UNIT – 1
PHYSICAL OPTICS
1.1.1 INTERFERENCE
Interference in light waves occurs whenever two or more waves overlap at a given point.
1.1.2 TYPES OF INTERFERENCE
Interference of light waves can be either constructive interference or destructive interference.
Figure 1: Types of interference
We know that the superposition of two mechanical waves can be constructive or destructive. In constructive interference, the amplitude of the resultant wave at a given position or time is greater than that of either individual wave, whereas in destructive interference, the resultant amplitude is less than that of either individual wave.
Light waves also interfere with each other. Fundamentally, all interference associated with light waves arises when the electromagnetic fields that constitute the individual waves combine.
If two light bulbs are placed side by side, no interference effects are observed because the light waves from one bulb are emitted independently of those from the other bulb. The emissions from the two light bulbs do not maintain a constant phase relationship with each other over time. Light waves from an ordinary source such as a light bulb undergo random phase changes in time intervals less than a nanosecond.
Therefore, the conditions for constructive interference, destructive interference, or some intermediate states are maintained only for such short time intervals. Because the eye cannot follow such rapid changes, no interference effects are observed. Such light sources are said to be incoherent.
Two sources are said to be coherent when the waves emitted from them have the same frequency and constant phase difference.
Interference from such waves happen all the time, the randomly phased light waves constantly produce bright and dark fringes at every point. But we cannot see them since they occur randomly. A point that has a dark fringe at one moment may have a bright fringe at the next moment. This cancels out the effect of the interference effect, and we see only an average brightness value. The interference is not said to be sustained since we cannot observe it.
1.1.3 CONDITIONS FOR SUSTAINED INTERFERENCE
In a sustained interference pattern, the position of maximum and minimum intensity regions remains constant with time. To obtain the sustained interference, the following conditions are required:
The phenomenon of interference may be grouped into two categories:
1.1.4 DIVISION OF WAVEFRONT
Under this category, the coherent sources are obtained by dividing the wavefront, originating from a common source, by employing mirrors, biprisms or lenses. This class of interference requires essentially a point source or a narrow-slit source. The instruments used to obtain interference by division of wavefront are the Fresnel biprism, Fresnel mirrors, Lloyd's mirror, lasers, etc.
Some of the standard arrangements used to observe two-wave interference by division of the wavefront are shown in Figure.
Figure 2: Two-wave interference; (a) Young’s two-slit arrangement, (b) Fresnel’s
biprism, (c) Fresnel’s two mirrors, (d) Loyd’s mirror
In Figure 2(a) The narrow slits S1 and S2 in Young’s double slit arrangement intercept portions of the spherical wavefront and act as real mutually coherent point sources, diffracting light in the forward direction. Interference among the diffracted waves can be observed anywhere in the region of overlap (shaded portion) behind the plane of the slits. Historically, the interference produced in Young’s double slit experiment was not accepted as a conclusive proof of the wave nature of light because it was thought that the fringes in Young’s experiment could arise due to some unexplained interaction of light with the edges of the slits.
In Figure 2(b) Fresnel’s biprisms arrangement, however, established the wave nature of light beyond any doubt. Fresnel’s biprism consists of two small angles (a few degrees) prisms in a manner that the incident spherical wavefront is split by the two prisms. The split wavefronts travel in different directions, eventually overlap and produce interference. S1 and S2 act as virtual but mutually coherent point sources.
Figure 2(c) shows the two-mirror arrangement, also devised by Fresnel. Here, the portions of the spherical wavefront reflected by the two mirrors overlap to produce interference. The source images S1 and S2, formed by the two mirrors, are virtual but mutually coherent.
In Figure 2(d) shows Lloyd’s single mirror arrangement, interference is produced by the portion of the wavefront reflected by the mirror and the portion which propagates directly to the region of superposition. In this case, the point source S and its virtual image S’ act as mutually coherent point sources.
1.1.5 FRESNEL’S BI-PRISM
The Fresnel double prism or biprism consists of two thin prisms joined at their bases, as shown in Figure 3. A single cylindrical wavefront impinges on both prisms. The top portion of the wavefront is refracted downward, and the lower segment is refracted upward. In the region of superposition, interference occurs. Here, again, two virtual sources S1 and S2 exist, separated by a distance a, which can be expressed in terms of the prism angle, where D>>d. The expression for the separation of the fringes is the same as before.
where (D = a + b) is distance of the sources from the eyepiece.
Figure 3: Fresnel biprism
Light from monochromatic source is made to fall on a thin slit mounted vertically on a rigid optical bench fitted with a scale. The biprism and the screen (in this case an eye piece) are also mounted vertically. The eye piece can be moved in the plane perpendicular to the axis of bench using a micrometer based translation stage.
1.1.6 APPLICATIONS OF FRESNEL'S BIPRISM
Fresnel biprism can be used to determine the wavelength of a light source (monochromatic), thickness of a thin transparent sheet/ thin film, refractive index of medium etc.
Determination of wave length of light
As expression for fringe width is
Biprism can be used to determine the wavelength of given monochromatic light using the expression.
Determination of fringe width β:
When the fringes are observed in the field view of the eyepiece, the vertical cross-wire is made to coincide with the centre of one of the bright fringes. the position of the eyepiece is read on the scale, say x°. The micrometer screw of the eyepiece is moved slowly and the number of the bright fringes N that pass across the cross-wire is counted. The position of the cross-wire is again read, say xN. The fringe width is then given by
Determination of (d):
The value of d can be determined as follows. The deviation δ produced in the path of a ray by a thin prism is given by
where α is the refracting angle of the prism. For the figure, it is seen δ=θ∕2. Since d is very small, we can also write d = aθ
Interference Fringes with white light:
In the biprism experiment if the slit is illuminated by white light the interference pattern consists of a central white fringe flanked on its both sides by a few coloured fringes and general illumination beyond the fringes. The central white fringe is the zero-order fringe.
With monochromatic light all the bright fringes are of the same colour and it is not possible to locate zero-order fringe. Therefore, in order to locate the zero-order fringe the biprism is to be illuminated by white light.
Lateral Displacement of Fringes
The biprism experiment can be used to determine the thickness of given thin sheet of transparent material such as glass or mica. If a thin transparent sheet is introduced in the path of one of the two interfering beams, the fringe system gets displaced towards the beam in whose path the sheet is introduced. By measuring of displacement, the thickness of the sheet can be determined. Suppose S1 and S2 are the virtual coherent monochromatic sources. The point O is equidistant from S1 and S2, where we obtain the central bright fringe. Therefore, the optical path S1O=S2O. Let a transparent plate G of thickness t and refractive index n be introduced in the path of one of the beams (see figure 4). The optical path lengths S1O and S2O are now not equal and the central bright fringe shits to P from O. The light waves from S1to P travel partly in air and partly in the sheet G: the distance travelled in air is (S1P –t) and that in the sheet is t.
