UNIT I LESSON - 1 CONTENTS

UNIT I

LESSON - 1

CONTENTS

1.0 Aims and Objectives

1.1. Convex and Concave Lenses 1.1a. Refraction through a thin lens

1.1b. Equivalent Focal Length of Two Thin Lenses Separated by a distance 1.2. Aberrations in Lenses

1.2.1 Introduction

1.2.1. Spherical aberration:

1.2.2.Chromatic Aberration in a Lens:

1.2.3. Condition for Achromatism of Two Thin Lenses placed in contact 1.2.4.Condition for Achromatism of Two Thin Lenses separated by a finite distance

1.3.Coma

1.3.1.Astigmatism and its minimization 1.3.2

1.4 Let us sum up 1.5 Check your Progress 1.6 Lesson end Activities 1.7 Points for Discussion 1.8 References

1.0 Aims and Objectives

From this lesson you will understand about the principle employed in convex and concave lenses and study about different types of defects and to minimize them.

1.1. Convex and Concave Lenses

A convex lens is a transparent refracting medium bounded by two spherical surfaces. The line joining the centres of curvature of the two surfaces is called the principal axis. A section of the lens through its principal axis is called its principal section.

Principal Focus and Focal Planes

AB represents the principal axis of a lens (Fig. 1).

Fig 1

First Principal Focus (F1) : It is that point on the principal axis of the lens, the rays

starting from which (convex lens) or appear to converge at which (concave lens) become parallel to principal axis after refraction from the lens.

The plane passing through F1 and perpendicular to the principal axis is called first focal plane.

Second Principal Focus (F2) :Second principal focus is that point on the principal axis at

which the rays parallel to principal axis converge (convex lens) or appear to diverge (concave lens) after refraction from the lens.

The plan passing through F2, and normal to the principal axis is called the second focal plane.

1.1a. Refraction through a thin lens

Consider a thin lens of refractive index n2 placed in a medium of refractive index nl (Fig. 1.1.) Let R1, and R2 be the radii of the first and second surfaces respectively. Consider a point object O placed on the principal axis. Let I' be the position of the image formed by refraction at the first surface.

Fig.1.1

Let V be the distance of image I' and u the distance of the object from the first surface.

Then 1 1 2 1 ' 2 R n n u n V n     …(1)

Now I' acts as the virtual object for the second surface. The final image I is formed at a distance V from the second surface.

Since the rays now pass from a medium of refractive index n2 into a medium of refractive index n1 we have 2 1 2 2 2 1 2 1 R n n R n n V n V n        …(2)

Adding (I) and (2),





_

          2 1 1 2 2 1 1 1 R R n n u n V n Dividing throughout by n1





_                               2 1 2 1 1 2 1 1 1 1 1 1 1 1 1 1 R R n u V or R R n n u V …(3)

First and second principal Foci: The first principal focus (F1) is the position of an object on the principal axis for which the image is at infinity. The distance of the first focus from the lens is the first focal length ƒ1 (Fig. 1.a).

If V = ∞, then u= ƒ1and substituting in Eq. (3), we get,





_                      1 1 2 1 1 2 1 1 ) 1 ( 1 1 1 1 1 1 R R n f R R n f …(4)

The second principal focus (F2 ) is the position of the image on the principal axis for which the object is ·al infinity (u =). The distance of the second focus from the lens is the second focal length, ƒ2 [Fig. 1.b].  V = ƒ2, when u = ∞ . Substituting in Eq. (3),





_                     2 1 2 2 1 2 1 1 ) 1 ( 1 1 1 1 1 1 R R n f R R n f …(5)

Thus the first and second focal lengths of a lens are numerically equal when the lens is placed in a uniform medium. The general convention is to call the second principal focal length as the focal length ƒ of the lens.





_            2 1 1 1 1 1 1 1 R R n u f  …(6)

This formula is known as Lens makers formula.

Power of a lens: Power of a lens is its ability to converge or diverge the rays of light. The power of a lens is measured by the reciprocal of its focal length, P = l/ƒ. The unit of power is dioptre. One dioptre is the power of a lens whose focal length is 1 metre. Convex lenses have positive power and concave lenses, negative power.

Focal length of combination of two thin lenses in contact: When two thin lenses of focal lengths ƒ1, and ƒ2 are placed in contact with each other,

then the focal length of the combination is given by

2 1 1 1 1 f f f   Power of the combination = P = P1 + P2

1.1b. Equivalent Focal Length of Two Thin Lenses Separated by a distance: Let L1, and L2, be two thin lenses of focal lengths ƒ1, and ƒ2 placed in air coaxially a distance ‘a’ apart. Consider a ray PA incident on L1 parallel to the axis at a height h1, above the axis

[Fig. 1.2]. This ray after refraction through the first lens is directed towards D which is the second principal focus of Ll.

Fig 1.1b. Then, deviation produced by first lens =₁ h₁ f₁

The emergent ray from the first lens strikes the lens L2, at a height h2,. The lens L2 deviates it further through an angle 2. Finally the ray meets the axis at F2 which is the second principal focus of the lens system.

Deviation produced by the second lens =₂ h₂ f₂

PA and ,F2B are produced to cut at E2. Then a single convex lens placed in the position E2P2 and having focal length P2F2 is equivalent to the leas system. Thus P2F2 = ƒ is the equivalent local length. Then,

Deviation produced by the equivalent lens = h₁ f Now =1+2

2 2 1 1 1 f h f h f h    …(1) Now _               1 1 1 1 1 1 1 2 2 2 1 f a h f ah h a h BK K O B O h 

Substituting the value of h2 in Eq(1), _         1 2 1 1 1 1 1 f a f h f h f h

or 2 1 2 1 1 1 1 f f a f f f    …(2)        1 2 2 1 2 1 f f a f f f f f …(3)

Here,a



f1 f2



is known as the optical separation or optical interval between the two lenses.

Let us find the position of the equivalent lens, i.e., the distance O2P2. The triangles BF2,O2, and E2F2P2 are similar.

                   1 1 2 1 2 2 1 2 2 2 2 2 2 2 1 2 1 1 f a h h f f a F P h h F orO F P F O h h  Now, 1 1 2 2 2 2 2 2 1 f fa f a f f F O F P P O _           a f f af f a a f f f f             2 1 2 1 2 1 2 1

Let O2P2=-(P2 lies to the left of the lens L2). 1 2 1 2 f fa a f f af       …(4)

Similarly, consider a ray parallel to the axis incident from the right hand side [Fig. 1.1(c)]. Then we can find the position of F1, the point where the ray intersects the

principal axis after refraction through the lens system. E1P1 is the first principal plane. P1 is the first principal point of the lens system. The distance of the first principal point from the first lens is =O1P1.

