Acoustic phenomena and Speed of light

The physical phenomena associated with the production, transmission of sound and its effects is called acoustic phenomena.

Pressure amplitude

It is maximum fluctuation in the pressure at a point in a medium when a longitudinal wave travels through the medium.

 Pressure at certain point is given by

P(x,t)  = Pmaxcos(wt – kx)

Where, P­max = Kka =Kawv$\frac{\mathrm{K}\mathrm{a}\mathrm{w}}{\mathrm{v}}$ is called pressure amplitude.

Musical Sound and Noise              

The sound which gives pleasing effect to our ears is called musical sound. Music is pleasing to ear. It is due to the periodic and regular vibration. Wave form does not show sudden change in wavelength and amplitude. Frequency is generally high. It is for long duration.

Characteristics of Musical sound

Pitch: The pitch of the sound depends upon the frequency of vibration of air and relative motion between the sounding body and the listener. The sound with low frequency is identified as low pitch and high frequency is identified as high pitch. The high pitch sound is called shrill and low pitch sound is called grave of flat.

Loudness or intensity: The intensity I of the sound is defined as the amount of energy crossing per unit area per second and the amount of intensity determines the loudness of the sound. Factors determining the loudness of the sound:

Loudness of the sound depends upon:

 Amplitude of sounding body: Greater the amplitude larger the intensity of sound.

Surface area of the vibrating source: Greater the surface area loudness of sound is also larger. eg, sound from the large drum is of larger intensity.

Presence of other bodies: If there is sound reflecting object near to the source the loudness of the sound increases. That why we hear much louder sound inside the room than outside.

Frequency of sound: The loudness of the sound of vibrating body is directly proportional to the square of the frequency of the vibrating body.

Distance between the source and listener: The loudness of the sound decreases as distance increases

Density of the medium: The loudness of the sound is directly proportional to the density of medium through which it passes.

Quality or timber: It is the factor that determines the characteristics of the sound having same pitch and loudness. It depends on the number and intensity of the harmonics.

Threshold of hearing

It is the lowest intensity of sound that can be heard by our ears is called threshold of hearing.

Intensity and intensity level

Intensity of the sound is defined as the rate of transfer of sound through one square meter area held normally at the place where the sound travels.

Let I be the intensity of the sound, E be the energy and A be the area and t be the time then the intensity is given by

Intensity (I)= energypassingarea∗time=EA∗t$\frac{\mathrm{e}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{g}\mathrm{y}\mathrm{p}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}}{\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\ast \mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}}=\frac{\mathrm{E}}{\mathrm{A}\ast \mathrm{t}}$

Intensity level: it is define as the logarithm of the ratio of intensity of the sound to an arbitrarily chosen intensity i.e. L= log10 (I/I0)

Where I be the intensity of the wave and I0 be the arbitrary intensity taken as 10-12Wm-2 and the minimum intensity of the sound that can be heard by our ear is called threshold hearing. It is denoted by I0 and the value is equal to 10-12Wm-2.

Relation between I and L

The loudness of sound is defined as the degree of sensation of sound produced in the ear. The loudness of sound depends on its intensity but the relationship is not linear. It states that - the magnitude of any sensation is proportional to the logarithm of the physical factor which produces it. This is called the Weber-Fechner law.. Thus, according to this law,

I ∝ log I or

L = K log I ----------- 1 Where L is the loudness, I is the intensity and K is a constant of proportionality. The minimum intensity of sound is denoted by I0 .whose value is chosen to be 10-12Wm-2. Let the loudness corresponding to this threshold of audibility be l0. Then equation (1) can be written as

L0 = K log I0 ---------- 2

Subtracting equation 2 equation 1 we get

L - L0= K log I - K log I0

= K (log I - log I0)

=K log (I/I0)

(L – L0) indicates as to how much the loudness of a given sound is above the minimum value for hearing. Therefore, it is called the intensity level L.

L=K log …………………………..3

If the value of the constant K is chosen to be 1, then L is measured in a unit called bel

if I = 

then, L =Log Or L = 1 bel ………….4

Hence, the intensity level of a sound is said to be 1 bel, if its intensity is ten times the threshold intensity.

If the value of the constant K is chosen to be 10, then L is measured in decibel.

Thus L(in db)=10 log(110)$\frac{(}{1}10)$……………..5

Bel and decibel both are the unit of loudness and the loudness of the sound is said to be one bel if its intensity (I) is 10 times the intensity of the sound at the threshold of hearing i.e. L=1 bel if I=10I0 .we have I=log10(10)=1bel. Bel is the biggest unit of loudness and the smallest unit of the loudness is decibel and it is defined as 1/10th of bel. i.e.1decibel=bel/10

Inverse square law

It states that “the intensity of the sound at a point is directly proportional to the square of the distance of the point from the source of the sound “.i.e. 