Figure 4: Lateral Displacement of Fringes
The optical path
S1P = (S1P -t) + nt = S1P +(n-1) t
The optical path
S2P = S2 P
The optical path difference at P is
S1P - 
S2P =0.
since in the presence of the thin sheet, the optical path lengths S1P and S2P are equal and central zero fringe is obtained at P.
S1P = 
S2P
[S1P +(n-1)t] = S2 P
S2P – S1P = (n-1) t
But according to the relation
S2P – S1P = Xd/D
Where x is the lateral shift of the central fringe due to the introduction of the thin sheet.
(n-1)t = Xd/D
Hence, the thickness of the sheet is
t = Xd/ D(n-1)
Key takeaways
1.2.1 DIVISION OF AMPLITUDE
In this method, the amplitude of the incident beam is divided into two or more parts either by partial reflection or refraction. Thus, we have coherent beams produced by division of amplitude. These beams travel different paths and are finally brought together to produce interference. The effects resulting from the superposition of two beams are referred to as two beam interference and those resulting from superposition of more than two beams are referred to as multiple beam interference. The interference in thin films, Newton's rings, and Michelson's interferometer are examples of two beam interference and Fabry-Perot's interferometer is an example of multiple beam interference.
In case of division by wavefront, we considered two-wave interference when different portions of the wavefront are made to propagate in different directions and then recombined. Now, we discuss two-wave interference when a quasi-monochromatic wave is incident on a thin transparent film as shown in Figure. Here we will discuss interference by division of amplitude.
The wave is partly reflected and partly transmitted at each interface. Amplitudes of successively reflected and transmitted waves diminish rapidly for films of low reflectivity. The amplitude transmission coefficient for passage of the wave from the medium of refractive index n1 to the medium of refractive index n2 is t and t’ is the corresponding amplitude transmission coefficient for passage in the reverse direction.
1.2.2 NEWTON ‘S RINGS
Newton’s Rings are the circular interference pattern first discovered by physicist Sir Isaac Newton in 1704. It is cosists of concentric bright and dark rings with the point of contact of lens and the glass plate as centre,
The fringes obtained by interference of light waves by using the following arrangement
When a Plano convex lens with large radius of curvature is placed on a plane glass plate such that its curved surface faces the glass plate, a wedge air film (of gradually increasing thickness) is formed between the lens and the glass plate. The thickness of the air film is zero at the point of contact and gradually increases away from the point of contact.
Figure 5: Newton Ring Assembly
If monochromatic (means light with single wavelength) light is allowed to fall normally on the lens from a source 'S', then two reflected rays R1 (reflected from upper surface of the film) and R2 (reflected from lower surface of the air film) interfere to produce circular interference pattern. This interference pattern has concentric alternate bright and dark rings around the point of contact. This pattern is observed through traveling microscope.
MATHEMATICAL ANALYSIS OF NEWTON’S RING
Figure 6: Mathematical analysis of newton’s ring
(OL)2 =(O’M)2-(ML)2 ………. (1)
R2=(R-t)2 +rn2
R2=R2 +t2-2Rt +rn2
Radius is large as compared to the thickness
so t2 is neglected as t2<< R2
R2=R2 +-2Rt +rn2
2Rt =rn2
Thickness of the film t =rn2 /2R ………. (2)
THEORY OF FRINGES:
The effective path difference between the two reflected rays R1 and R2 for a wedge-shaped film from equation
∆ = 2μtcos(r+θ) +λ/2 ………. (3)
If the light is incident normally on the lens,
r = 0 and near to point of contact θ is small;
Therefore, near point of contact, (r+θ) approaches to 0 and cos(r+θ) =cos0=1
Therefore
∆ = 2μt+λ/2 ………. (4)
Also, At point of contact t = 0 therefore the effective path difference ∆ = λ/2
Which is odd multiple of λ/2 Therefore the Central fringe is dark.
BRIGHT FRINGE: CONDITION OF MAXIMA
For the condition of maxima, the effective path difference
∆ = ±nλ
Using equation (4) ∆ = 2μt+λ/2 we have
2μt+λ/2= ±nλ
2μt = ± (2n-1) λ /2 ………. (5)
DIAMETER OF BRIGHT RINGS
we know by equation (2) t =rn2 /2R substitute in equation (5) we have
2μ (rn2 /2R) = ± (2n-1) λ /2
rn2 = ± (2n-1) λR /2μ
We know diameter D=2r and for nth fringe Dn=2rn
so, we have Dn2=± 2(2n-1) λR /μ
Dn=
The medium enclosed between the lens and glass plate is if air therefore, 
=1. The diameter of nth order bright fringe will be
D=
n=0,1,2,3,4……. ………. (6)
The diameter of bright ring is proportional to square root of odd natural numbers
DARK FRINGE: CONDITION FOR MINIMA
For the condition of minima, The effective path difference
∆ =± (2n+1) λ /2
2μt+λ/2 =± (2n+1) λ /2
2μt= ±nλ ………. (7)
it is clear that for particular dark or bright fringe t should be constant.
Every fringe is the locus of points having equal thickness. Hence the fringes are circular in shape.
DIAMETER OF DARK RINGS
we know by equation (2) t =rn2 /2R substitute in equation (7) we have
2μ (rn2 /2R) = nλ
rn2 = nλR/ μ
We know diameter D=2r and for nth fringe Dn=2rn
so, we have Dn2= 4nλR/ μ
Dn= 
The medium enclosed between the lens and glass plate is if air therefore, 
=1. The diameter of nth order bright fringe will be
Dn= 
n=0,1,2,3,4……. ………. (8)
The diameter of dark ring is proportional to square root of natural numbers.
SPACING BETWEEN FRINGES
The Newton’s rings are not equally spaced because the diameter of ring does not increase in the same proportion as the order of ring and rings get closer and closer as ‘n’ increases.
For example, the diameter of dark ring is given by Dn= 
where n=0,1,2,3,4…….
D3 - D2 =
- 
= (
-
)
= 0.635
D7 – D6 =
- 
= (
-
)
= 0.392
D10– D9 =
- 
= (
-
)
= 0.324
From above result we conclude that the fringe width reduces with increase in n.
1.2.3 APPLICATIONS
1. Determination of Wavelength of Light
The diameter of the nth order ring is calculated by subtracting the left and right-side position of the microscope. As we know that the square of diameter of nth dark ring is
D2n = 4 n λ R
Therefore, the square of diameter of (n+p) th ring is
D2n + p = 4(n+p) λR
Subtracting both the above equation
D2n + p - D2n = 4(n+p) λR - 4 n λ R-----------------(9)
Therefore
λ = D2n + p - D2n / 4pR -----------(10)
In order to determine the refractive index of liquid the Newton’s ring experiment is first performed for the air medium and the difference in the square of the diameter of (n+p) th and nth dark ring is found as discussed above.
( D2n + p - D2n) air= 4pλR …………. (11)
After this few drops of liquid of μ refractive index is placed on the glass plate. The plano-convex lens is then placed on the glass plate, as a result a film of liquid is formed between the lens and the plate.