2 2 1 1 1 1 f af a f f af P O     

The first principal focus F1, is situated at a distance ƒ towards the left of the point P1. If P1 and P2 are the powers of the component lenses and P the power of the combination, then, P=P1+P2-aP1P2

1.2 ABERRATIONS IN LENSES:

1.2.1 Introduction: The deviations in the size, shape, position and colour in the actual images produced by a lens in comparison to the object are called aberrations produced by a lens. Chromatic aberrations are distortions of the image due to the dispersion of light in the lenses of an optical system when white light is used. The defect of coloured image formed by a lens with white light is called chromatic aberration. If monochromatic light is used, then such defects are automatically removed. Besides these defects, there are defects which are present even when monochromatic light is used. Such defects are called monochromatic aberrations. These aberrations are the result of (i) the large aperture of the optical system, (ii) the large angle subtended by the rays with the principal axis and (iii) the large size of the object. As a result of these aberrations, (i) a point is not imaged as a point, (ii) a plane is not imaged as a plane and (iii) equidistant points are not imaged as equidistant points. Following are the monochromatic aberrations: (i) Spherical aberration, (ii) Astigmatism, (iii) Coma, (iv) Curvature of field and (v) Distortion.

Spherical Aberration in a Lens : This aberration is due to large aperture of the lenses. The lens of large aperture may be thought to be made up of lanes.

The marginal and paraxial rays form the images at different places. Fig. 1.3. shows that a monochromatic point source S on the axis is imaged as SP, and Sm. Here, SP, and Sm are the images formed by marginal and paraxial rays respectively. Thus the point object is not imaged as a point. Similarly the focus of marginal and paraxial rays do not coincide. The distance SP Sm on the axis measures longitudinal spherical aberration.

1.2.2. Spherical aberration:

The failure of a lens to form a point image of a point object on the axis is called spherical aberration.

Fig.1.2.1(a) Fig.1.2.1(b)

For rays parallel to principal axis, the distance between the foci of marginal and paraxial rays gives the extent of longitudinal spherical aberration. In Fig. 1.2.1(a) Fp and Fm are the focii for the paraxial and the marginal rays respectively. Spherical aberration of a convergent lens is taken to be positive as the distance (ƒp – ƒm), measured along the axis. The spherical aberration of a diverging lens is negative (Fig. 1.2.1b ).

The following methods are used to reduce spherical aberration.

(i) By using stops: By using stops, we can reduce the lens aperture. We can use either paraxial or marginal –rays. Here, circular discs, called the stops, are used to cut off the unwanted rays. It eliminates marginal rays or paraxial rays. But the use of stops reduce the intensity of the image and the resolving power of the instrument.

(ii) By using the two lenses separated by a distance: When two convex lenses separated by a finite distance are used the spherical aberration is minimum when the distance between the lenses is equal to the difference in their focal lengths. In this arrangement, the total deviation is equally shared by the two lenses, Hence the spherical aberration is minimum.

(iii)By using a crossed lens: The radii of curvature R1, R2 of a thin lens satisfy the following relation:





_          2 1 1 1 1 1 R R n f

It, there fore , shows that , spherical aberration depends upon (i) the refractive index of the lens medium (n) and (ii) the shape factor , which is determined by the ratio

=R1/R2. If the refractive index of material of the lens is 1.5, the spherical aberration will be minimum when=R1/R2 = - 1/6, A convex lens whose radii of curvatures bear the said ratio is called as a crossed lens. It is essential to divide the deviation on two surfaces equally. The axial and marginal rays of light come to focus with minimum of spherical aberration.

Condition for Minimum Spherical Aberration of Two Thin Lenses- Separated by a distance:

Spherical aberration may be minimized by using, two Plano-Convex lenses separated by a distance equal to the difference in their focal lengths.

Let two plano convex lenses L1 and L2 of focal lengths ƒ1 and ƒ2 be placed coaxially separated by a distance a (Fig.1.3.1(c)). Consider a ray OA, parallel to principal axis, incident on lens L1 at height h1 above the principal axis.

Fig.1.2.2(c) The deviation 1 produced by the lens L1 is given by,

The refracted ray AB is incident at B at a height h2 from the axis on lens L2.

The deviation 1 produced by the lens L2 is given by,

The ray AB produced meets the axis at F1 which is the principal focus of lens L1. Hence CF1=ƒ1.

For minimum spherical aberration, the deviation produced by both the lenses should be equal, i.e., 1=2

From similar triangles ACF1 and BDF1 we get

Comparing Eqs (3) and (4), we get

1 1 1 f h   2 2 2 f h   2 1 2 1 2 2 1 1 f f h h or f h f h   a f f h h or CD CF CF DF CF BD AC      1 1 2 1 1 1 1 1 (2) (1) (3) (4)

2 1 1 2 1 1 2 1 f f a f f or a f f f f      

This is the condition for minimum spherical aberration for two lenses separated by a distance.

1.2.3.Chromatic Aberration in a Lens: The focal length of a lens is given by





_          2 1 1 1 1 1 R R n f

Since n changes with the colour of light, ƒ must be different for different colours. This change of focal length with colour is responsible for chromatic aberration. It is classified into two types: (a) Longitudinal chromatic aberration, (b) Lateral chromatic aberration. a) Longitudinal chromatic aberration: A beam of white light is incident on a convex lens parallel to the principal axis (Fig.1.3.2). The dispersion of colours takes place due to

prismatic action of the lens. Violet is deviated most and red the least. Red rays are brought to focus at a point farther than the violet rays. Evidently ƒr> ƒv. The

difference ƒr - ƒv is a measure of the axial chromatic aberration of a lens for parallel rays.

Fig.1.2.3 Expression for Longitudinal chromatic aberration The focal length of a lens is given by





_          2 1 1 1 1 1 R R n f

Let ƒv, ƒr and ƒy be the focal lengths of the lens for violet, red and yellow colours respectively. Also let nv, nr,and ny, be the respective refractive indices. Then





_          2 1 1 1 1 1 R R n f_v v ..(1)





_          2 1 1 1 1 1 R R n f_r r …(2)





_          2 1 1 1 1 1 R R n f_y y …(3) subtracting Eq.(2) from Eq.(1),





_           2 1 1 1 1 1 R R n n f f_{v r} v r or





_             2 1 1 1 1 1 n R R n n n f f f f y y r v r v v r

Now  



n_vn_r





n_y 1



=dispersive power of the material of the lens; f_vf_r  f_y2

 fr  fv fy …(4)

(b) Lateral chromatic aberration: Fig. 1.2.3(a) shows a convex lens and an object AB placed in front of the lens. The lens forms the image of white object AB as BvAv and BrAr in violet and red colours respectively. The images of other colours lie in between the two. Evidently, the size of red image is greater than the size of violet image (BrAr > BvAv). The difference (BrAr - BvAv) is a measure of lateral or transverse chromatic aberration.

