 μm and with

polarization marked in the figure) are incident on a two-prism assembly and emerge

along x -direction, as shown. The prisms are of identical material and o n and e n are the refractive indices of the o -ray and e -ray, respectively. Use  and n_o = 

(a) Find the value of θ and n_e . (b) If the right hand side prism starts sliding down with the vertical component of the velocity u_y = 1μm/ s , what would be the minimum time after which the state of polarization of the emergent beam would repeat itself?

(a) The beam will propagate as an o − ray in the first prism (because the vibrations are

perpendicular to the optic axis) and will see the refractive index n_o . When this beam

enters the second prism, it will become an e − ray and will see the refractive index e n . For calcite n_o > n_e and the ray will bend away from the normal. Since the optic axis is

normal to the plane of, the refracted ray will obey Snell’s laws, and the angle of refraction will be given by n₀sin30^o = n_esinθ We next consider the incidence of a second beam. The beam will propagate as an e − ray in the first prism and as an o −ray in the second prism. The angle of refraction is now given by n_esin30^o = n_osinθ 

⇒  ⇒

From 1st equation n₀sin30^o = n_esinθ ⇒

sin²θ = 3/4 ⇒ sinθ = √3/2 ⇒ θ = 60^o

Hence n_e =

(b) The state of polarization will repeat if change in phase difference (δ ) is π .

The relation between phase difference and path difference (Δ) is δ  = 2π/λx Δ

Where, Δ = (n_o - n_e)d = (n_o - n_e) x 10^-6x t

Thus δ  = 2π/λx Δ =  = π ⇒ t = .

While n_o - n_e = =

Thus t = =  = 0.85 sec

Analysis of Polarized Light

• Linearly polarized
• Circularly polarized
• Elliptically polarized
• Unpolarized
• Mixture of linearly polarized and unpolarized
• Mixture of circularly polarized and unpolarized
• Mixture of elliptically polarized and unpolarized light

If we introduce a Polaroid in the path of the beam and rotate it about the direction of

propagation, then one of the following three possibilities can occur:

• If there is complete extinction at two positions of the polarizer, then the beam is linearly polarized.
• If there is no variation of intensity, then the beam is unpolarized or circularly polarized or a mixture of unpolarized and circularly polarized light. We now put a quarter wave plates on the path of the beam followed by the rotating Polaroid. If there is no variation of intensity, then the incident beam is unpolarized. If there is complete extinction at two positions, then the beam is circularly polarized (this is so because a quarter wave plate will transform a circularly polarized light into a linearly polarized light). If there is a variation of intensity (without complete extinction), then the beam is a mixture of unpolarized and circularly polarized light.
• If there is a variation of intensity (without complete extinction), then the beam is elliptically polarized or a mixture of linearly polarized and unpolarized or a mixture of elliptically polarized and unpolarized light. We now put a quarter wave plate in front of the Polaroid with its optic axis parallel to the pass axis of the Polaroid at the position of maximum intensity. The elliptically Polarized light will transform to a linearly polarized light. Thus, if one obtains two positions of the Polaroid where complete extinction occurs, then the original beam is elliptically polarized. If complete extinction does not occur and the position of maximum intensity occurs at the same orientation as before, the beam is a mixture of unpolarized and linearly polarized light. Finally, if the position of maximum intensity occurs at a different orientation of the Polaroid, the beam is a mixture of elliptically polarized and unpolarized light.