The difference in the square of the diameter of (n+p) th and nth dark ring is again calculated in the same manner for the liquid medium.
( D2n + p - D2n)liquid = 4pλR/ μ ----------------------------------(12)
Dividing equation 11 by 12, we get
μ= ( D2n + p - D2n) air/ ( D2n + p - D2n)liquid
…….…. (13)
2. Newton’s ring with white light
If the monochromatic source is replaced by the white light, dark and bright fringes are not produced. Because the diameter of the rings depends upon wavelength and it is proportional to the square root of wavelength.
If …. (The monochromatic source is replaced by the white light superposition of rings take place due to different wavelength. Few coloured rings are seen around dark centre later illumination is seen in the field of view. As shown in below figure.
Figure 7: Newton’s ring with white light
3. Determination of the Radius Curvature of a Plano-Convex Lens
From the knowledge of wavelength of light and diameters of various dark rings in the Newton’s rings, we can easily determine correct value of radius of curvature of the given plano-convex lens by the following equation.
R = 
Example: In a Newton’s rings experiment the diameter of the 15th ring was found to be 0.59 cm and that of the 5th ring is 0.336 cm. If the radius of curvature of the lens is 100 cm, find the wave length of the light.
Solution:
The given data are
Diameter of Newton’s 15th ring (D15) = 0.59 cm = 0.59×10–2 m
Diameter of Newton’s 5th ring (D5) = 0.336 cm = 0.336 × 10–2 m
Radius of curvature of lens (R) = 100 cm = 1 m
Wave length of light (λ) =?
Here m is difference between rings = 15-5=10
λ = D2n+m - D2n / 4mR
λ = (0.59×10–2 )2- (0.336 × 10–2)2 / 4 x 10 x 1
λ = 0.3481 x10-4 – 0.112896 x10-4 / 40
λ =0.00588 x 10-4 m
λ =5880 x 10-10 m
λ =5880Å
Key takeaways
λ= ( D2n + p - D2n)/ 4pR
Determination of Refractive Index of liquid
s
1.3.1 INTERFEROMETERS
Interferometry is a word used to study and exploit interference of light in optical devices in which two or more beams originate from the same source and after passing through different paths, they interfere with each other. The interference pattern can be analyzed to get information about the objects or phenomenon being studied. This idea involves the principle of division of amplitude.
An interferometer is designed to make the use of two-beam interference (Michelson Interferometer) and multiple-beam interference (Fabry-Perot Interferometer) of light.
There are two types of interferometers:
(1) Wavefront division interferometers in which the different portions of the same wavefront of a coherent beam of light is used to produce interference (Young's double slit, Lloyd's mirror, Fresnel Biprism, etc.) and
(2) Amplitude division interferometers in which a beam splitter is used to divide the initial coherent beam into two parts.
Basic Principle of Interferometers
When a beam of light is incident on a beam splitter or half-transparent mirror (a glass plate partially coated with sliver acts as a partially reflective surface), the incident light beam splits into two beams; one is directed towards a fixed mirror and the other is transmitted towards another movable mirror. The two beams reflected from these mirrors are subsequently recombined and produce interference pattern due to phase or path difference between them. An optical device that works on the above said principle is called an interferometer.
The Michelson Interferometer
Figure 8: Michelson Interferometer Experiment
Principle
Construction:
Figure 9: Michelson interferometer
Working
Thus, the telescope receives two reflected beams which are originally derived from the same source so that these are mutually coherent. Hence, the two beams M1A T and M2AT will interfere and produces interference fringes in the field of view of the telescope. This process is known as interference by the division of amplitude.
When one looks into the beam splitter from the telescope, there appears both the mirror M1 and a virtual image M2’ of M2 formed in G1 which are separated by a distance d = x1 
x2. Thus, the two interfering beams reflected by the mirrors M1 and M2 appears as reflected from M1 and M2. Hence, the Michelson interferometer is equivalent to a thin air film enclosed between M1 and M2’, see Figure 9.
The path difference between the reflected beams depends on the separation between M1and M2’ and the inclination of M1 and M2’ respectively. Therefore. the interference fringes may be straight, circular, and parabolic is formed.
Circular fringes:
Figure 10: The circular fringe interference pattern produced by a Michelson interferometer
To view these fringes with monochromatic light, the mirrors must be almost perfectly perpendicular to each other. The conditions for constructive and destructive interference pattern can be understood from Figure 11
Figure 11: Virtual images from the two mirrors created by the light source and the beam splitter in the Michelson interferometer
Let S1 and S2 be the virtual images of a source S, see Figure 11. If the distance between M1 and M2’ is d, then the distance between the virtual images S1 and S2 will be 2d. The geometrical path difference 
between the virtual images S1 and S2 can be obtained from the 
S1S2B (see Figure 11)
= S2B = 2dcosθ …….... (1)
where, d = x1 
x2 and θ is the angle that the ray from virtual source makes with the normal S2S1. When viewed obliquely into the interferometer, the line OS2 makes an angle θ with the axis. If the light beam reflected from mirror M2 undergoes an abrupt phase change of π or path difference of λ/2, then the optical path deference is:
= 2dcosθ + λ/2 ……... (2)
The condition for destructive interference (dark fringes) is:
2dcosθ + λ/2= (2m +1) λ/2
or 2dcosθ = mλ., m = 0, ±1, ±2, ... the order of interference ……... (3)
When viewed along the axis θ = 0,
then we have:
2d = mλ, m = 0, ±1, ±2,
The condition for constructive interference (bright fringes) is:
2dcosθ + λ/2= mλ m = 1, ±1, +2, ... ……... (4)
2dcosθ = (2m +1) λ/2 m = 0, ±1, ±2, ...
Expanding the cosine function, it can be shown that the radii of bright fringes are proportional to the square root of integers.
The formation of circular fringes for different conditions is shown in Figure 12.
• When M1 and M2’are separated by a few centimeters, the fringe system has a general appearance as shown in Figure 12(a). If M1is moved towards M2, the distance d decreases, and from Equation (4), as dcosθ is constant, the order of the fringes m decreases, and the fringe radius decreases. Thus, the rings shrink and vanish at the center.
For a ring disappearing each time, 2d decreases by λ or d by λ/2.
2d = m λ since cosθ = 1
• When M1 approaches M2’ r, the rings are more widely spaced as shown in Figure 12 (b).
• If the mirror M1 is moved further, the distance between M1 and M2 again increases to give widely spaced rings (Figure 12 (d)), and finally when the distance is large in centimeters, we get rings with smaller radius (Figure 12 (e)).