Chromatic aberration is eliminated by :

(i) keeping two lenses in contact with each other and (ii) keeping two lenses out of contact.

Achromatic Combination of Lenses:

When two or more lenses are combined together in such a way that the combination is free from chromatic aberration, then such a combination is called achromatic combination of lenses.

The minimization or removal of chromatic aberration is called achromatisation. Chromatic aberration cannot be removed completely. Usually, achromatism is achieved for two prominent colours.

1.2.4. Condition for Achromatism of Two Thin Lenses placed in contact : The focal length of a thin lens is given by





_          2 1 1 1 1 1 R R n f …(1) Here, ƒ is the focal length, n the refractive index, R1 and R2 are radii of curvature of the two surfaces of the lens. Now we know that ƒvaries with n.

Therefore, differentiating Eq. (1),



















2 1 2

1

1

R

R

dn

f

df











n



f dn R R n n dn 1 1 1 1 1 1 _{1 2}_          or







   1 n dn f df …(2) Here ω is the dispersive power of the material of the lens.

Let ƒ1 and ƒ2 be the focal lengths of the two lenses in contact and ω1 and ω2 w2 their dispersive powers.

_         2 1 1 1 1 f f f …(3) Differentiating Eq.(3) 2 2 2 1 1 1 2 2 2 2 1 1 2 1 1 f f df f f df f df f df f df       but 2 2 2 1 1 1       f df and f df 2 2 1 1 2 1 f f f df     

If the combination is to be achromatic, ƒ should be the same for all colours or dƒ=0. 0 2 2 1 1    f f   or 2 1 2 1     f f ….(4) In order, therefore, to design an achromatic doublet of focal length ƒ, the focal lengths of the constituent lenses must satisfy Eq. (3) and Eq. (4). Knowing ω1, ω2 and ƒ, the magnitudes of ƒ1 and ƒ2 can be found by solving Eqs. (3) and (4).

Since ω1, and ω2 are positive, ƒ1 and ƒ2 must be of opposite signs. That is, if one lens is convex, the other should be concave. Since the achromatic doublet is to behave as a converging lens, ƒ1 must be less than ƒ2. Consequently, ω1, < ω2. The converging lens is, therefore, made of crown glass (smaller dispersive power) and the diverging lens, of flint glass (larger dispersive power).( Fig. 1.2.4)

1.2.5. Condition for Achromatism of Two Thin Lenses separated by a finite distance:

Let us consider two convex lenses of focal lengths ƒ1,ƒ2 separated by a distance a (Fig. 1.2.5).

Fig.1.2.5 The focal length of the combination is

_          2 1 2 1 1 1 1 f f a f f f …(1) Differentiating Eq.(1),              1 2 2 2 2 2 1 1 2 2 2 2 1 1 2 f f df f f df a f df f df f df Now 2 2 2 1 1       f df and f df              2 1 2 2 1 1 2 2 1 1 2 f f f f a f f f df    

Since both the lenses are or the same material ω1 =ω2

2 1 2 2 1 1 2 2 f f a f f f df        

For an achromatic combination, the focal length ƒ should not change with colour. :.dƒ=0. Hence, 0 2 2 1 2 2 1 1 _{  } f f a f f    or 2 1 2 1 2 1 1 f f a f f  

2 2 1 f f a  …(2)

That is, the distance between the two coaxial lenses must be equal to half the sum of their focal lengths.

1.3.Coma :

When a lens is corrected for spherical aberration, it forms a point image of a point object situated on the axis. But if the point object is situated off the principal axis, the lens, even corrected for spherical aberration, forms a comet-like image in place of point image. This defect in the image is called coma.

Consider an off axis point A in the object (Fig. 1.3) The rays leaving A and passing through the different zones of the lens such as 11,22,33 are brought to focus at different points B1,B2,B3, gradually nearer to the lens. The radius of these circles go on increasing with increase in radius of zone. Thus the resultant image is comet like.

Fig.1.3.

Removal of coma. The comatic aberration may be eliminated as follows: 1. By using a stop before the lens and so making the outer zones ineffective.

2. By properly choosing the radii of curvature or the lens surfaces. For example, for an object situated at infinity, the comatic aberration may be minimized by taking a lens of n = 1.5 and

3. Abbe sine condition. Abbe showed that coma may be eliminated if each zone of the lens satisfies the Abbe sine condition

9 1 2 1 _  R R k

n1 h1 sinθ1 = n2h2 sin θ2,

Here, n1 and n2 are refractive indices of the object and image regions respectively. h1 and h2 are the heights of the object and the image. θ1 and θ2 are the angles which the incident and the conjugate emergent rays make with the axis. (Fig.l.4a)

Fig 1.3a If this condition is satisfied, the lateral magnification

2 2 1 1 1 2 sin sin   n n h h 

will be same for all the rays of light, irrespective of the angles θ1 and θ2. Therefore, coma will be eliminated.

1.3.1.Astigmatism and its minimization:

Consider a point B situated off the axis in a line object which is vertically below the axis of the lens. When the cone of rays from B falls on full circumference of the lens, then after refraction all the rays do not meet at a single point ( Fig 1.4.1)

a) The rays lying in the vertical plane BMN (called meridional plane) form the image as a horizontal line P.

(b) The rays lying in the horizontal plane BRS (called sagittal plane) form the image as a vertical line S. The circle of least confusion C lies in between P and S. The best image for the object point is obtained here. This defect is called astigmatism. The distance between P and S is a measure of astigmatism and is called the astigmatic difference.

Fig1.3.1

The astigmatic difference in the concave lens is in opposite direction to that produced by a convex lens. Hence astigmatism may be reduced by suitable combination of concave and convex lenses. Such a combination of lenses is called anastigmatic combination. It is used in the construction of objective lens in a photographic camera.

1.4 Let us sum up

In this lesson you learned about the principle employed in convex and concave lenses and studied about the different types of defects and how to minimize them.