 Beats

Consider two waves from two different sources of nearly same frequency travelling in the same direction. At any point let’s say A the two waves arrives in phase and intensity becomes maximum at that point. The phase difference between the two waves increases until the compression from one of the sources and rarefaction from the other arrives at the same time. At this point B, the intensity becomes minimum and no sound is produced. Each rise and fall in the intensity constitutes one beat and the corresponding time interval is called beat frequency.

Graphical representation of beat

 Figure

When two sound waves of nearly the same frequency are travelling in the same medium in the same direction, the intensity of the resultant sound rises and falls regularly a number of times a second. These periodic variations in the intensity of sound due to superposition of two sound waves of slightly different frequencies are called beats and hence in this way they are formed. The number of beats heard per second is called beat frequency

Analytical method for the formation of beat

Suppose two waves of the frequency f1 and f2 and each of amplitude are traveling in a medium in the same direction. The equation of the waves are:

Y1=asin(ωt – k1x)…………..1

And from the superposition principle, the resultant displacement at that point is given by

= asin(ω1t – k1x)+ asin(ω2t – k2x)

When two point are superimpose at a point x=0 then we have

1. asin(ω1+ω22)tcos(ω1−ω22)t$\mathrm{a}\mathrm{s}\mathrm{i}\mathrm{n}\left(\frac{{\omega }_1+{\omega }_2}{2}\right)\mathrm{t}\cos \left(\frac{{\omega }_1-{\omega }_2}{2}\right)\mathrm{t}$

=Where ω1=2πf1 and ω2=2πf2

= 2acos2 π f1−f22$\frac{{\mathrm{f}}_1-{\mathrm{f}}_2}{2}$t. sin2 π f1−f22$\frac{{\mathrm{f}}_1-{\mathrm{f}}_2}{2}$t

Let A=2a cos2πf1−f22$\frac{{\mathrm{f}}_1-{\mathrm{f}}_2}{2}$tThen the above equation become

Condition for maximum.

If the resultant amplitude A become maximum when become maxium

i.e.cos2 π f1−f22$\frac{{\mathrm{f}}_1-{\mathrm{f}}_2}{2}$t=±1

or,cos2 π f1−f22$\frac{{\mathrm{f}}_1-{\mathrm{f}}_2}{2}$t=cosnπ where n=0, 1, 2, 3, ……..

or,2π f1−f22$\frac{{\mathrm{f}}_1-{\mathrm{f}}_2}{2}$t=nπ

or, t=nf1−f2$\frac{n}{{\mathrm{f}}_1-{\mathrm{f}}_2}$ cos2πf1−f2s$\frac{{\mathrm{f}}_1-{\mathrm{f}}_2}{s}$tor, putting the value of n we get

t=0,1f1−f2$\frac{1}{{\mathrm{f}}_1-{\mathrm{f}}_2}$, 2f1−f2$\frac{2}{{\mathrm{f}}_1-{\mathrm{f}}_2}$, 3f1−f2$\frac{3}{{\mathrm{f}}_1-{\mathrm{f}}_2}$, ………..

or, T = (1f1−f2)−0,T=(3f1−f2)−(2f1−f2)=(1f1−f2)$\left(\frac{1}{{\mathrm{f}}_1-{\mathrm{f}}_2}\right)-0,\mathrm{T}=\left(\frac{3}{{\mathrm{f}}_1-{\mathrm{f}}_2}\right)-\left(\frac{2}{{\mathrm{f}}_1-{\mathrm{f}}_2}\right)=\left(\frac{1}{{\mathrm{f}}_1-{\mathrm{f}}_2}\right)$

or, T = (1f1−f2)$\left(\frac{1}{{\mathrm{f}}_1-{\mathrm{f}}_2}\right)$

Condition of minima.