Figure 12: Shape of circular fringes by varying the movable mirror M1
When the separation between M1 and M2’ decreases, the fringes will move to the left across the field of the telescope, and new fringes crossing the centre of the telescope if d changes by λ/2. If M1 and M2’ are exactly cross with each other, fringes become straight as shown in Figure 13(b), where a wedge-shaped air film is formed. Further movement of M1, as shown in Figure 13(c), the fringes are again curved but in opposite direction. When the distance between M1 and M2’; is too large, the fringes cannot be observed. We conclude that due to the variation of path difference between the reflected light from the mirrors M1 and M2 by changing the distance, the fringe pattern observed contains fringes of equal thickness.
Figure 13: The localized fringe interference pattern–(a) and (c) are depictions of curved fringes, and (b) shows straight, parallel fringes due to wedge-shaped air thin film
White light source
To avoid spurious interference fringes, filters can be used. In the case of monochromatic light, maximum contrast fringes can be obtained with precisely defined focus. Therefore, the white light source will be required in the interferometers. Normally, if white light is used as a source. no fringes will be observed for large path difference. However, for small path differences of the order of few wavelengths, coloured fringes are observed.
For observing white light interference, the mirrors are tilted slightly as for localized fringes discussed above and the position of M1 is such that it intersects M2’. At the same time, two important conditions need to be satisfied.
• The position of the zero-order interference fringe must be independent of wavelength, i.e., a dark fringe at the centre of the interference pattern.
• The spacing of the interference fringes must be independent of wavelength.
Overall, the position of interference fringes is independent of order number and wavelength. Generally, in a white light interferometer, only the first condition is satisfied. The white light interference pattern contains a central dark fringe and coloured fringes on either side of the dark fringe. When M1 and M2’coincide, the path difference between the light beams is zero and hence the central fringe is dark in white light fringes. After 8 to 10 fringes from the central dark fringe, so many colours are present at a given point so that it appears white. The white light fringe pattern is shown in Figure 14.
Figure 14: The formation of white light fringes with a dark fringe at the center
1.3.2 APPLICATIONS
(i) Determination of wavelength of light:
For the determination of the wavelength of light, first, the Michelson interferometer is set for circular fringes with a central bright spot. The mirror M2 is kept fixed and when the mirror M1 is moved, each fringe gets displaced to itself in the field of view of the telescope. If the mirror is moved over a distance d, let m fringes cross the field of view of the telescope, then we have,
2d = mλ ……...... (5)
The wavelength of the light source
λ =2d/m ……...... (6)
By measuring the values of d and m, the wavelength λ can be determined.
(ii) Determination of refractive index of the thin film:
Consider a thin film of thickness t whose refractive index μm is to be measured will be introduced in one of the paths of interferometer beams, say in the path of beam going towards mirror M2. The thin film causes an additional path difference between the two interfering beams. If m be the number of fringes cross the centre of the field of view, then we have,
2(μm— 1) t = mλ
If the reflective index μm and the wavelength λ are known, we can determine the thickness of the thin film
t =
Or
If the thickness of the thin film t and the wavelength λ are known, then the refractive index μm of the thin film can be determined.
μm = 
+1 ……...... (7)
(iii) Resolution of spectral lines:
When a source produces two nearby wavelengths, say λ1 and λ2, then it gives two sets of the interference pattern. When the mirror is advanced, the two sets get in step or out of step alternatively. When a bright fringe of λ1 falls on a bright fringe of λ2, there will be a maximum intensity in the field of view of the telescope. On the other hand, when the bright fringe of λ1 falls on the dark fringe of λ2, there will be a. minimum intensity in the field of view of the telescope. If the distance corresponding to the change from maximum intensity to the minimum intensity is measured as d, then
= λ2 - λ1 = 
……...... (8)
If λ= (λ1 + λ2)/2, then the spectral resolution is given by the ratio λ /
.
Key Takeaways
2dcosθ = mλ., m = 0, ±1, ±2, ... the order of interference
2dcosθ = (2m +1) λ/2 m = 0, ±1, ±2, ...
= λ2 - λ1 = 
and λ= (λ1 + λ2)/2
1.4.1 DIFFRACTION: INTRODUCTION
When the light falls on the obstacle whose size is comparable with the wavelength of light then the light bends around the obstacle and enters the geometrical shadow. This bending of light is called diffraction.
Diffraction is the light bending of light as it passes around the edge of an object. The amount of bending depends on the relative size of the wavelength of light to the size of the opening. If the opening is much larger than the light's wavelength, the bending will be almost unnoticeable. However, if the two are closer in size or equal, the amount of bending is considerable, and easily seen with the naked eye.
Figure 15: Diffraction of light
In the atmosphere, diffracted light is bent around atmospheric particles -- most commonly, the atmospheric particles are tiny water droplets found in clouds. Diffracted light can produce fringes of light, dark or colored bands. An optical effect that results from the diffraction of light is the silver lining sometimes found around the edges of clouds or coronas surrounding the sun or moon. The illustration above shows how light (from either the sun or the moon) is bent around small droplets in the cloud.
Optical effects resulting from diffraction are produced through the interference of light waves. To visualize this, imagine light waves like water waves. If water waves were incident upon a float residing on the water surface, the float would bounce up and down in response to the incident waves, producing waves of their own. As these waves spread outward in all directions from the float, they interact with other water waves. If the crests of two waves combine, an amplified wave is produced (constructive interference). However, if a crest of one wave and a trough of another wave combines, they cancel each other out to produce no vertical displacement (destructive interference).
Figure 16: Diffraction
This concept also applies to light waves. When sunlight (or moonlight) encounters a cloud droplet, light waves are altered and similarly interact with one another as the water waves described above. If there is constructive interference, (the crests of two light waves combining), the light will appear brighter. If there is destructive interference, (the trough of one light wave meeting the crest of another), the light will either appear darker or disappear entirely.
1.4.2 TYPES OF DIFFRACTION
There are two types of diffractions
Figure 17: Types of diffractions Fresnel Diffraction & Fraunhofer Diffraction
From the above figure, we observe that the source is located at a finite distance from the slit, and the screen is also at a finite distance from the slit. The source and the screen are not very far from each other. So, this is Fresnel diffraction. Here, if suppose the ray of light comes exactly at the edge of the obstacles, the path of the light is changed. So, the light bends a little and meets the screen.
A beam of width α travels a distance of α2/λ, called the Fresnel distance before it starts to spread out due to diffraction. But when the source and the screen are far away from each other, and when the source is located at the infinite position, then the ray of light coming from that infinite source are parallel rays of light. So, this is Fraunhofer diffraction.
Here we have to make use of the lens. But why do we use the lens? Because in Fraunhofer diffraction, the source is at infinity so the rays of light that pass through the slit are parallel rays of light.
So, to make these rays parallel to focus on the screen, we, make use of the converging lens. The zone which we get in front of the slit in the central maxima. On either side of the central maxima, there is a bright zone i.e., 1st maxima.