1.5 Check your Progress

1) What is meant by principal focus and focal planes in convex and concave lenses 2) Define spherical and chromatic aberrations in lenses

3) What is abbe sine condition in a convex lens 4) What is coma and astigmatism

1.6 Lesson end Activities

1) An achromatic telescope objective of 1.5 m focal length consists of two thin layers in contact with each other and their dispersive power are 0.05 and 0.070 respectively. Calculate their focal lengths.

2) Two thin converging lenses of focal lengths, 10cm and 20cm are separated by a distance ‘a’. Calculate the effective focal length of the combination.

1.7Points for Discussion

1. What do you mean by spherical and chromatic aberration of a lens? Explain how they are caused. How would you correct chromatic aberration in the case of lens system in contact?

2. Obtain an expression for the dispersive power of a lens in the condition for for achromatism of a combination of two thin coaxial lenses (i) when in contact and (ii) when separated by a difference.

1.8 References

1) Optics and spectroscopy - A. Murugesan 2) Geometrical optics – Brijlal and Subramanyam

LESSON – 2

CONTENTS

2.0 Aims and Objectives 2.1 Eye-pieces

2.1.1 Hygen’s Eye-piece 2.1.2 Ramsden’s Eye-piece

2.1.3 Abbe’s Homogeneous Oil Immersion Objective 2.2 Dispersion by A Prism

2.2.1 Refraction through A Prism

2.2.2 Angular and Chromatic Dispersions 2.2.3 Cauchy’s Dispersion Formula 2.2.4 Dispersive Power 2.3 Rainbow 2.3.1 Primary Rainbow 2.3.2 Secondary Rainbow 2.3.3 General Discussion 2.4 Let us Sum UP

2.5 Check your Progress 2.6 Lesson end Activities 2.7 Points for Discussion 2.8 References

2.0 Aims and Objectives

From this lesson you will learn about the usage of convex and concave lenses in different optical instruments as eye pieces. Also you will study about the light reflection through a prism and also about the angular and chromatic dispersions. Also you will learn about the formation of different types of rainbows.

2.1 EYE-PIECES :

An eye-piece is a combination of lenses designed to magnify the image already formed by the objective of a telescope and microscope. An eyepiece consists of two plano-convex lenses. F is called the field lens and E the eye lens (Fig. 1.5). The field lens has large aperture to increase the field of view. The eye lens mainly magnifies the image. To reduce the spherical aberration, the lenses taken are plano-convex lenses. Further the focal lengths of the two lenses and their separation are selected in such a way as to minimize the chromatic and spherical aberrations.

A combination of lenses is used in an eyepiece of a simple lens magnifier for the following reasons:

(i) The field of view is enlarged by using two or more lenses. (ii) The aberrations can be minimized.

Fig 1.3.5. Fig.2.1. 2.1.1. Huygens' Eye-piece:

Construction: It consists of two plano- con vex lenses of focal lengths 3ƒ(field lens) and ƒ (eye lens), placed a distance 2 ƒ apart [Fig.2.1). They are arranged with their convex faces towards the incident rays. The eye-piece satisfies the following conditions of minimum spherical and chromatic aberrations.

(i) The distance between the two lenses for minimum spherical aberration is given by a = ƒl – ƒ2. In Huygen's eyepiece, a= 3 ƒ - ƒ = 2 ƒ. Hence this eye-piece satisfies the

(ii) For chromatic aberration to be minimum a



f₁ f₂



2. In Huygens' eyepiece,



f f



f

a 3  ₂ 22 . Hence this eyepiece satisfies the condition of minimum chromatic aberration.

Fig 2.1.1

Working: An eye-piece forms the final image at infinity. Thus the field lens forms the image I2 the first Field Lens focal plane of eye· lens, i.e., at a distance ƒ to the left of eye-lens. Now the distance between the field lens and eye-lens is 2ƒ. Therefore, the image I2 lies at a distance f to the right of field lens. The image I1 formed by the objective of microscope or telescope acts as the virtual object for the field lens. Thus we treat I1 as the virtual object for the field lens, and I2 as the image of I1 due to it (Fig.2.1.1a) or ν=ƒ., F=3ƒ, u=? We have

F u 1 1 1    or f u 3f 1 1 1   2 3f u  

i.e. I1 should be formed at a distance 3/2 ƒ from the field lens. Therefore the rays coming from the objective which converge towards I1 are focussed by the field lens at I2. The rays starting from I2 emerge from the eye-lens as a parallel beam.

Fig.2.1.1(a) Cardinal Points of Huygens Eyepiece

The equivalent focal length F of this eyepiece is f f f f f f f f a f f F 3 2 3 2 1 3 1 1 1 1 2 1 2 1          F 3f 2

The second principal point is at a distance  from the eye lens.

f f f f f f f f a f f a f             2 2 2 3 2 2 2 1 2 

The first principal point is a distance α from the field lens.

f f f f f f f f a f f a f 3 2 6 2 3 2 3 2 2 1 1 _{ }         

The position of the principal points P1 and P2 and the principal foci F1 and F2 are shown in Fig.2.1.1b Since the system is in air, the nodal points coincide with the principal points.

Fig.2.1.1b 2.1.2.Ramsden's Eyepiece

Construction: It consists of two plano convex lenses each of focal length ƒ. The distance between them is (2/3) ƒ [Fig.2.1.2]. For achromatism, the distance between the

two lenses should be









.

2 2 2 1 f f f f f a      But here a =(2/3)ƒ Fig.2.1.2

Thus in this eyepiece the chromatic aberration is only partly reduced. Similarly, for minimum spherical aberration, a f₁ f₂  f  f 0. Hence the spherical aberration is not at all reduced. This is a demerit of this eyepiece.

Working: I1 is the image formed by the objective of the microscope or telescope. It serves as an object for eyepiece. The eyepiece is adjusted such that the image I2 formed

by the field lens lies in the first focal plane of the eyelens [Fig.2.1.2a]. Then the eyepiece

forms the final image at infinity. Since the focal length of the eye lens is ƒ and a = (2/3) ƒ, I2 is at a distance ƒ/3 from the field lens. Now, the image I1 due to objective

serves as the object for field lens. I2 is the image of I1 due to field lens. Or,V f 3,F  f, to find u we have

f u or f u f or F u V 4 1 1 1 3 1 1 1 1        4 f u   Fig. 2.1.2(a)

Thus the eyepiece its so adjusted that the image (I1) formed by the objective of telescope or microscope lies at a distance ƒ/4 towards the left of field lens. The crosswire is placed at I1. I1 serves as the object for field lens and its image is formed at I2.