If the resultant amplitude A become minima whencos2π(f1−f22)$\mathrm{c}\mathrm{o}\mathrm{s}2\pi \left(\frac{{\mathrm{f}}_1-{\mathrm{f}}_2}{2}\right)$t is minima. i.e.

            cos2π(f1−f22)t=0$\mathrm{c}\mathrm{o}\mathrm{s}2\pi \left(\frac{{\mathrm{f}}_1-{\mathrm{f}}_2}{2}\right)\mathrm{t}=0$

Or,cos2π(f1−f22)t=cos(2n+1)π2$\mathrm{O}\mathrm{r},\mathrm{c}\mathrm{o}\mathrm{s}2\pi \left(\frac{{\mathrm{f}}_1-{\mathrm{f}}_2}{2}\right)\mathrm{t}=\frac{\cos \left(2\mathrm{n}+1\right)\pi }{2}$        where n=0, 1, 2, 3, …………

Or, t = 12(2n+1f1−f2)$\frac{1}{2\left(\frac{2\mathrm{n}+1}{{\mathrm{f}}_1-{\mathrm{f}}_2}\right)}$

Onputting the value of the n we get

            t=12(1f1−f2),12(3f1−f2),12(5f1−f2)$\mathrm{t}=\frac{1}{2\left(\frac{1}{{\mathrm{f}}_1-{\mathrm{f}}_2}\right)},\frac{1}{2\left(\frac{3}{{\mathrm{f}}_1-{\mathrm{f}}_2}\right)},\frac{1}{2\left(\frac{5}{{\mathrm{f}}_1-{\mathrm{f}}_2}\right)}$

or, T = 12(3f1−f2)−12(1f1−f2)=(1f1−f2)$\frac{1}{2\left(\frac{3}{{\mathrm{f}}_1-{\mathrm{f}}_2}\right)}-\frac{1}{2\left(\frac{1}{{\mathrm{f}}_1-{\mathrm{f}}_2}\right)}=\left(\frac{1}{{\mathrm{f}}_1-{\mathrm{f}}_2}\right)$

Therefore for both maxima and minima we have T= 1f1−f2$\frac{1}{{\mathrm{f}}_1-{\mathrm{f}}_2}$. Therefore the beat frequencyis equal to the frequency to the difference of the frequencies of sound waves

Doppler effects

The apparent change in frequency of the sound heard by observer is due to the relative motion of the source or observer or both are called Doppler effect of sound.

Apparent frequency is calculated by f’  = V+VrV+Vs$\frac{\mathrm{V}+\mathrm{V}\mathrm{r}}{\mathrm{V}+{\mathrm{V}}_{\mathrm{s}}}$ * fs

Relative motion of the receiver

If a source is stationary, it will emit sound waves that propagate out from the source as shown below.

As the receiver moves towards the source, it will detect the sound coming from the source but each successive sound wave will be detected earlier than it would have if the receiver were stationary, due to the motion of the receiver .Thus the frequency that each successive wave front would be detectedwould be changed by this relative motion where:

Δf=Vrλo$\mathrm{\Delta }f=\frac{Vr}{\lambda o}$

λo is the original wavelength of the source

Δf$\mathrm{\Delta }\mathrm{f}$ is the change in observed frequency

Vr is the velocity of the receiver

Since the original frequency of the source can be expressed in terms of the wavelength where

fo= vλo$\frac{v}{\lambda o}$, the observed frequency becomes:

f’=fo +Δf$\mathrm{\Delta }\mathrm{f}$

f’=fo(v+Vr)v$\frac{fo\left(v+Vr\right)}{v}$ this equation works when Vr is moving towards the source.

Note that this equation only works if the relative velocity of the receiver, Vr is towards the source. If the motion is away from the source, the relative velocity would be in the opposite direction and the equation would become:

f’=fo(v−Vr)v$\frac{fo\left(v-Vr\right)}{v}$

Relative motion of the source

If the source is moving towards the receiver, the spacing between the successive wave fronts would be less as seen in the figure below. This would be expressed as:

Δ$\mathrm{\Delta }$λ=Vs/fo

‘Vs’ is the relative velocity of the source.

Observed frequency f’=vλ+λo$\frac{v}{\lambda +\lambda o}$

Or, f’=fovv−Vs$\frac{fov}{v-Vs}$

This is only when the source is moving towards the receiver. If the source is moving away it becomes f’=fovv+Vs$\frac{fov}{v+Vs}$

Limitations of Doppler effect in sound

The Doppler Effect is not applicable in following conditions:

a. If the velocity of sound of the source is greater than that of the sound because   the wave gets distorted due to which no change in frequency will be observed.

b. If the velocity of the sound of the observer is greater than that of the sound.

 Applications of Doppler effects

Infrasonic and ultrasonic sound

The sound that is produced below the audible frequency is called infrasonic .It range is below 20KHz. It is produce by large object eg.earthquake waves.