1.4.3 COMPARISON
Fraunhofer diffraction | |
1 If the source of light and screen is at a finite distance from the obstacle, then the diffraction is called Fresnel diffraction. | 1 If the source of light and screen is at an infinite distance from the obstacle then the diffraction is called Fraunhofer diffraction. |
2 The corresponding rays are not parallel. | 2 The corresponding rays are not parallel. |
3 The wavefronts falling on the obstacle are not plane. | 3 The wavefronts falling on the obstacle are planes. |
4 No convex lens is needed to converge spherical wavefronts. | 4 Plane diffracting wavefronts are converged by means of a convex lens to produce a diffraction pattern. |
5 Fresnel diffraction patterns on flat surfaces. | 5 Fraunhofer diffraction patterns on spherical surfaces. |
6 Diffraction patterns Change as we propagate them further ‘downstream’ of the source of scattering. | 6 Diffraction patterns Shape and intensity of a Fraunhofer diffraction pattern stay constant. |
7 To obtain Fresnel diffraction, zone plates are used. | 7 To obtain Fraunhofer diffraction, the single-double plane diffraction grafting is used. |
8 Diffraction pattern moves along the corresponding shift in the object. | 8 Diffraction pattern remains in a fixed position. |
Key Takeaways
DIFFRACTION: FRAUNHOFFER DIFFRACTION DUE SINGLE SLIT
Diffraction
When the light falls on the obstacle whose size is comparable with the wavelength of light then the light bends around the obstacle and enters in the geometrical shadow. This bending of light is called diffraction.
FRAUNHOFFER DIFFRACTION AT SINGLE SLIT
The adjacent figure represents a narrow-slit AB of width ‘e’. Let a plane wavefront of monochromatic light of wavelength '
' is incident on the slit. Let the diffracted light be focused by means of a convex lens on a screen. According to Huygen Fresnel, every point of the wavefront in the plane of the slit is a source of secondary wavelets. The secondary wavelets traveling normally to the slit i.e., along OP0 are brought to focus at -+
az P0 by the lens. Thus, P0 is a bright central image. The secondary wavelets traveling at an angle '
' are focused at a point P1 on the screen.
The intensity at the point P1 is either minimum or maximum and depends upon the path difference between the secondary waves originating from the corresponding points of the wavefront.
Figure 18: Fraunhoffer diffraction at single slit
Theory:
In order to find out the intensity at P1, draw a perpendicular AC on BR.
The path difference between secondary wavelets from A and B in direction θ is BC i.e,
∆ = BC = AB sin 
= e sin 
So, the phase difference,
= 2π/ λ x ∆ = 2π/λ (e sin 
Let us consider that the width of the slit is divided into ‘n’ equal parts and the amplitude of the wave from each part is ‘a’.
So, the phase difference between two consecutive points
∂=1/n {2π/λ (e sin 
) } ----------------------(1)
Then the resultant amplitude R is calculated by using the method of vector addition of amplitudes
Figure 19: Mathematical analysis of Fraunhoffer diffraction at single slit
The resultant amplitude of n number of waves having same amplitude 'a' and having common phase difference of '
' is
R = a sin (n∂/2)/ sin(∂/2) -------------------------(2)
Substituting the value of 
in equation (2)
R = a 
................... (3)
Substituting
= 
.esinθ in equation (3)
R = a sin α/ sin(α/n) sin α/n -> α/n
As 
is small value; sin 
R = n
and na = A
Therefore
R =A
................... (4)
Therefore, the Intensity is given by
I =R2 = A2 
................... (5)
Case (i): Principal Maximum:
Eqn (4) takes maximum value for
= 0
= 
.esinθ = 0
sinθ = 0 or θ=0
The condition
The condition θ=0 means that this maximum is formed by the secondary wavelets which travel normally to the slit along OP0 and focus at P0. This maximum is known as “Principal maximum”.
Intensity of Principal maxima
Rmax = lim α->0 A sin α/α = A lim α->0 sin α/α
Rmax = A.1 = A
Therefore
Imax = R2max = A2
Case (ii): Minimum Intensity positions:
Equation (3) takes minimum values for sin
The values of '
' which satisfy sin
are
where 
………... (6)
In the above equation (6) n = 0 is not applicable because corresponds to principal maximum. Therefore, the positions according to equation (6) are on either side of the principal maximum.
Case (iii): Secondary maximum:
In addition to principal maximum at 
= 0, there are weak secondary maxima between minima positions. The positions of these weak secondary maxima can be obtained with the rule of finding maxima and minima of a given function in calculus. So, differentiating equation (5) and equating to zero, we have
= 
(A2 
) =0
= 2A2 
( 
)(
=0
A2 
0; 
=0
Because 
=0 correspond to minima positions
=0
………... (7)
The values of '
' satisfying the equation (7) are obtained graphically by plotting the curves 
and 
on the same graph.
The points of intersection of the two curves gives the values of 
which satisfy equation (7).
The points of intersections are
Figure 20: Variation of intensity
But 
, gives principal maximum, substituting the values of 
in equation (5), we get
and so on.
From the above expressions, Imax, I1, I2, I3… it is evident that most of the incident light is concentrated at the principal maximum.
INTENSITY DISTRIBUTION GRAPH
A graph showing the variation of intensity with '
' is as shown in the adjacent figure
Figure 21: Intensity Distribution Graph
Key Takeaways
1.6.1 THEORY OF DIFFRACTION GRATING
A set of a large number of parallel slits of the same width and separated by opaque spaces is known as a diffraction grating.
Diffraction gratings are much more effective than prisms for dispersing light of different wavelengths so they are used almost exclusively in instruments designed to detect and identify characteristic spectral lines. There is nothing mysterious about these devices.
Fraunhofer used the first grating consisting of a large number of parallel wires placed side by side very closely at regular separation. Now the gratings are constructed by ruling the equidistance parallel lines on a transparent material such as glass with a fine diamond point. The ruled lines are opaque to light while the space between the two lines is transparent to light and acts as a slit.
Figure 22: Plane Transmission Grating
Let ‘e’ be the width of the line and ‘d’ be the width of the slit.
Then (e + d) is known as the grating element.
If N is the number of lines per inch on the grating, then
(𝑒+𝑑) =1𝑖𝑐ℎ=2.54𝑐𝑚
(𝑒+𝑑) =2.54𝑐𝑚/𝑁
Commercial gratings are produced by taking the cost of actual grating on a transparent film like that of cellulose acetate. The solution of cellulose acetate is poured on a ruled surface and allowed to dry to form a thin film, detachable from the surface. This film of the grating is kept between the two glass plates.
1.6.2 GRATING SPECTRUM -DETERMINATION OF WAVELENGTH
Figure 23: Grating spectrum
The principle maxima in a grating are formed in direction given by
(e + d) sinθ =nλ
where (e + d) is the grating element, ‘n’ the order of the maxima and 
the wavelength of the incident light.
1) For a given wavelength 
the angle of diffraction 
is different for principal maxima of different orders.
2) For white light and for a particular order n, the light of different wavelengths will be diffracted in different directions.
The longer the wavelength, greater is the angle of diffraction. So, in each order, we will get the spectra having as many lines as the wavelength in the light source.