Cardinal points: The focal length F of the equivalent lens is

4 3 3 4 3 2 1 1 1 1 1 2 2 1 2 1 f F f f f f f f f a f f F         







2 3



2 2 3 2 2 1 2 f f f f f a f f a f          







2 3



2 2 3 2 2 1 1 f f f f f a f f a f          

The positions of the principal points Pl and P2 and the principal foci F1 and F2 are shown in Fig.2.1.2(b). Since the system is in air, the nodal points coincide with the principal points.

Fig.2.1.2(b).

Distance of the first principal focus from the field lens of the eyepiece = F1L1 = F1P1 - = 3ƒ/4 - ƒ/2=ƒ/4. Similarly the distance of the second principal focus from the eye lens is L2F2 = P2F2 - = 3ƒ/4 - ƒ/2=ƒ/4.

Comparison of Eyepieces

Huygens' eyepiece Ramsden's eyepiece

1

2. 3. 4.

The image of the object formed by the objective falls in between the two lenses. Therefore, no cross wires can be used. For this reason. it is called a negative eyepiece.

It satisfies the condition for minimum spherical aberration.

It satisfies the condition for achromatism.

It is generally used for biological observations where no measurements are required.

The image of the object formed by the objective lies in front of the field lens. Therefore, cross wires can be used. For this reason, it is called a positive eyepiece.

It does not satisfy the condition for minimum spherical aberration.

It does not satisfy the condition for achromatism.

It is used with instruments meant for physical measurements.

2.1.3 ABBE’S HOMOGENEOUS OIL IMMERSION OBJECTIVE:

In a microscope the objective is a lens system corrected for chromatic and spherical aberrations. In a microscope designed to magnify 500 times or more, the objective alone produces a magnification of 50 or more. The microscope objective is a hemispherical lens L1 of radius R with its plane surface directed towards the object A. The plane surface is in contact with cedar wood oil having the same refractive index as that of the lens. Under such conditions no refraction can take place except at the spherical surface of the lens L1. If the object is placed at A at a distance



R

from C, then the image formed at I1 at a distance  R from C (Fig.2.1.3). Then the image will be free from spherical aberration. The magnification produced is

2 1 _     R R AC CI Fig.2.1.3

If the second lens L2 has the centre of its concave surface at I1 then the rays emerging from the hemispherical lens L1 will fall on its lower concave surface normally and will pass undeviated. Then the rays fall on the convex surface of L2. The radius of the convex surface of the lens L2 is so selected that I1 is one of the planatic points. The rays appear to diverge from J" which is the other a planatic point. This gives added magnification without introducing spherical aberration.

This property of the lens L2 holds true only for rays from the point A and not for points adjacent to it.

The object is immersed in oil such as cedar wood oil and the hemispherical lens with its plane face is also immersed in the oil. This is known as Abbe's homogeneous oil immersion objective.

2.2. DISPERSION BY A PRISM

A beam of white light, when it passes through a prism is split up into constituent colours and that is called dispersion of light. The image thus formed on a screen is called a. spectrum.

Fig 2.2

The spectrum consists of visible and invisible regions. In the visible region the order of the colours is from violet to red. The principal colours are given by the word VIBGYOR (Violet, Indigo, Blue, Green, Yellow, Orange and Red). The deviation produced for the violet rays of light is maximum and for red rays of light it is minimum. Fig.2.2 represents the dispersion of a white ray of light by a prism in the visible region. The region of the spectrum, of wavelengths shorter than violet is called ultra-violet and the region of wave-lengths longer than red is called infra-red. In the present chapter, the discussion relates only to the visible region of the spectrum.

The refractive index for the material of a prism (or a lens) is different for different wavelengths (or colours). The deviation and hence the refractive index is more for blue rays of light than the corresponding values for red rays of light. The deviation and the

refractive Index of the yellow constituent are taken as the mean values. If the dispersion through a prism does not follow the order given by VIBGYOR, it is said to be anomalous dispersion.

2.2.1.REFRACTION THROUGH A PRISM:

The refractive index of the material of a prism is given by

2 sin 2 sin A D A  

where A is the angle of the prism and D is the angle of minimum deviation. For a small angled prism            2 2

where  and  refer to the angle of the prism and the angle minimum deviation (for small values of , the angle  is also small and the sines of the angles are taken equal to the angles).









  1

Fig. 2.2.1.represents the angles of deviation _b, and _r, produced in the blue, mean yellow and red rays of light.

Fig.2.1.1 The deviations _b, and _rcan be written as:

 (1) for mean yellow light. _b (_b 1)for blue light and

_r (_r 1) for red light

The difference in deviation between two colors is called angular dispersion. _b_r (_b1) (_r 1)  (



_b 



_r )



dividing ) 1 ( ) ( ) 1 ( ) (                  b r b r b r

where b and r are the refractive indices for the blue and red rays of light and  is the refractive index for the mean yellow rays of light. The expression _

         1 1     _{b r} d is called the dispersive power of the material of the prism. It is constant for two colours (or wavelengths] chosen end is represented by ω,

1         b r

The reciprocal of ω is called the Constringence. It is also customary to represent b,  and

r where F, D and C which correspond to the Fraunhoffer lines (dark lines) in the solar spectrum. The P, D and O lines lie in the blue, yellow and red regions of the spectrum and their wavelengths are 4861Å, 5893 Å and 6563 Å respectively. (l Å =l Angstrom unit=10-8 cm).

2.2.2.ANGULAR AND CHROMATIC DISPERSIONS: The refractive index  of the material of a prism is given by

2 sin 2 sin A A    

where A is the angle of the prism and θ is the angle of minimum deviation. 2 sin 2 sin A   A 2 sin 1 2 cosA  2 2 A 

If dθ is the difference in the angle of deviation between two spectral lines of wavelengths

λ and λ +d λ, then

 

d d

is called the angular dispersion between the wavelengths λ and λ +dλ .

Differentiating equation with respect to  on the left hand side and θ by right hand side.

   d A A d 2 sin 2 cos 2 1   dividing by d λ

     d d A A d d 2 sin 2 2 cos        d d A A d d 2 cos 2 sin 2       d d A A 2 sin 1 2 sin 2 2 2     d d

is called chromatic dispersion of the material of the prism. The angular dispersion of the material of a prism depends on the angle of the prism and refractive index of the material of the prism. Using a spectrometer and the given prism, a graph is drawn between  and λ ( along the Y-axis and λ along the X-axis). The tangent to the curve at any point measures the chromatic dispersion

 

d d

for that particular wavelength. Substituting this value of

  d d in equation (iv),   d d can he calculated. 2.2.3. CAUCHY'S DISPERSION FORMULA:

When an electromagnetic wave is incident on an atom or a molecule, the periodic electric force of the wave sets the bound charges into vibratory motion. The frequency with which these charges are forced to vibrate is equal to the frequency of the wave. The phase of this motion as compared to the impressed electric force will depend on the impressed frequency. It will vary with the difference between the impressed frequency and the natural frequency of the charges.