The sound wave of frequency above the audible (i.e. above 20 KHz) are called ultrasonic. It is produce by small objects. Eg. quartz crystal. Ultrasonic are the longitudinal mechanical waves of frequency greater than 20 KHz and the wavelength very short. Supersonics are those objects which travel faster than sound i.e. faster than 332 m/s. Jet planes or waves are supersonics.

 Noise pollution

An unwanted sound which causes unpleasant effect in our ears is called noise pollution.

Causes of sound pollution

Transportation systems – including cars, trains and airplanes – are one of the most common sources of noise pollution since they can be particularly loud and unrelenting in certain areas. In general, people who live in urban centers are more likely to be exposed to noise pollution due to population density and the increased presence of the transportation systems described above. Living in an urban city may also mean that people are more exposed to the noises of construction, which is another major source of noise pollution.

While those in larger cities are more likely to experience noise pollution, those in rural settings may also experience this problem as well. A key example of rural noise pollution is farming, which may include a variety of machines that produce harsh or loud noises.

Effects of noise pollution

It may cause deafness.

People may suffer from headaches and migraines.

It decreases the working efficiency.

Controlling noise pollution

Rooms and wall should be covered with sound absorbing materials.

Loud speaker, radio and other music system should be played at low volume.

Industries that produces nose should be build far from the residential areas.

Speed of light

Wave optics is the branch of opticswhich 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 109 Hz. Radio waves emitted by radio stations. Radio waves are also emitted by stars and gases in space.

Microwaves: Frequency of these waves ranges from109 Hz to 3.0 *1011Hz.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 *1011Hz to 109 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 *1014 Hz to7.5 * 1014 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 * 1014 Hz to 5.0 *1015 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 *1015Hz to3.0* 1018 Hz.The wavelength of X-ray is 60nm to 10-8nm. Hot gases in the Universe also emit X-rays.

Gamma ray: Frequency of these waves rangesfrom 3.0*1018 Hz to3.0 * 1022 Hz. The wavelength of Gamma rays is 0.1nm to 10-5nm. 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 I1, I2, I3 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. Sinisinr$\frac{\mathrm{S}\mathrm{i}\mathrm{n}\mathrm{i}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{r}}$=μ$\mu$ 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=BCAC$\frac{\mathrm{B}\mathrm{C}}{\mathrm{A}\mathrm{C}}$=ctAC$\frac{\mathrm{c}\mathrm{t}}{\mathrm{A}\mathrm{C}}$…………..1

Sinr = ADCD$\frac{\mathrm{A}\mathrm{D}}{\mathrm{C}\mathrm{D}}$ =vtAC$\frac{\mathrm{v}\mathrm{t}}{\mathrm{A}\mathrm{C}}$……….2

 From 1 and 2 sinisinr$\frac{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{i}}{\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{r}}$ = cv$\frac{\mathrm{c}}{\mathrm{v}}$ =μ$\mu$ 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 450to 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+d$\frac{\mathrm{b}}{\mathrm{a}+\mathrm{d}}$

or, II’=(a+d)xb$\frac{(\mathrm{a}+\mathrm{d})\mathrm{x}}{\mathrm{b}}$…………………………6

From 4 and6 we have

ɵ= (a+d)x2bd$\frac{(\mathrm{a}+\mathrm{d})\mathrm{x}}{2\mathrm{b}\mathrm{d}}$………………7

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

C = 8πnbd2(a+d)x$\frac{8\pi \mathrm{n}\mathrm{b}{\mathrm{d}}^2}{(\mathrm{a}+\mathrm{d})\mathrm{x}}$

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 (M1), concave mirror (M2) & plane mirror (M3). Light from the source (S) incident at an angle of 450 on one face of octagonal mirror (M1) then, the reflected light from this face falls on a distant concave mirror (M2).

With the help of plane mirror (M3)placed at a center of curvature of concave mirror (M2), the light is returned back and falls on the face of mirror (M1) again at angle of 450 . The light reflected from this face is collected by telescope and observed by eye. If light returning from the mirror (M2) will not in general incident at an angle 450light is not observed by telescope.

The rotation of mirror (M1) 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 (M1) to (M2) and returning back to (M1). Then image of source will be reappearing.

If‘d’ be the distance between mirror (M1) and (M2) and ‘c’ be the velocity of light, then time taken by light to travel from (M1) to (M2) and returning back to (M1) 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=θ2πn$\frac{\theta }{2\pi \mathrm{n}}$

2dc=θ2πn$\frac{2d}{c}=\frac{\theta }{2\pi \mathrm{n}}$

c= 4πndθ$\frac{4\pi \mathrm{n}\mathrm{d}}{\theta }$

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.