At centre (n = 0, zero order) 
gives the maxima of all wavelengths. So here different wavelengths coincide to form the central image of the same colour as that the light source.
Similarly, the principal maxima of all wavelengths corresponding to n = 1 will form the first order spectrum, the principal maxima of all wavelengths corresponding to n = 2, will form the second order spectrum and so on.
Figure 24: Spectrum
From this we conclude that
Important characteristics of grating spectra:
1) Spectra of different orders are situated symmetrically on both sides of zero order.
2) Spectral lines are almost straight and quite sharp.
3) Spectral colours are in the order from Violet to Red.
4) Spectral lines are more dispersed as we go to higher orders.
5) Most of the incident intensity goes to zero order and rest is distributed among the other orders.
Maximum number of orders formed by a Grating:
The principal maxima in a grating satisfy the condition
(e+d) sin 
= n λ
Or
n=(e+d) sin 
/ 
The maximum angle of diffraction is 90o, hence the maximum possible order is given by
Nmax = (e+d) sin 90 / λ = (e+d)/λ
Consider a grating having grating element which is less than twice the wavelength of the incident light, then
(e+d) < 2λ
Nmax < 2λ/λ<2
i.e., only the first order is possible.
Key Takeaways
1.7.1DISPERSIVE POWER OF GRATING
Dispersive power of a grating is defined as the ratio of the difference in the angle of diffraction of any two neighbouring spectral lines to the difference in the wavelength between the two spectral lines.
Or
It can also be defined as the diffraction in the angle of diffraction per unit change in wavelength.
The diffraction of the nth order principal maximum for a wavelength λ is given by the equation,
(a + b) sin θ = nλ (1)
Differentiating this equation with respect to θ and λ
a + b is constant and n is constant in a given order.
(a + b) cos θ dθ = n dλ
(2)
In equation (2) dθ/dλ is the dispersive power,
n is the order of the spectrum,
N’ is the number of lines per cm of the grating surface and
θ is the angle of diffraction for the nth order principal maximum of wavelength λ.
From equation (2), it is clear, that the dispersive power of the grating is directly proportional to the number of lines per cm and inversely proportional to cos θ.
Thus, the angular spacing of any two spectral lines is double in the second-order spectrum in comparison to the first order. Secondly, the angular dispersion of the lines is more with a grating having a larger number of lines per cm. thirdly, the angular dispersion is minimum when θ = 0. If the value of θ is not large the value of cosθ can be taken as unity approximately and the influence of the factor cosθ in the equation (2) can be neglected.
Neglecting the influence of cosθ, it is clear that the angular dispersion of any two spectral lines (in a particular order) is directly proportional to the difference in wavelength between the two spectral lines. A spectrum of this type is called a normal spectrum.
If the linear spacing of two spectral lines of wavelength λ and λ + dλ is dx in the focal plane of the telescope objective or the photographic plate, then,
dx = fdθ
Where ƒ is the focal length of the objective. The linear dispersion
The linear dispersion is useful in studying the photographs of a spectrum.
1.7.2 RESOLVING POWER.
Resolving power of diffraction grating
It is defined as the capacity of a grating to form separate diffraction maxima of two wavelengths that are very close to each other
It is measured by λ/dλ where dλ is the smallest difference in two wavelengths which are just resolvable by grating and λ is the wavelength of either of them or mean wavelength.
Figure 25: Resolving Power of Diffraction Grating
Let AB represent the surface of a plane transmission grating having grating element (e+d) and N total number of slits.
Let a beam of light having two wavelengths by λ and dλ is normally incident on the grating. Let P1 is an nth primary maximum of a spectral line of wavelength λ at an angle of diffraction 
and P2 is the nth primary maximum of wavelength λ+dλ at diffracting angle θ+dθ.
According to the Rayleigh criterion, the two wavelengths will be resolved if the principal maximum λ+dλ of nth order in a direction θ+dθ falls over the first minimum of nth order in the same direction θ+dθ.
Let us consider the first minimum of l of nth order in the direction θ+dθ as below.
The principal maximum of 
in the θ direction is given by
(e+d) sinθ =nλ ……………… (1)
The equation of minima is
N(e+d) sinθ = mλ ……………… (2)
Where m takes all integers except 0, N, 2N, …, nN, because, for these values of m, the condition for maxima is satisfied.
Thus, first minimum adjacent to the nth principal maximum in the direction θ+dθ can be obtained by substituting the value of ‘m’ as (nN+1).
Therefore, the first minimum in the direction of θ+dθ is given by
N(e+d) sin(θ+dθ) = (nN+1) λ
(e+d) sin(θ+dθ) = (n+
)λ ……………… (3)
The principal maximum of λ+dλ in direction θ+dθ is given by
(e+d) sin(θ+dθ) = n(λ+dλ) ……………… (4)
Dividing eqn (3) by equation (4), we get
(n+
)λ = n(λ+dλ)
nλ +
= nλ + ndλ
= ndλ
= nN
Resolving Power = 
= nN ……………… (5)
Thus, the resolving power is directly proportional to
(i) The order of the spectrum ‘n’
(ii) The total number of lines on the grating ‘N
Example: A monochromatic light with wavelength of 500 nm (1 nm = 10-9 m) strikes a grating and produces the second-order bright line at an 30° angle. Determine the number of slits per centimetre.
Solution:
Wavelength (λ) = 500.10-9 m = 5.10-7 m
θ = 30o
n = 2
Distance between slits:
d sin θ = n λ
d (sin 30o) = (2) (5.10-7)
d (0.5) = 10.10-7
d = (10.10-7) / 0.5
d = 20.10-7
d = 2.10-6 m
Number of slits per centimetre:
x = 1 / d
x = 1 / 2.10-6 m
x = 0.5.106 / 1 m
x = 0.5.106 / 102 cm
x = 0.5.104 cm
x = 5.103 /cm
x = 5000 /cm
Example: A monochromatic light with wavelength of 500 nm (1 nano = 10-9) strikes a grating and produces the fourth-order bright line at an 30° angle. Determine the number of slits per centimetre.
Solution:
Wavelength (λ) = 500.10-9 m = 5.10-7 m
θ = 30o
n = 4
Distance between slits:
d sin θ = n λ
d (sin 30o) = (4) (5.10-7)
d (0.5) = 20.10-7
d = (20.10-7) / 0.5
d = 40.10-7
d = 4.10-6 m
Number of slits per centimetre:
x = 1 / 4.10-6 m
x = 0.25.106 / m
x = 0.25.106 / 102 cm
x = 0.25.104 / cm
x = 25.102 per cm
x = 2500 per cm
Key Takeaways
The phenomena of interference and diffraction confirm the wave nature of light hence neither of these phenomena is able to indentify the nature of waves that light is. These phenomena do not tell whether the oscillation of light waves are longitudinal or transverse in nature. This is accompalished by the study of polarization.
A light wave that is vibrating in more than one plane is referred to as unpolarized light.