Dispersion can be explained with the concept of secondary waves that are produced by the induced oscillations of the bound charges. When a beam of light propagates through a transparent medium (solid or liquid), the amount of lateral scattering is extremely small. The scattered waves travelling in a lateral direction produce destructive interference. However, the secondary waves travelling in the same direction as the incident beam superimpose on one another. The resultant vibration will depend on the phase difference

between the primary and the secondary, waves. This super-imposition, changes the phase of the primary waves and this is equivalent to a change in the wave velocity. Wave velocity is defined as the speed at which a condition of equal phases is propagated. Hence the variation in phase due to interference, changes the velocity of the wave through the medium. The phase of the oscillations and hence that of the secondary waves depends upon the impressed frequency. It is clear, therefore that the velocity of light in the medium varies with the frequency of light. Also refractive index depends upon the velocity of light in the medium. Therefore the refractive index of the medium varies with the frequency (wave length) of light.

The relative permittivity of the medium in the case of dynamic polarizability is given by



     i i e r f m Ne X _{2 2} 0 2 1 1    

Here N is the number of electrons per unit volume, e the charge and m the mass of the electron, ε0 permittivity of free space, ƒi oscillator strengths of the substance, ωi is the angular frequency of the electromagnetic spectrum of the substance, ω is the impressed angular frequency.

Also



f_i 1

and relative permeability in majority of substances that transmit electromagnetic Waves is equal to 1.

r



 

 2

Assuming that there is only one atomic frequency ωo where

ω < < ωo



2 2



0 0 2 2 1         m Ne

using the binomial expansion,





2 1 2 2 0 0 2 1 _             m Ne

1 2 0 2 2 0 0 2 1 2 1                         m Ne                    ₂ 0 2 2 0 0 2 1 2 1     m Ne As,     2 c and 0 0 2     c                  ₂ 2 0 2 0 2 1 2 2 1       c m Ne _{2 2} 0 2 4 0 2 2 0 2 2 0 2 8 8 1         mc Ne mc Ne    Taking A mc Ne   ₂ 0 2 2 0 2 8 1    And B mc Ne  2 0 2 4 0 2 8   ₂    A B

Equation represents Cauchy's dispersion formula. A and B are called Cauchy's constants. The values of A and B depend on the medium. From equation (vi) it is evident

that the refractive index of the medium decreases with increase in wavelength of light. If a graph is plotted between  and 1₂

 it will he a straight line.



2 2



0 0 2 2 1        m Ne

Fig 2.2.3

The intercept OP on the Y-axis gives value A. the slope of the line PC gives the value of B (Fig.2.2.3)

DISPERSIVE POWER

According to Cauchy's dispersion formula,

₂    A B Differentiating equation 2₃    B d d     d d

is the dispersive power of the medium. Therefore, dispersive power is inversely proportional to the cube of the wavelength of light. At λ=4000 Å (violet) the dispersive power is about eight times the dispersive power at λ =8000 Å (red). It. means the spectral lines are more dispersed near the violet end of the spectrum that at the red end.

2.3. RAINBOW

Rainbows are formed by sunlight falling on raindrops. Sometimes two rainbows are seen. The common rainbow known as the primary rainbow is a coloured band, having red on the outside and violet on the inner side. It is formed due to two refractions and one reflection of light falling on raindrops. The other rainbow called the secondary rainbow is formed due to two refractions and two reflections of the sunlight falling on the raindrops. The rainbows are visible only when the altitude of the sun is less than 42°. No rainbows are seen when the altitude is more than 42°. A complete rainbow can be seen in an aeroplane flying at high altitudes.

2.3.1. PRIMARY RAINBOW

Consider a ray of sunlight incident at the point B of a raindrop (Fig. 2.3.1). The ray AB after refraction travels along BC and is reflected along CD and finally comes out along DE. The deviation of the ray AB after refraction is (i-r). The deviation of the ray BC after refraction at C is (180-2r) and the deviation of the ray CD after refraction at D is (i-r). Therefore, total deviation = 2( i-r)+(180-2r)

 1802i4r For the angle of deviation to be maximum or minimum, the differential coefficient of 

with respect to i must be zero.

 Differentiating equation (i), di dr di d 4 2   But 0 di d  24 0 di dr or 2 1  di dr Also, r i sin sin   or sin r = sin i Differentiating, i di dr r cos cos   or r i di dr cos cos   Equating 2 1 cos cos  r i  or i r 4 cos cos 2 2 2  



1-sin



- sin r cos 4 2 i2 2r 2 2 2

But  sinr sini

 4cos2 i 2-sin2r 1 ) cos r (sin -cos 3 2 i2 2  2i    i 3 1 cos 2    Taking refractive index of water for red light =1.329,

i=59.6° and =137.2° 180-137.2=42.8°

i=58.8° and =139.2° 180-139.2=40.8°

These angles 42.8° and 40.8° are shown in Fig.2.3.1(a)

Fig 2.3.1(a)

It is to be remembered that sunlight strikes the raindrop at different angles of incidence and undergoes different deviations. Only those rays produce a rainbow which have the angle of incidence corresponding to minimum angles of deviation. All such rays produce concentrated effect of light in the formation of a rainbow. Also, as shown in the diagram, in the primary rainbow, the angle of inclination of red light is more on the eye than the violet. Therefore, the outside of the rainbow appears red and the inner violet. The other spectral colours lie in between violet and red in their order.

2.3.2.SECONDARY RAINBOW

In the case of secondary rainbow, there are two reflections and two refractions. Suppose a

ray AB from the sun strikes the drop at B end after refraction goes along BC. It is reflected at C and D and after refraction at E, finally emerges along EG (Fig. 2.3.2)

Fig.2.3.2

The total deviation

=2(i-r)+2(180-2r)=360+2i-6 The rays are concentrated around the directions of minimum deviation.

Differentiating  with respect to i

di dr di d 6 2   But 0 di d 0 6 2  di dr or 3 1  di dr Also, r i sin sin   or sin r = sin i

Differentiating, i di dr r cos cos   r i di dr cos cos   Equating 3 1 cos cos  r i  or 9 cos cos 2 2 2 r i  

or 9cos2i 2(1sin2r)22sin2r But sin r = sin i

9cos2i 2sin2i ) cos (sin cos 8 2i 2  2i 2i 8 1 cos 2    i

Taking  of water for red light=1.329, the angle of deviation =360-129.2=230.8. The acute angle = 230.8-180= 50.8° (Fig.2.3.2) Taking  of water for violet light = 1.342, the angle of deviation = 360 -126.48 = 234.52°. The acute angle = 234.62-180 = 64.62°.

In Fig. 2.3.2(a), it is shown that the angle of inclination for violet rays is more than for red rays.

2.3.3.GENERAL DISCUSSION

The primary rainbow is formed by light from the sun undergoing one internal reflection and two refractions and emerging at, minimum deviation. The inner violet edge subtends an angle of 40.8° and the outer red edge subtends an angle of 42.8°. The secondary rainbow is formed by light from the sun undergoing two internal reflections and two refractions and also emerging at minimum deviation. This rainbow is fainter than the primary one and inner red edge subtends an angle of 50.8° and the outer violet edge subtends an angle of .54.52°. Therefore the colours in the secondary rainbow appear in the reverse order compared to those of the primary rainbow. All the spectral colours are present in order between violet and red (Fig.2.3.3). Between the two rainbows no bows are viewed because for this the angle of deviation should be lese than the minimum in both the cases, which is not possible.

Sometimes other bows are observed which are near the inner edge of the primary bow or near the outer edge of the secondary bow. These are known as supernumerary bows and depend upon the size of the raindrops. These are due to diffraction just similar to diffraction at narrow slits.

2.4 Let us sum up

From this lesson your learned about the usage of lenses as eyepices and also you learned about the light reflection through a prism and also about angular and chromatic dispersions. Finally you learned about the different types of Rainbows.

2.5 Check your Progress

1) What are the differences between Huygens and Ramsden’s eyepieces ? 2) What are cardinal points

3) Explain angular dispersion

4) State the importance of the Cauchy’s dispersion formulae 5) Define dispersive power

2.6 Lesson end Activities

1) The focal length of an eyepiece is 0.015 m. Calculate the focal length of the two lenses used in the Huygen’s eyepiece. Also calculate the chromatic aberration. (Dispersive power is 0.03)

2) The angle of the prism is 60 and its refractive index for green light is 1.5. If the green light passes through it find the deviation.

2.7 Points for Discussion

1. Give the theory of formation of a (i) primary rainbow (ii) secondary rainbow. Explain in what order are the colours arranged.

2. (i) Derive Cauchy’s dispersive formulae/

(ii) Write a note on ABBE’s homogeneous oil immersion objective. 3. With a neat diagram explain

(i) Huygen’s eyepiece and its cardinal points (ii) Ramsden’s eyepiece and its cardinal points 2.8 References:

1) Light by Brijlal and Subramanyam

UNIT – II

LESSON - 3

CONTENTS

3.0 Aims and Objectives 3.1 Introduction

3.1.1. Interference due to reflected light( Thin Film) 3.1.2 Interference due to Transmitted Light (Thin films) 3.2. Wedge-Shaped Film –(Air Wedge)

3.2.1. Experiment to measure the Diameter of a thin wire 3.2.2. Applications of Air Wedge

3.3. Newton’s Rings

3.3.1. Expression for the radii of the rings:

3.3.2.Determination of Wavelength of Sodium Light by Newton’s Ring. 3.3.3.Determination of Refractive index of a Liquid by Newton's Rings 3.4 Let us sum up

3.5 Check your progress 3.6 Lesson end activities 3.7 Points for discussion 3.8 References

3.0 Aims and Objectives

In this lesson you will study about the principles of interference and its characteristics in thin films for reflected and transmitted light. Also you will learn about the wedge shaped film. You will also learn about Newton’s rings and its application in determining the wavelength of sodium light and also the determination of  of liquid.

3.1 INTRODUCTION

Interference is the phenomenon of the superimposition of one light source over the other. Due to the superimposition of any two light sources, the resultant energy is redistributed into a position of maximum intensity and of minimum intensity. If the crest of one wave falls on the crest of the other, constructive interference is produced. If the crest of one

wave falls on the trough of the other, destructive interference is produced. When a soap film or an oil film is viewed from a reflected light or transmitted light, it exhibits different colours. It is due to the interference pattern produced in thin films

3.1.1. INTERFERENCE DUE TO REFLECTED LIGHT

Consider a transparent film of thickness t with refractive index  Let AB be the upper surface of the film and CD be its lower surface. Let a ray of light PQ be incident at an angle of incidence i on the upper surface of the film. Some part of the incident beam is reflected by the upper surface along QR and the remaining part is transmitted through the film along QS. This transmitted beam undergoes reflection by the lower surface of the film CD. SU is the reflected beam and ST is the transmitted beam from the interface CD. The ray SU undergoes reflection at the interface AB. UV is the transmitted beam and UW is the reflected beam from the interface AB. (Fig 3.1.1.)

Consider the rays PQR (ray I) and PQSUV (ray II). Ray II has traveled greater distance compared to ray 1. The path difference between the two rays can be determined by drawing a perpendicular line from the point U to the line QR. UN is the normal to the line QR drawn from the point U.

The path difference between the two rays is given by,

x = (QS+SU) – QN …..(1) Let r be the angle of refraction . (In Fig2.1,) consider the triangle QSM

QS SM r  cos r t r SM QS cos cos  

similarly, from le SMU,

r t SU cos  …(2)

Fig 3.1.1 Therefore, r t r t SU QS cos cos    …(2)

from le QSM, QS QM r  sin QM=QS sin r r r t r QS QM QU sin cos 2 sin 2 2    …(3)

Consider the le QUN,

QU QN i 

sin

QN QUsini …(4) From Eqs (3) and (4),

r r i t QN cos sin sin 2  …(5) From Snell’s law,

r i sin sin   ...(6)  sin i= sin r From Eqs (5) and (6), we get

r r r t QN cos sin . sin 2   …(7) From Eqs(1), (2) and (7), we get

r r t r t x sin2 cos 2 cos 2     r t r r t cos 2 ) sin 1 ( cos 2 2     

The ray QR is the reflected ray and the ray UV is the transmitted ray. Due to reflection, there is a phase change of  degree or an additional path difference of

2



will be introduced. So, the total path difference is written as, 2

cos

2 

 t r x

To get the condition for bright fringe, the path difference is equated to an integral multiple of λ, (i.e., n=1,2,3…) i.e., t r n 2 cos 2 i.e.,          2 1 2 cos 2 t r n …(8) To get the condition for dark fringe, the path difference is equated to half integral

multiple of λ, i.e., (2 1) 2 2 cos 2t r  n  or 2tcosr n …(9) Equations (8) and (9) give the conditions for bright and dark fringes produced by thin films due to interference of light.

