**Wave optic**s is the branch of optics which deals with interference, diffraction, polarization, and other phenomena of light.

**Electromagnetic spectrum**

Electromagnetic spectrum is the distribution of electromagnetic radiation of all possible frequencies and wavelengths.

Electromagnetic spectrum is classified are as follows:

**Radio waves:** Frequency of these waves ranges from few Hz to 10^{9 }Hz. Radio waves emitted by radio stations. Radio waves are also emitted by stars and gases in space.

**Microwaves: **Frequency of these waves ranges from10^{9 }Hz to 3.0 *10^{11}Hz.The wavelength of microwaves is greater than 1mm and less than 30cm.It is used byastronomers to learn about the structure of nearby galaxies.

**Infrared:** Frequency of these waves ranges from 3.0 *10^{11}Hz to 10^{9 }Hz. The wavelength of infrared is 1nm to 700nm. In space; infrared light helps us map thedust between stars.

**Visible:** Frequency of these waves ranges from 4.3 *10^{14} Hz to7.5 * 10^{14 }Hz.The wavelength of visible light is 400nm to 700nm.Our eyes detect visible light. Fireflies, light bulbs, and stars all emit visible light.

**Ultraviolet:** Frequency of these waves ranges from 7.5 * 10^{14 }Hz to 5.0 *10^{15} Hz .The wavelength of ultraviolet rays is 400nm to 60nm Ultraviolet radiation is emitted by the Sun and are the reason skin tans and burns. “Hot” objects in space emit UV radiation as well.

**X-ray:** Frequency of these waves ranges from 5.0 *10^{15}Hz to3.0* 10^{18 }Hz.The wavelength of X-ray is 60nm to 10^{-8}nm. Hot gases in the Universe also emit X-rays.

**Gamma ray:** Frequency of these waves rangesfrom 3.0*10^{18} Hz to3.0 * 10^{22 }Hz. The wavelength of Gamma rays is 0.1nm ^{to} 10^{-5}nm. Doctors use gamma-ray imaging to see inside body. The biggest gamma-ray generator of all is the Universe.

**Wavelets**

Wavelets are the disturbance of the point source but the point source is taken in the primary wave front. Wavelets formed by the locus of the virtual source.

Wave lets are of two types:

a. Primary wavelets and

b. Secondary wavelets.

**Wave front**

During the propagation of the wave, all the particles of the medium which are located at the same distance from the source receive the disturbance simultaneously and vibrate in the same phase. Thus, a wave front of light at any instance is the locus of all particles of the medium vibrating in the same phase at that time. The shape of wave front depends on the nature of source and the disturbance of the wave front from the source. That is wave front. Wave front is the disturbance of the point source. Wave front formed by the locus of the real source. Wave fronts are of three types:

a. Spherical wave front

b. Cylindrical wave front

c. Plane wave front.

**Huygens principle**

Huygens’s principle states that:

a. Each point on the primary wave front acts as a source of secondary wave lets, sending out disturbance in all direction in a similar manner as the original source of the light does.

b. The new position of the wave front at any instant is given by the forward envelope of the secondary wavelets at that instant.

Consider a point source of light. Let XY be the section of the spherical wave front at any time t. supposed we are interested in finding the new position of the wave front at time t+∆t. to do so, a number of points a,b,c,d are the point taken on the primary wave front. These point acts as the source of secondary wavelets. In time ∆t light will travel a distance c∆t. taking the point a,b,c,d,…as the centre of sphere each of radius c∆t are drawn. The forward enveloped X’Y’ of these spheres give the position of wave front at ∆t +t and called secondary wave front.

Figure.

**Laws of reflection on the basis of wave theory**

Let us consider a plane wave front PQ incident on the reflecting surface AB at an angle of incidence i as shown in figure.