Example: Light emitted by the sun, by a lamp in the classroom, or by a candle flame is unpolarized light.
These unpolarized light is created by electric charges that vibrate in a variety of directions, this concept of unpolarized light is difficult to visualize. For simplicity it is helpful to visualize unpolarized light as a wave that has an average of half its vibrations in a horizontal plane and half of its vibrations in a vertical plane. It is possible to transform unpolarized light into polarized light.
Polarized light waves are light waves in which the vibrations occur in a single plane. The process of transforming unpolarized light into polarized light is known as polarization.
Figure 26: Polarization
We know that Light is a transverse electromagnetic wave, but natural light is generally unpolarized, all planes of propagation being equally probable.
Linearly Polarized: Light in the form of a plane wave in space is said to be linearly polarized. The electric field of light is limited to a single plane along the direction of propagation.
Circularly Polarized If light is composed of two plane waves of equal amplitude by differing in phase by 90°, then the light is said to be circularly polarized. if the thumb of your right hand were pointing in the direction of propagation of the light, the electric vector would be rotating in the direction of your fingers.
Elliptically Polarized If two plane waves of differing amplitude are related in phase by 90°, or if the relative phase is other than 90° then the light is said to be elliptically polarized. If the thumb of your right hand were pointing in the direction of propagation of the light, the electric vector would be rotating in the direction of your fingers.
Figure 27: Showimg Linear, Circular and Elliptical Polarization
Key Takeaways
1.9.1 PRODUCTION OF POLARIZED LIGHT BY REFRACTION
Polarization can also occur by the refraction of light. Refraction occurs when a beam of light passes from one material into another material. At the surface of the two materials, the path of the beam changes its direction. The refracted beam acquires some degree of polarization.
Birefringence, an optical property in which a single ray of unpolarized light entering an anisotropic medium is split into two rays, each traveling in a different direction. One ray (called the extraordinary ray) is bent, or refracted, at an angle as it travels through the medium; the other ray (called the ordinary ray) passes through the medium unchanged.
The ordinary ray and the extraordinary ray are polarized in planes vibrating at right angles to each other. The refractive index of the ordinary ray is constant in all directions; the refractive index of the extraordinary ray varies according to the direction because it has components that are both parallel and perpendicular to the crystal’s optic axis. An extraordinary ray can move either faster or slower than an ordinary ray.
Figure 28: production of polarized light by refraction
The Figure shows the phenomenon of double refraction through a calcite crystal. An incident ray is seen to split into the ordinary ray CO and the extraordinary ray CE upon entering the crystal face at C. If the incident ray enters the crystal along the direction of its optic axis, however, the light ray will not become divided.
The polarization occurs in a plane perpendicular to the surface. The polarization of refracted light is demonstrated by a unique crystal that serves as a double-refracting crystal.
Iceland Spar (calcite mineral), refracts incident light into two different paths. The light is split into two beams upon entering the crystal.
Figure 29: Refracted rays passing through Iceland spar crystal
That mean if we see an object through an Iceland spar crystal, two images will be observed. These two images are the result of the double refraction of light. Both refracted light beams are polarized - one in a direction parallel to the surface and the other in a direction perpendicular to the surface
Since these two refracted rays are polarized with a perpendicular orientation, a polarizing filter can be used to completely block one of the images. If the polarization axis of the filter is aligned perpendicular to the plane of polarized light, the light is completely blocked by the filter; meanwhile the second image is as bright as can be. And if the filter is then turned 90-degrees in either direction, the second image reappears and the first image disappears.
Negative and Positive Crystals:
Suppose vo and ve be the velocities of the ordinary and extraordinary rays. The crystals in which vo<ve are termed as negative and in which vo>ve are termed as positive. Calcite is negative crystal and quartz is an example of positive crystal.
Uniaxial and Biaxial crystals:
Uniaxial crystal: A crystal which has only one optic axis is called uniaxial crystal. The refractive index of the ordinary ray is constant for any direction in the crystal. Examples: - Calcite, Quartz, Ice, Tourmaline etc.
Biaxial crystal: A crystal which has only two optic axis is called biaxial crystal. The refractive index of the extraordinary ray is variable and depends on the direction. Examples: - Mica, Topaz, Borax etc.
All transparent crystals except those of the cubic system, which are normally optically isotropic, exhibit the phenomenon of double refraction: in addition to calcite, some well-known examples are ice, mica, quartz, sugar, and tourmaline. Other materials may become birefringent under special circumstances.
For example, solutions containing long-chain molecules exhibit double refraction when they flow; this phenomenon is called streaming birefringence. Some isotropic materials like glass may even exhibit birefringence when placed in a magnetic or electric field or when subjected to external stress.
1.9.2 DOUBLE REFRACTION BY NICOL PRISM
This was invented by William Nicol in the year 1828. Nicol prism is made from a double refracting calcite crystal.
A Nicol Prism is a type of polarizer, an optical device used to produce a polarized beam of light.
It is made in such a way that it eliminates one of the rays by total internal reflection, i.e., the ordinary ray is eliminated and only the extraordinary ray is transmitted through the prism.
Construction of Nicol prism:
The refractive index of Canada balsam (1.55) is in between the refractive indices of ordinary (1.658) and extraordinary (1.486) rays in calcite crystal.
Figure 30: Production of plane polarized light using Nicol prism
Working
Figure 31: (a) Parallel Nicols; (b) Crossed Nicols
Difference between Positive Crystal and Negative Crystal
Positive Crystal | Negative Crystal |
|
|
2. The crystals in which vo>ve are termed as positive. In positive crystals E-ray travels slower than o-ray in all directions except along the optic axis. | 2. The crystals in which vo<ve are termed as negative. In negative crystals O-ray travels slower than E- ray in all directions except along the optic axis. |
3. According to Huygen's ellipse corresponding to e-ray is contained within the sphere corresponding to o-ray | 3. According to Huygen's ellipse corresponding to e-ray lies outside the sphere corresponding to O-ray |
4. Birefringence or amount of double refraction of a crystal is defined as ∆μ=μe-. μ for ∆μ is positive quality for positive crystals. | 4. ∆μ Negative for negative crystals. |
5. Example: Quartz | 5. Example: calcite |
Key Takeaways
A wave plate or retarder is an optical device that alters the polarization state of a light beam travelling through it. There are many types of retarders. Some common retarders are quarter-wave (λ/4) plates and half-wave (λ/2) plates.
A half-wave plate functions as a polarization rotator for linearly polarized light. It rotates the polarization of a linearly polarized light by twice the angle between its optic axis and the initial direction of polarization, as shown in Figure 32. It introduces a phase difference of radians between the
A typical wave plate is simply a birefringent crystal or a double refracting plastic foil with a carefully chosen thickness. If a beam of parallel light strikes perpendicularly a wave plate the light beam is splitted into two components due to its double refracting properties. The two components have planes of oscillation perpendicular to each other and slightly different phase velocities.