3.1.2 Interference due to Transmitted Light(Thin films):

Consider a thin transparent film of thickness t and refractive index . A ray SA after refraction goes along AB. At B it is partly reflected along BC and partly refracted along BR. The ray BC after reflection at C, finally emerges along DQ. Here at B and C reflection takes place at the rarer medium ( medium-air interface). Therefore, no phase change occurs. Draw BM normal to CD and DN normal to BR. The optical path difference between DQ and BR is given by,

BN CD BC x(  ) also, orBN MD MD BN r i      sin sin

Fig.3.1.2 In Fig.3.1.2 BPC=r and CP=BC=CD  BC+CD=PD  x(PD)(MD)(PDMD)PM In the  BPM. PM BP r BP PM r or .cos cos   But, BP=2t  PM=2t cos r  xPM 2tcosr (i) For bright fringes, the path difference xn

 2t cos r=nλ

where n=0,1,2,3,…etc.,

(ii) For Dark fringes, the path difference

2 ) 1 2 (    n x  2 ) 1 2 ( cos 2t r  n  where n=0,1,2,3,…etc.,

In the case of transmitted light, the interference fringes obtained are less distinct because the difference in amplitude between BR and DQ is very large. However, when the angle of incidence is nearly 45, the fringes are more distinct.

3.1.3 Production of Colours in thin films:

When white light is incident on a thin film, the light rays are reflected from the top and bottom surfaces of the film. The condition for brightness or darkness depends upon (i) , (ii) t and (iii) r. Here t and r are constants. White light contains a number of wavelengths( colours). Condition for brightness in the reflected system is

2 ) 1 2 ( cos 2t r  n 

The colour for which this condition is satisfied will be present with maximum intensity, other colours in the neighborhood will have lesser intensity. Colours for which the condition or darkness



tcosr n

2

is satisfied wil be absent. Hence the film will appear in colored. 3.2. WEDGE-SHAPED FILM

Consider a wedge-shaped film of refractive index  enclosed by two plane surfaces OP and OQ inclined at an angle θ (Fig.2.2). The thickness of the film increases from O to P. When the film is illuminated by a parallel beam of monochromatic light, interference occurs between the rays reflected at the upper and lower surfaces of the film. So equidistant alternate dark and bright fringes are observed. The fringes are parallel to the line of intersection of the two surfaces. The interfering rays are AB and DE, both

Fig 3.2.

Expression for the fringe width : The condition for a dark fringe is 2nt cos r = mλ. Here for air n = I. For normal incidence cos r = cos θ =1. Suppose the mth dark fringe is formed where the thickness of the air film is tm ,(Fig 3.2a). Then,

2 1 tm 1 =mλ or 2tm = mλ …(1)

Suppose the (m + 1) th dark fringe is formed where the thickness of the air film is tm+1 Then,

2 tm+1= (m + 1)λ …(2) Subracting (1) from (2),

2 (tm+1-tm)=λ …(3)

Let xm+1 and xm be the distances of the (m+1) th and mth dark fringes from O. d=diameter of the wire; L=distance between O and the wire. Then

Fig.3.2a       L d x t x t m m m m 1 1 ; 1 1     _m x_m L d t _m x_m L d t 

substituting these values in Eq(3), we get,

    ) ( 2 x ₁ x L d m But x_m1x -fringe width.

or  

L d

    2 2    d L

d, λ and L are constants. Therefore, fringe width β is constant. Similarly if we consider

two consecutive bright fringes, the fringe width β will be the same.

3.2.1.Experiment to measure the Diameter of a thin wire:

An air wedge is formed by inserting the wire between two glass plates. Monochromatic light is reflected vertically downwards on to the wedge by the inclined glass plate G (Fig.3.2.1). A traveling microscope M with its axis vertical is placed above G. The microscope is focused to get clear dark and bright fringes. The fringe width (β) is measured. The length (L) of the wedge also is measured. Knowing λ, the diameter (d) of the wire is calculated using the formula,

Fig.3.2.1   2 L d 

Testing a surface for planeness: A wedge shaped air film is formed between an optically plane glass plate OP and the surface under test (OQ). The fringes will be straight if the surface under test is perfectly plane. If the surface OQ is not perfectly plane, the fringes will be irregular in shape. In practice, perfectly plane surfaces are

produced by polishing the surfaces and testing them from time to time, until the fringes are straight. In testing for planeness, an extended source of light should be used.

3.2.2. Applications of Air Wedge (i) Testing of flat surfaces

The air wedge experiment is used to test the planeness of a glass plate. To test the planeness of a glass plate the given glass plate is placed over an optically plane glass plate. Then the interference fringes are formed. If the fringes are of equal thicknesses and of straight lines then the given glass plate is optically flat.

The condition for bright fringe is 2tcosr (2n1) 2

For normal incidence, cos r =1 and for air =1. Hence, the above equation becomes 2 ) 1 2 ( 2t  n 

The interference fringes are characterized for a particular value of n. For a particular value of n, the thickness of the air film and the flatness of the given glass plate are constant. Therefore, straight fringes are obtained. If the fringe obtained for a constant value of n is a colored one, it represents the varying thickness of the film and it represents that the given glass plate is not flat. The flatness of a given glass plate can be tested up to one tenth of the wavelength of the light used.

ii) Thickness of a thin sheet of paper

A monochromatic source of light is passed through a convex lens and is made to fall on a glass plate inclined at an angle of 45° as shown in Fig. 3.2.2. The rays of light are reflected in the downward direction and are used to illuminate an air wedge set up. An interference pattern is formed due to the superposition of the rays emerging from the upper glass plate and the rays reflected from the bottom glass plate.

Alternate bright and dark fringes are viewed through the microscope. Anyone bright or

dark fringe is taken as the

nth fringe and the

corresponding to n, n + 5, n + 10…n + 50 fringes are taken. From the observation, the value of the fringe width β is determined. Then, using the relation ,

2 

l

t  the thickness of the paper is determined.

Fig.3.2.2 3.3. Newton’s Rings:

A plano-convex lens of large radius of curvature is placed with its convex surface in contact with a plane glass plate. Then, an air film is formed between the lower surface of the lens AOB and the upper surface of the plate POQ (Fig.2.3). The thickness of the air film is zero at the point of contact O and gradu