The I_{1}, I_{2}, I_{3} incident ray on the wave front PQ which are also perpendicular to the wave the wave front PQ. NP is normal to the reflecting surface AB. Point P of the wave front reaches the reflecting surface at time t=0. By the time , point Q, of the wave front reaches at point P’(t=t), the secondary wavelets from P spread out in the form of sphere having radius PQ’=QP’=ct, where c is the velocity of the light .now draw the tangent to the sphere from P’ point then P’Q’ become reflected wave front. Similarly, the wavelets from R reach point S and from S reach to the point T of the reflected wave front in the time t. reflected rays after strike the reflecting surface must be at right angle to the wave front P’Q’. in figure reflected rays are represented by I’_{1}, I’_{2}, I’_{3}.

Now draw P’N’ normal to the reflecting surfaceAB. Then∠Q’P’P=r, where r is the angle of reflection. From figurewe have right angle triangle PQP’ and PQ’P’, we have

PQ’=PA’

∠PQ’P’= ∠PQP’ = 90° and

PP’ is the common. Therefore the two triangles are congruent.

So, ∠QPP’=∠Q’P’P therefore i=r

This shows that the angle of incident is equal to the angle of reflection which is the first law of reflection.

**Laws of refraction on the basis of wave theory**

The laws of refraction are:

The ratio of the sine of the angle of the incidence to the sine angle of the refraction is constant for any two medium. SinisinrSinisinr=μμ which is the refractive index of medium in a air

The incident rays, refracted rays and the normal at the point of incident on the refracting surface lie on the same plane. These laws can be verified as:

Here each point of the wave front AB acts as secondary wavelets.

Figure

If t is time taken by the secondary wavelets to reach at time point C from B then BC=ct. In the same, time wave front originating from A has traveled a distance of vt =AD in denser medium

Similarly wave front from P reaches at Q. if we draw the sphere of radius AD=vt with tangent to the sphere CD emanating from points A and C then CD is the new wave front in denser medium.

Let us draw NA normal to the surface XY

Then ACD = r

In rt angled triangle ABC

Sini=BCACBCAC=ctACctAC…………..1

Sinr = ADCDADCD =vtACvtAC……….2

From 1 and 2 sinisinrsinisinr = cvcv =μμ which proves the Snell’s law

Further the incident ray refracted ray and the normal to the surface of incident all lie on the same plane. This verifies the laws of refraction.

**Foucault’s method for the determination of speed of light**

The experimental arrangement of Foucault’s method are given below

Experimental arrangement:

The rays of light from a bright source S are allowed to fall on convex lens L, which will bring them to focus at point I in the absence of plane mirror XY, which is capable of rotation about an axis through the point Q. The plane mirror make the ray to meet at point P, the pole of the concave mirror M such that PQ=IQ=d say. When the plane mirror is stationary, the rays of the light after reflection from concave mirror retrace their path and finally image coincident with S if the glass plate is placed at 45^{0}to the optical axis of the lens, then the returning light is reflected from it so as to produce the image I’( instead the image coincident with S)

Theory

When the plane mirror is rotated about its axis though the point Q,the intermittent image of the source is seen through the eyepiece. It is because the light falls on the concave mirror for a small fraction of revolution. As the speed of revolution is increased slowly, a stage comes, when image is seen continuously due to persistence of vision. It happens when mirror is rotated at the speed of more than 10r.p.s. as the distance between the plane mirror and the concave mirror is very small, negligible is compared to the velocity of light therefore , the light returning to the plane mirror ( after the reflection from the concave mirror)will find it practically in the same position . As a result, when plane mirror is rotating at low speed, the image will be seen still at I’

Now, suppose the speed of rotation of the plane mirror is increased. The light reflected from the plane mirror in the position XY, on returning from the concave mirror will find it in position X’Y’ i.e. displaced through, say angle ɵ the reflected ray will turn through angle 2ɵ . To the eye, the rays will appear to diverge from I and the image will be seen to shift to position I”. The displacement II’’ can be measure with the micrometer attached to the eyepiece.

Let C be the velocity of the light d be the distance between the plane mirror and concave mirror and n be the no. of revolutions made per second by the rotating mirror.

The time taken by the light to cover distance 2d i.e. from Q to P and back to Q is given by

T=2d/c…………………….1

As the plane mirror make n rotation per second. It covers an angle 2∏n in one second. Therefore, time taken by the mirror to rotate thought angle ɵ is given by

T=ɵ/2∏n………………2

From 1 and 2 we get

C=4∏nd/ɵ…………………3

To find ɵ

Let a and b be the distance of the plane mirror and the source of the light from the optical centre of the convex lens.

Now angle between two reflected rays QI and QI’ is 2ɵ therefore

2ɵ=II’/d

Or, II’=2ɵd……………….4

The image S and S’ formed by the lens are the images of I and I’

From the relation

Size of image/size of object=distance of image/ distance of object

Then we have, SS’/II’=OS/OI…………5

Where OS=b and OI=a+d

Now SS’=I’I”

Let the displacement I’I” in the image be x then, SS’=x

Then from equation 5 we have

x/II’=ba+dba+d

or, II’=(a+d)xb(a+d)xb…………………………6

From 4 and6 we have

ɵ= (a+d)x2bd(a+d)x2bd………………7

on putting the value of ɵ in equation in 3 we get

C = 8πnbd2(a+d)x8πnbd2(a+d)x

This equation gives the speed of light in term of speed rotation. According to the Newton’s corpuscular, the velocity of the light in water should be greater than air. By using the Foucault method, It was found that s’/s is greater i.e. the velocity of light in water is smaller than velocity of light in air. Also the ratio of c/c’ c’ being speed of light in water, equal to the refractive index supporting the wave theory of light.

The advantages of Foucault’s method are:

a. It can be performed in laboratory as it covers the small area.

b. Speed of the light in any optical medium can be determined.

c. It justified the validity of wave theory of light as velocity of light in water found to be less than velocity of light in air.

The disadvantages of Foucault’s method are:

a. The image obtained is very faint due to reflection and refraction of light at various surfaces thus makes the observation difficult.

b. Due to small displacement of the image accurate measurement cannot be obtained.

**Michelson method for the determination of speed of light**

The experimental arrangement of Michelson method is as shown in figure. It consists of three mirrors such as octagonal mirror (M_{1}), concave mirror (M_{2}) & plane mirror (M_{3}). Light from the source (S) incident at an angle of 45^{0} on one face of octagonal mirror (M_{1}) then, the reflected light from this face falls on a distant concave mirror (M_{2}).

With the help of plane mirror (M_{3})placed at a center of curvature of concave mirror (M_{2}), the light is returned back and falls on the face of mirror (M_{1}) again at angle of 45^{0} . The light reflected from this face is collected by telescope and observed by eye. If light returning from the mirror (M_{2}) will not in general incident at an angle 45^{0}light is not observed by telescope.

The rotation of mirror (M_{1}) is so adjusted that, the face 1 of mirror occupies exactly the same position as was occupied by face 2 during the time light travels from (M_{1}) to (M_{2}) and returning back to (M_{1}). Then image of source will be reappearing.

If‘d’ be the distance between mirror (M_{1}) and (M_{2}) and ‘c’ be the velocity of light, then time taken by light to travel from (M_{1}) to (M_{2}) and returning back to (M_{1}) is

given by, t=2d/c

if ‘m’ be the no. of faces of polygonal mirror and ‘n’ be the no. of revolution per second. Then angle rotated by mirror during the time t i.e. ɵ=2π/m

and time taken by mirror to rotate by angle θ i.e.

t=θ2πnθ2πn

2dc=θ2πn2dc=θ2πn

c= 4πndθ4πndθ

Since, ɵ=2π/m

Then, ɵ=2dnm.

The advantages of Michelson’s method are:

The distance between the two stations is very large.

Images obtained are very bright so that position can be determined accurately.

There is no measurement of the displacement of image.

The disadvantages are:

It is very difficult to maintain the high speed rotation of mirror.

High speed of the rotation of the mirror can break the mirror