For a quarter-wave plate the thickness of the foil is chosen in such a manner that the light component whose electric field vector oscillates in parallel to the rotation lever lags by a λ/4 behind other perpendicular oscillating light component. For a half-wave plate the thickness is chosen so that the created phase difference has the amount of λ/2. In this experiment monochromatic light falls on a quarter-wave and half-wave plate. The polarization of the emergent light is investigated at different angles between the optic axis of the wave plates and the direction of the incident light.
Figure 32: optic axis
Half wave plates
A half-wave plate functions as a polarization rotator for linearly polarized light. It rotates the polarization of a linearly polarized light by twice the angle between its optic axis and the initial direction of polarization, as shown in Figure 33. It introduces a phase difference of radians between the two components of the electric field vectors.
Suppose a plane-polarized wave is normally incident on a wave plate, and the plane of polarization is at an angle ϕ with respect to the fast axis, as shown. After passing through the plate, the original plane wave has been rotated through an angle 2ϕ.
Figure 33: Half wave plate
A half-wave plate is very handy in rotating the plane of polarization from a polarized laser to any other desired plane (especially if the laser is too large to rotate). Most large ion lasers are vertically polarized. To obtain horizontal polarization, simply place a half-wave plate in the beam with its fast (or slow) axis 45° to the vertical. The l/2 plates can also change left circularly polarized light into right circularly polarized light or vice versa. The thickness of half waveplate is such that the phase difference is 1/2 wavelength (l/2, Zero order) or certain multiple of 1/2-wavelength [(2n+1) l/2, multiple order].
Quarter Wave Plate
A quarter-wave plate is used to convert linear polarization to circular polarization and vice-versa. In order to obtain circular polarization, the amplitudes of the two E vector components must be equal and their phase difference must be 90 or 270 degrees. A quarter-wave plate is capable of introducing a phase difference of 90 degrees between the two components of incident light. However, this is not enough to produce circular polarization. The direction of polarization of the incident light with respect to the optic axis of the quarter-wave plate is equally important. It is obvious that the amplitudes of the two components will be equal only when the incident linear polarization direction makes an angle of 45 degrees with respect to the optic axis of the crystal (i.e., α= 45degrees). Figure 33 illustrates the conversion of linearly polarized light into circularly polarized light. Note that conversely, if circularly polarized light is incident on a quarter-wave plate, the resulting polarization will be linear, at an angle of 45 degrees with respect to the optic axis. Again, the reason is that a phase difference of 90 degrees is introduced between the two components of the electric field vector, which traces a helix-like path as the circularly polarized wave propagates.
Figure 34: Quarter Wave Plate
Key Takeaways
1.11.1 POLARIMETER
Polarimeter – it is an optical instrument used to measure the angle of rotation of plane polarized light when it is passed through an optically active substance.
By measuring the angle, the specific rotation of an optically active substance can be determined.
Two types of polarimeters are generally used in the laboratory now days Laurent’s Half Shade Polarimeter and Biquartz Polarimeter.
1.11.2 BIQUARTZ POLARIMETER
It consists of a white light source, a convex lens as before, which renders the light into a parallel beam, the polarizer changes the incident beam into a plane polarized beam, then a biquartz plate, then the experimental tube containing the active substance and a polarizer working as an analyser as before. A telescope is fitted with a circular scale and is telescope is fitted with a circular scale and is focused on the biquartz plate.
Biquartz plate consists of two semi-circular plates of quartz. One of left-handed quartz Land other of right-handed quartz R, each of thickness about 3.75 millimetres. Both are cut perpendicular to the optic axis. This means that propagation here is along the optic axis now. They are joined together along the diameter PQ.
Figure 35: Biquartz Polarimeter
When the plane polarized white light passes through a biquartz plate normally, along the optic axis the phenomena of rotary dispersion occurs because the planes of vibrations of different colours are rotated through different angles. Remember, we have seen that the amount of rotation is proportional to 1 upon lambda square and rotation will be in one sense for the left-handed portion and other direction for the right-handed portion. The amount of rotation is maximum for violet which has the minimum wavelength and least for red
The sense of rotation is opposite in the two halves. The amount of rotation also depends on the thickness. For a thickness of 3.75 millimeters, the rotation of the depends on the thickness. For a thickness of 3.75 millimeters, the rotation of the plane of polarization for yellow light is about 90 degrees. Hence YOY is a straight line.
If the pass direction of the analyzer is parallel to POQ, the yellow light will not be transmitted through the analyzer, Malus Law and the appearance of two halves will be similar. The two halves will have a grayish violet tint called the tint of the passage. When the analyzer is rotated to one side from this position, one half of the field of view appears blue, while the other half appears red.
When the analyzer is rotated in the opposite direction, the colours are interchange the first half which was bluish earlier now appears red and the second half which was reddish earlier now appears blue.
The position dependence of the tint of passage is very sensitive and is used for accurate determination of the angle of optical rotation
1.11.3 LAURENT’S HALF SHADE POLARIMETER
Construction
Figure 36: Laurent’s half shade polarimeter
Working
To understand the working of this device we have to know the need of a Laurent half shade device. For this we assume that half shade device is not present.
In this basic setup (without the half-shade A) you are looking for the maximum and minimum brightness, which then tells you that the analyzer A is precisely aligned with the output rotation.
When tube is empty placed the analyzer in such a position that the view of field is completely dark. Note the position of the analyzer with the help of circular scale.
Now filled the tube with optically active solution and it is set in its proper position. The optically active solution rotates the plane of polarization of the light emerging out of the polariser P by some angle. So, the light is transmitted by analyzer A and the field of view of telescope becomes bright.
Now the analyzer is rotated by a finite angle so that the field of view of telescope again become dark. This will happen only when the analyzer is rotated by the same angle by which plane of polarization of light is rotated by optically active solution.
The position of analyzer is again noted. The difference of the two readings will give you angle of rotation of plane of polarization.
But here the difficulty associated with this procedure that when analyzer is rotated for the total darkness, then it is attained gradually and hence it is difficult to find the exact position correctly for which complete darkness is obtained.
Figure 37: Laboratory Set Up of Polarimeter
To overcome above difficulty half shade device is introduced between polariser P and glass tube T.
Half Shade Device
Figure 38: Half Shade Device
Now if the Principal plane of the analysing Nicol is parallel to OP, then the light will pass through glass half passable. Hence glass half will be brighter than quartz half or we can say that glass half will be bright and the quartz half will be dark. Similarly, if principal plane of analysing. Nicol is parallel to OQ then quartz half will be bright and glass half will be dark.
When the principal plane of analyzer is along AOB then both halves will be equally bright. On the other hand, if the principal plane of analyzer is along DOC. then both the halves will be equally dark.
Thus, it is clear that if the analysing Nicol is slightly disturbed from DOC, then one half becomes brighter than the other. Hence by using half shade device, one can measure angle of rotation more accurately.
Mathematically
The specific rotation is given by S=θ/lC where l is length of tube and C is concentration.
Key Takeaways
References:
