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Second Year Physics study material 2015 Wave Optics 1. Describe Various theories of light? A) Descartes or Newton corpuscular theory : 1. According to this theory light is made up of stream of weightless particles called as corpuscles. 2. Different colors’ of light are due to different size of corpuscles of light. 3. By means of corpuscle theory Newton explained Reflection and Refraction of light. Reflection is due to force of repulsion between corpuscles and reflecting surface. 4. According to Newton , light travels faster in the denser medium compared to rarer medium. But this theory was proved wrong by Focault that speed of light in rarer medium is greater than denser medium. Other properties of light like Interference, Diffraction , Polarization can not be explained based on corpuscular theory of light. Huygens Wave theory of light : 1. According to Huygens wave theory , light is travelled in the form of wave in a medium called luminiferous ether. 2. According to Huygen light is travelling in the form of longitudinal wave like sound waves. But later it was proved that light is a transverse wave. ா

3. The speed of light is very high. so according to equation c=ටఘ. For such materials density is less, Elasticity is more. Such type of materials will never exist. more importantly that type of matter is never detected. 4. This theory failed to explain rectilinear propagation of light. Later this concept of Ether is rejected. Maxwell Electromagnetic theory:

So we can not consider this theory as final theory.

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Electromagnetic waves require no material medium to propagate. But this theory cannot explain photoelectric effect and Compton effect.

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According to Maxwell light is an electromagnetic wave which consists of mutually perpendicular electric and magnetic fields which are also perpendicular to the direction of propagation of light.

Second Year Physics study material 2015 Planck’s quantum theory : According to this theory light is propagated in the form of packet of energy called quanta. The quanta of energy of light is called photon. The energy of photon is given by E = hν. where ν is frequency of radiation. Photoelectric effect and Compton effect was successfully explained by this theory of light. It failed to explain interference , diffraction and polarization. Debroglie dual nature of Matter : In some phenomenon light behaving as wave , in some phenomenon light behaving as a particle. Two properties are not created in the same experiment. so light has dual nature. In the same way , Debroglie showed that matter also behaves as a wave. 2. Define Wave front? Explain various types of Wave front? A) Wave front is defined as locus of all points in the medium vibrating in the same phase at a given instant. The shape of wave front depends on the source of disturbance. 1. Spherical wave front (if the source of disturbance is a point source) 2. Cylindrical wave front (if the source of disturbance is a slit) 3. Plane wave front (When spherical wave front is at infinity, part of spherical wave front appears as plane wave front) Note : Phase difference between two points situated on the same wave front is zero. 3. State and Explain Huygens principle? A) According to Huygens principle 1. Each point on a wave front is the source of new disturbance called secondary source that emits secondary wavelets. 2. The disturbance travels with the speed of light in that medium

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In figure (a), if we know the shape of wavefront at time t =0, Huygens principle allows us to determine shape of wavefront after time t = τ . From the old wave front now draw a sphere of radius ‘c τ’ from each point on the spherical wavefront. where C represent speed of waves in medium. If we now draw common tangents to all these secondary wavelets , we obtain a wavefront at new position after time t = τ. The new wave front

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3. The envelop of all the secondary wavelets draws in the direction in which the wave travels is the position of new wave front.

Second Year Physics study material 2015 is again spherical with point source. Similarly we can construct wavefront for planar wavefront. 4. Explain the phenomenon of Refraction using Huygens wave front? A) By using Huygens principle we can able to derive the laws of refraction. Let PP’ represents surface separating medium 1 and medium 2. Let v1 and v2 represents speed of light in medium 1 and medium 2. Let us assume the plane wavefront AB propagating as shown in figure. Let ‘i’ be angle of incidence by this plane wavefront at the interface. Let τ be the time taken by wavefront to travel distance ‘BC’ , then BC = v1τ. To determine the shape of refracted wavefront, draw a sphere of radius v2τ from the point ‘A’ in the second medium. Let CE represents a tangent plane drawn from a point C on to the sphere. Then AE = v2τ and CE would represents refracted plane wavefront. At point ‘A’ draw a normal. In the triangle ABC , AB is perpendicular to incident ray AA’ and AC is perpendicular to AN. In triangle AEC , ∟A + ∟E + ∟c = 180 (90-r) +90 + ∟c = 180 ,

∟c = ∟r

If we now consider the triangles ABC and AEC , we obtain sin i = BC/AC = v1τ/AC and sin r = AE/AC = v2τ/AC where i and r are angle of incidence and angle of refraction respectively. Then sin i/sin r = v1/v2 If ‘C’ represents the speed of light in vaccum, then n1 = c/v1 and n2 = c/v2 are known as refractive indices of medium 1 and medium 2 respectively n2/n1 = v1/v2 , Hence sin i/sin r = n2/n1 therefore

n1sin i = n2 sin r

This is known as snells law of refraction.

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A) Consider plane wave A’B’ incident at an angle ‘i’ on reflecting surface PQ. If ‘v’ represents the speed of wave in the medium and if ‘τ’ represents the time taken by the wavefront to advance from the point B to C then the distance BC = vτ. In order to construct a reflected wave front we draw a sphere of radius vτ from the point ‘A’ as

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5. Explain the phenomenon of Reflection using Huygens wave front?

Second Year Physics study material 2015 shown. Let CE represent the tangent plane drawn from point ‘C’ to this sphere. Obviously AE = BC = vτ and CE represents reflected plane wavefront. In triangle ABC , AB is perpendicular to A’A and AC is perpendicular to AN. In ∆AEC , ∟A + ∟E+∟C = 180 (90 –r) + 90 + ∟C = 180 , ∟C =r From triangle ABC , sin i = BC/AC and from triangle AEC , sin r = AE/AC = BC/AC (as AE = BC) since sin i = sin r , therefore ∟i = ∟r This is known as law of reflection. 6) Explain briefly Refraction and Reflection of Plane wavefronts? A) When a plane wavefront passes through the thin prism , Speed of light waves is less in glass , the lower portion of the incoming wavefront will get delayed resulting in tilt of emerging wave front When a plane wave front passes through the thin convex lens , the central part of incident plane wave traverses the thickest portion of lenses and is delayed the most. The emerging wavefront has a depression at the center. Therefore wavefront becomes spherical and converges to the point F which is known as focus. When a plane wave incident on concave mirror and on reflection we have a spherical wave converging to the focal point .

7. Write a short note on Doppler effect in light? A) When ever there is a relative motion between the source of light and the observer, the apparent frequency of light observed is different from the actual frequency of light emitted by the source of light. This effect is known as Doppler effect.

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Now if a source of light moves away from stationary observer , the time taken by the wavefronts to reach the observer from the source will increase. Hence, the frequency of the light received by the observer will be less than the actual frequency of light. How ever , if the source of light moves towards the stationary observer, the frequency of light received by observer will be more than the actual frequency of light.

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Suppose a source of light is stationary with respect to an observer. The frequency of light received by the observer is given by ν =c/λ , where c = 3 x 108 m/s.

Second Year Physics study material 2015 Frequecy of light received when source and observer are in relative motion is known as apparent frequency. When speed of light is very less, we use the same formula used for sound waves. The fractional change in frequency ∆ν/ν = -Vradial/C where Vradial is the component of source velocity along the line joining the observer to the source relative to the observer. Vradial is considered positive when the source moves away from the observer. Applications : 1. It is used to measure the speed of stars and galaxies 2. Doppler effect is used in RADAR and SONAR 8. Define Superposition principle? A) According to this principle “ The resultant displacement of the particle at any instant is the vector sum of individual displacements caused to the particle by the two or more waves” let y1 and y2 be the displacements of particle respectively at any instant caused by the two waves. Then the resultant displacement of particle at that instant is given by y = y1 +- y2 The same principle can be applied for two or more waves. They are two types of superposition 1. Constructive superposition : When two waves of same wavelength are superimposed with each other in phase, then the superposition is constructive. y = y1 + y 2 2. Destructive superposition : When two waves of same wavelength are superimposed with each other out of phase, then the superposition is destructive. y = y1 - y 2 9. Define Coherent addition of source of light waves?

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coherent sources are generally obtained from the single parent source.

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A) Two sources of light are said to be coherent if they emit waves of same frequency (or wavelength) and are either in phase or have a constant initial phase difference.

Second Year Physics study material 2015 Slits A and B can be coherent sources because the light waves passing through them are derived from single wavefront that illuminates the slits so they have same frequency and may have constant initial phase difference or no phase difference. The phenomenon of coherence can be explained by means of two vibrating needles. These needles are allowed to oscillate in phase. These needles generate two waves , each of amplitude A at any instant. Phase difference between their displacement does not change with time. consider a point P such that S1P = S2P . At this point , resultant displacement is the sum of individual displacements y1 and y2 of the two waves respectively. y = y1 + y2 But y1 = A sinωt and y2 = A sinωt Therefore y = A sinωt + A sinωt = 2A sinωt Since the intensity is proportional to square of amplitude , the resultant intensity will be given by I = 4I0 Where I0 is intensity produced by individual sources. Consider another point P’ , where S1P’ ≠ S2P’ S2P’ - S1P’ = λ , where λ is wavelength of wave Hence y1 = A sinωt λ is 2π radian)

and y2 = A sin(ωt + 2π) (since phase difference corresponding to

On substituting these values in equation y = y1 + y2 , we get y =2A sinωt . Amplitude of resultant wave is double the amplitude of single wave. Hence Intensity in increased and superposition is constructive. Now consider one more point P’’ , such that S2P’’ - S1P’’ = λ/2 y1 = A sinωt , y2 = A sin(ωt + π) = - A sinωt

Condition for constructive superposition (brightness) can be generalized as

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y =0. , giving zero

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On substituting these values in equation y = y1 + y2 , we get amplitude or destructive superposition.

Second Year Physics study material 2015 S2P’’ - S1P’’ = n λ , where n = 0,1,2…. Condition for destructive superposition (darkness) can be generalized as S2P’’ - S1P’’ = (n + ½) λ , where n = 0,1,2…..

Note : Sources having different frequencies or the sources having same frequency but no stable phase difference are known as incoherent sources. 10. Define Interference of light and Give some examples of Interferenceof light ? A) The phenomenon of redistribution of light energy due to the superposition of light waves from two coherent sources is known as interference of light. When two sources of light having same frequencies and constant phase difference then the redistribution of light energy is not uniform. At certain points the Intensity of light energy is maximum and at certain points Intensity of light energy is minimum. Thus the light energy is redistributed and this phenomenon is called Interference of light. Examples of Interference : 1. When oil spills on water we observe coloured streaks. These are formed due to interference of light reflected from the oil film on water. 2. Soap bubbles appear coloured in sunlight due to interference of light reflected from the soap bubble. 11. What is the relation between phase difference and Path difference of light? A) Phase difference = (2π/wavelength) X Path difference 12. Explain Briefly the Theory of Youngs Double slit Experiment? A) The phenomenon of interference in light was demonstrated by Thomos young. Dark and bright bands in the interference pattern is called interference fringes.

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Light waves emitted from the slits s1 and s2 reach point O on the screen after travelling equal distances. So path difference and hence phase difference between these waves is zero. Therefore they meet at O in phase and hence constructive interference takes place. Thus O is the position of central bright fringe.

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Consider two coherent sources s1 and s2 separated by a small distance d. Let D be the distance between screen and the plane of slits s1 and s2.

Second Year Physics study material 2015 Let the waves emitted by s1 and s2 meet at point P on the screen at a distance y from the central bright fringe. The path difference between these waves at P is given by ∆ x = s2P – s1P

------------------ (1)

From right angled triangle S2BP, S2P = [S2B2 +PB2]1/2 = [D2 + (y +d/2)2]1/2 = D [1 + (y +d/2)2/D2]1/2 -------------------(2) Using Binomial theorem , and neglecting terms containing high powers S2P = D [1 + (y +d/2)2/2D2] = D + (y +d/2)2/2D Similarly from right angled triangle S1AP , S1P = D + (y -d/2)2/2D----------------- (3) substituting 2 and 3 equations in 1 , we get ∆ x =( D + (y +d/2)2/2D) – (D + (y -d/2)2/2D) = 2yd/2D ∆x

= yd/D

Condition for bright fringes If the path difference is an integral multiple of λ , then bright fringe will be formed at P yd/D = mλ

or y = mλD/d

which is the position of mth bright fringe from central bright fringe. Fringe width (β) : The distance between any two successive bright fringes (or successive fringes) is called Fringe width. β = y2 –y1 = 2λD/d - λD/d = λD/d Condition for dark fringes : If the path difference is an odd multiple of λ/2 , then dark fringe will be formed at P yd/D = (m+1/2)λ , where m = 0,1,2…….

13. Explain briefly factors depending on fringe width?

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Fringe widths of bright and dark are equal.

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β = y2 –y1 = 3λD/2d - λD/2d = λD/d

Second Year Physics study material 2015 A) Fringe width directly proportional to wavelength of light. Larger will be the fringe width and vice versa. It means fringe width will be greater for red colour than violet colour. Fringe width directly proportional to D , Larger is the distance of the screen from the slits , greater will be the fringe width. Fringe width inversely proportional to d , smaller is the distance between the coherent sources , larger will be the fringe width. 14. List out some important observations regarding interference (general reference) A) If one of the slits is closed , Interference pattern will disappear. If instead of Monochromatic light , if we use white light we will get coloured fringes of various widths . But central spot is white. If two independent sources are used instead of coherent sources , Interference pattern is not observed. 15. What is Diffraction of light? A) The phenomenon of bending of light around the corners of an obstacle or an aperture into the region of geometrical shadow of obstacle is called diffraction of light. Diffraction of light is more pronounced when size of the obstacle/aperture should be the order of wavelength of wave. They are two types of diffraction 1. Fresnel diffraction ( Source or screen both are at finite distance from obstacle or aperture) 2. Fraunhoffer diffraction (Source or screen both are at infinite distance from obstacle or aperture)

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16. Explain theory of diffraction due to a single slit?

Second Year Physics study material 2015 A)

Light which is diverging from a monochromatic source S is made parallel after refraction through convex lens L1 . The refracted light from L1 is propagated in the form of wavefront WW’ . The plane wavefront WW’ in incident on the slit AB of width ‘d’ . According to Huygens principle, each point on the AB acts as a source of secondary disturbance or wavelets. Second lens L2 is used in converging the parallel beam. Now consider a point O on screen, which is placed at a distance D from the slit AB . Since point O is equidistance from A and B , therefore the secondary wavelets from A and B reach the point O in the same phase and the constructive interference takes place at O . In other words point O is the position of central maximum. Now Let the light is diffracted through an angle θ . So the secondary wavelets will also be diffracted through an angle θ . Let these wavelets meet the screen at point P. The point P will be of maximum or minimum intensity depending on the path difference between secondary wavelets (reaching point P) originating from corresponding points of wavefront. To find the path difference between secondary wavelets originating from the corresponding points of A and B of the same plane wavefront , draw AN perpendicular to BB’ . The path difference between these wavelets originating from A to B is BN. From ∆ BAN ,

BN/AB = sin θ

for minima :

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BN = dsinθ = dθ (if θ is small)

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or BN = AB sin θ

Second Year Physics study material 2015 If path difference is equal to one wavelength that is BN = dsinθ = λ , then divide the slit into two equal halves that is AC and CB . Now the path difference between the secondary wavelets originating from A and C is equal to λ/2. So these wavelets will meet at point P out of phase (phase difference π) and hence destructive interference will take place at P . similarly the path difference between the wavelets originating from C and B is λ/2 and hence these will also produce destructive interference at P. Thus Position P will be of minimum intensity. Hence for first minimum = dsinθ1 = λ sin θ1 = λ/d θ1 = λ/d ( if angle is very small) similarly , if BN = 2λ , then the slit AB can be imagined to be divided into four equal halves . Then the path difference between the secondary wavelets originating from the corresponding from of each half = (2λ/4) = λ/2 . Thus these wavelets produce destructive interference and point P is of minimum intensity. Thus for second minimum dsinθ2 = 2λ sinθ2 = 2λ/d θ2 = 2λ/d (for smaller angles) In general , for minima dsinθm = mλ , sinθm = mλ/d θm = mλ/d (for smaller angles) Where θm is the angle giving direction of the mth order minimum and m = 1,2,3---- an integer. For Maxima : If path difference BN = d sinθ is an odd multiple of λ/2 , then the constructive interference takes place at P , Hence point P is position of secondary maxima dsinθm = (m+1/2)λ θm = (m+1/2)λ/d

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Where θm is the angle giving direction of mth order secondary maximum and m = 1,2,3----

Second Year Physics study material 2015 Width of central maximum or principal maximum :

Central maximum occurs at O on screen, where angle θ is zero. that is path difference is zero so all points in the slit being in same phase lead to maximum brightness. Let f be the focal length of lens L2 (placed very close to slit) and the distance of first minimum on either side of the central maximum be x tan θ = x/f Since the Lens L2 is very close to the slit , so f = D , tan θ = x/D since θ is very small , so tan θ ≈ sin θ , sin θ = x/D

---------(1)

Also, for first minimum , d sinθ = λ or sinθ = λ/d -----------(2) from equations 1 and 2 x/D = λ/d

or x = λD/d

This is the distance of first minimum on either side from the center of central maximum. width of central maximum is given by 2x = 2λD/d 17) What are the factors effecting width of central maxima ? A) It is directly proportional to wavelength of light used. so it is small for violet colour and maximum for red colour.

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if width of slit (d) is small , width of central maximum is large and viceversa

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Width of central maximum is directly proportional to distance D between plane of slit and screen, so width will increase with increase in D

Second Year Physics study material 2015 18) What is the difference between Interference and diffraction of light? A) Interference It is the result of superposition of waves from two wavefronts originating from the same source Interference fringes are all of same width All Maxima are bright and all minima are perfectly black Over a large area on screen interference is observed

Diffraction It is due to superposition of wave from different points of the same wavefront Fringes are not of the same width Maxima are of decreasing intensity , minima are not perfectly dark Over a small region on screen diffraction is observed.

19) Define Resolving power of an optical instrument? A) The ability of an optical instrument to produce distinctly separate images of two objects located very closely to each other is called its Resolving power. Resolving power can also be defined as the reciprocal of limit of resolution. smaller is the limit of resolution , larger is its resolving power. 20) Explain briefly Resolving power of an Microscope? A) Let PQ be object of size d placed in front of objective lens AB of the microscope. The central maxima of the images of P and Q are P’ and Q’ respectively. As the size of object PQ decreased, d decreases , central maximum P’ moves closer to the central maximum Q’. According to Rayleigh criterion , two points P and Q are resolved if central maximum P’ coincides with the minimum N near to Q’. The corresponding d = dmin gives the resolving limit of microscope. Since N is the first minimum around Q’ so , (QA + AN) – (QB + BN) = 1.22λ or (QA – QB) + (AN – BN) = 1.22 λ -----------------(1) Also P’ coincides with N , so

using equation 2 in equation 1

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(AN – BN) = (PB – PA) --------------------- (2)

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PA + AN = PB + BN

Second Year Physics study material 2015 (QA – QB) + (PB – PA) = 1.22 λ ----------------- (3) if β is the semi angle subtended by the objective AB of the object PQ , then (QA – QB) + (PB – PA) =2 dmin sin β Hence equation 3 becomes 2 dmin sin β= 1.22 λ dmin = 1.22 λ /2 sin β Generally the high resolving microscope used in laboratories has liquid between the object and objective of microscope. If n is the refractive index of the liquid , then the wavelength in liquid becomes λ/n and the resolving limit of microscope is given by

dmin = 1.22 λ /2 nsin β where nsin β is called numerical aperture of the objective lens of the microscope. Resolving power of the microscope is defined as the reciprocal of the resolving limit of the microscope, therefore Resolving power of microscope = 2nsin β/1.22 λ 21) Explain briefly Resolving power of Astronomical telescope? A) The minimum angle subtended at the objective lens of the telescope by two distinct objects (Lying close to each other) whose images are just resolved is called limit of resolution of the telescope. The reciprocal of limit of resolution is called resolving power of telescope.

In ∆ABC , sin dθ = AC/AB

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The path difference of between BP2 and AP2 is given by , ∆x =AC

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Let dθ be minimum angle subtended at the objective lens L of the telescope by two distant objects O and O’. The images of these objects are formed at P1 and P2 respectively. According to Rayleigh criterion, the two images P1 and P2 will be just resolved if the central maximum of the image P1 of the object O coincides with the first minima of the image P2 of object O’ or vice -versa

Second Year Physics study material 2015 ∆x = AB sin dθ

if D (= AB) be the diameter of the objective lens L of the telescope then ∆x = D sin dθ The angular separation of the center of the central maximum and the first minima for a circular aperture ( of a lens) can be obtained ∆x = λ or

D sin dθ = λ

since dθ is very small angle , so sin dθ = dθ D(dθ) = λ dθ = λ/D But for a circular aperture , dθ = 1.22 λ/D Which is the limit of resolution of the telescope

it depends on diameter of the aperture of objective lens and wavelength of light used.

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= D/ 1.22 λ

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Resolving power of a telescope

Second Year Physics study material 2015 22) Write a short note on Validity of Ray optics ? A) Fresnal distance is defined as the distance of the screen from the slit or aperture when the spreading of light due to diffraction from the center of screen is equal to size (width) of the slit . It is represented by ZF . ZF = d2/λ Fresnal distance depends on Square of size of slit and wavelength of light 23) What is polarization of light ? Explain briefly experimental polarization of light ? A) The phenomenon of restricting the vibrations of light vector or electric field vector in a particular direction in a plane perpendicular to direction of propogation of light is called polarization of light. Tourmaline crystal is used to polarize the light and hence it is called Polarizer. Ordinary light is represented as

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Polarized light can be represented in two ways . One way where vibrations parallel to plane of paper and another way where vibrations perpendicular to paper.

Second Year Physics study material 2015 Consider two polaroids , Polarizer P and Analyzer A . Let both polaroids P and A are placed in such a way that their pass axes or polarizing directions are parallel to each other. When unpolarized light incident on Polaroid P , it is found tha intensity of light transmitted by the Polaroid is reduced to Half tohe original value of intensity of incident light. When ordinary light is incident on Polaroid P , electric vector vibrating parallel to the pass axis of Polaroid P passes through it , while the electric vector vibrating perpendicular to the pass axis of Polaroid is blocked. so the light transmitted through the Polaroid P has only electric field vector parallel to the axis. Now place the two polaroids P and A such that their axes are perpendicular to each other. when ordinary light is incident on the Polaroid P , it is found that the intensity of light transmtted thorugh the Polaroid is zero. 24) State and Explain Malus law ? A) This law states that “ Intensity of the polarized light transmitted through the analyzer varies as square of the cosine of the angle between the plane of transmission of the analyzer and the plane of the polarizer. Let E be the amplitude of the light transmitted through the analyzer and θ be the angle between the planes of polarizer and analyzer. Resolve E into two components 1) E cosθ along OP (parallel to plane of transmission of analyzer) 2. E sinθ along OV (perpendicular to plane of transmission of analyzer) only E cosθ component is transmitted through the analyzer. we know Intensity α (Amplitude)2 Intensity of transmitted light through the analyzer is given by I α (E cosθ)2 I = kE2 cos2θ

A) A Plane polarized light can be created by means of two methods

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25) State and Explain methods for creating Plane polarization light?

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I = Io cos2θ which is known as Malus law

Second Year Physics study material 2015 1. Polarization of light by scattering 2. Polarization of light by reflection Polarization of light by Scattering : Let a beam of unpolarized light be incident on a small dust particles or air molecules (scatterer) along X axis. Since light is transverse in nature , therefore all the possible directions of vibrations of electric vector E in the unpolarized light lie in Y-Z plane . Hence the electrons associated with the scatterer will be set with vibrations along Y and Z axis. A vibrating electrons behaves as an oscillating dipole and emits radiations in all directions expects its own line of vibration. when an observer observe the scatterer along Y axis then the light reaching the observer comes from the electrons vibrating along Z axis. This light is linearly polarized. Polarization of light by reflection : When ordinary light or unpolarized light falls obliquely on a glass slab , it is partially reflected and partly refracted. Both the reflected and refracted beams of light are partly polarized. The incident ordinary light has two components , one component of incident light lies in the plane parallel to the surface of the glass slab and this component is reflected. The other component lies in the plane perpendicular to the glass surface is refracted. Both the reflected and refracted components of light are partly polarized. At a particular angle of incidence , the reflected light is completely plane polarized. The angle of incidence at which the reflected light is completely plane polarized is known as polarization angle . This is also known as Brewster angle. when angle of incidence ordinary light is equal to polarization angle , then reflected and refracted beams of light are perpendicular to each other.

26) What is Brewster law ? A) The refractive index of refractive medium (n) is numerically equal to the tangle of angle of polarization.

from figure

∟BOY + ∟COY = 90

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Experimentally reflected component and refracted component of light are mutually perpendicular then ∟i = ∟iB

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n = tan iB

Second Year Physics study material 2015 (90 –r) + (90-r’) = 90 (90-IB) + (90-r’) = 90 r’ = 90- IB using n = sin i/sin r’ = sin iB/sin (90 –θP) = sin iB/cos θP n = tan iB

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Brewster law is used to calculate polarizing angle for any two media in contact when their refractive indices are already known.

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3. The speed of light is very high. so according to equation c=ටఘ. For such materials density is less, Elasticity is more. Such type of materials will never exist. more importantly that type of matter is never detected. 4. This theory failed to explain rectilinear propagation of light. Later this concept of Ether is rejected. Maxwell Electromagnetic theory:

So we can not consider this theory as final theory.

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Electromagnetic waves require no material medium to propagate. But this theory cannot explain photoelectric effect and Compton effect.

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According to Maxwell light is an electromagnetic wave which consists of mutually perpendicular electric and magnetic fields which are also perpendicular to the direction of propagation of light.

Second Year Physics study material 2015 Planck’s quantum theory : According to this theory light is propagated in the form of packet of energy called quanta. The quanta of energy of light is called photon. The energy of photon is given by E = hν. where ν is frequency of radiation. Photoelectric effect and Compton effect was successfully explained by this theory of light. It failed to explain interference , diffraction and polarization. Debroglie dual nature of Matter : In some phenomenon light behaving as wave , in some phenomenon light behaving as a particle. Two properties are not created in the same experiment. so light has dual nature. In the same way , Debroglie showed that matter also behaves as a wave. 2. Define Wave front? Explain various types of Wave front? A) Wave front is defined as locus of all points in the medium vibrating in the same phase at a given instant. The shape of wave front depends on the source of disturbance. 1. Spherical wave front (if the source of disturbance is a point source) 2. Cylindrical wave front (if the source of disturbance is a slit) 3. Plane wave front (When spherical wave front is at infinity, part of spherical wave front appears as plane wave front) Note : Phase difference between two points situated on the same wave front is zero. 3. State and Explain Huygens principle? A) According to Huygens principle 1. Each point on a wave front is the source of new disturbance called secondary source that emits secondary wavelets. 2. The disturbance travels with the speed of light in that medium

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In figure (a), if we know the shape of wavefront at time t =0, Huygens principle allows us to determine shape of wavefront after time t = τ . From the old wave front now draw a sphere of radius ‘c τ’ from each point on the spherical wavefront. where C represent speed of waves in medium. If we now draw common tangents to all these secondary wavelets , we obtain a wavefront at new position after time t = τ. The new wave front

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3. The envelop of all the secondary wavelets draws in the direction in which the wave travels is the position of new wave front.

Second Year Physics study material 2015 is again spherical with point source. Similarly we can construct wavefront for planar wavefront. 4. Explain the phenomenon of Refraction using Huygens wave front? A) By using Huygens principle we can able to derive the laws of refraction. Let PP’ represents surface separating medium 1 and medium 2. Let v1 and v2 represents speed of light in medium 1 and medium 2. Let us assume the plane wavefront AB propagating as shown in figure. Let ‘i’ be angle of incidence by this plane wavefront at the interface. Let τ be the time taken by wavefront to travel distance ‘BC’ , then BC = v1τ. To determine the shape of refracted wavefront, draw a sphere of radius v2τ from the point ‘A’ in the second medium. Let CE represents a tangent plane drawn from a point C on to the sphere. Then AE = v2τ and CE would represents refracted plane wavefront. At point ‘A’ draw a normal. In the triangle ABC , AB is perpendicular to incident ray AA’ and AC is perpendicular to AN. In triangle AEC , ∟A + ∟E + ∟c = 180 (90-r) +90 + ∟c = 180 ,

∟c = ∟r

If we now consider the triangles ABC and AEC , we obtain sin i = BC/AC = v1τ/AC and sin r = AE/AC = v2τ/AC where i and r are angle of incidence and angle of refraction respectively. Then sin i/sin r = v1/v2 If ‘C’ represents the speed of light in vaccum, then n1 = c/v1 and n2 = c/v2 are known as refractive indices of medium 1 and medium 2 respectively n2/n1 = v1/v2 , Hence sin i/sin r = n2/n1 therefore

n1sin i = n2 sin r

This is known as snells law of refraction.

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A) Consider plane wave A’B’ incident at an angle ‘i’ on reflecting surface PQ. If ‘v’ represents the speed of wave in the medium and if ‘τ’ represents the time taken by the wavefront to advance from the point B to C then the distance BC = vτ. In order to construct a reflected wave front we draw a sphere of radius vτ from the point ‘A’ as

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5. Explain the phenomenon of Reflection using Huygens wave front?

Second Year Physics study material 2015 shown. Let CE represent the tangent plane drawn from point ‘C’ to this sphere. Obviously AE = BC = vτ and CE represents reflected plane wavefront. In triangle ABC , AB is perpendicular to A’A and AC is perpendicular to AN. In ∆AEC , ∟A + ∟E+∟C = 180 (90 –r) + 90 + ∟C = 180 , ∟C =r From triangle ABC , sin i = BC/AC and from triangle AEC , sin r = AE/AC = BC/AC (as AE = BC) since sin i = sin r , therefore ∟i = ∟r This is known as law of reflection. 6) Explain briefly Refraction and Reflection of Plane wavefronts? A) When a plane wavefront passes through the thin prism , Speed of light waves is less in glass , the lower portion of the incoming wavefront will get delayed resulting in tilt of emerging wave front When a plane wave front passes through the thin convex lens , the central part of incident plane wave traverses the thickest portion of lenses and is delayed the most. The emerging wavefront has a depression at the center. Therefore wavefront becomes spherical and converges to the point F which is known as focus. When a plane wave incident on concave mirror and on reflection we have a spherical wave converging to the focal point .

7. Write a short note on Doppler effect in light? A) When ever there is a relative motion between the source of light and the observer, the apparent frequency of light observed is different from the actual frequency of light emitted by the source of light. This effect is known as Doppler effect.

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Now if a source of light moves away from stationary observer , the time taken by the wavefronts to reach the observer from the source will increase. Hence, the frequency of the light received by the observer will be less than the actual frequency of light. How ever , if the source of light moves towards the stationary observer, the frequency of light received by observer will be more than the actual frequency of light.

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Suppose a source of light is stationary with respect to an observer. The frequency of light received by the observer is given by ν =c/λ , where c = 3 x 108 m/s.

Second Year Physics study material 2015 Frequecy of light received when source and observer are in relative motion is known as apparent frequency. When speed of light is very less, we use the same formula used for sound waves. The fractional change in frequency ∆ν/ν = -Vradial/C where Vradial is the component of source velocity along the line joining the observer to the source relative to the observer. Vradial is considered positive when the source moves away from the observer. Applications : 1. It is used to measure the speed of stars and galaxies 2. Doppler effect is used in RADAR and SONAR 8. Define Superposition principle? A) According to this principle “ The resultant displacement of the particle at any instant is the vector sum of individual displacements caused to the particle by the two or more waves” let y1 and y2 be the displacements of particle respectively at any instant caused by the two waves. Then the resultant displacement of particle at that instant is given by y = y1 +- y2 The same principle can be applied for two or more waves. They are two types of superposition 1. Constructive superposition : When two waves of same wavelength are superimposed with each other in phase, then the superposition is constructive. y = y1 + y 2 2. Destructive superposition : When two waves of same wavelength are superimposed with each other out of phase, then the superposition is destructive. y = y1 - y 2 9. Define Coherent addition of source of light waves?

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coherent sources are generally obtained from the single parent source.

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A) Two sources of light are said to be coherent if they emit waves of same frequency (or wavelength) and are either in phase or have a constant initial phase difference.

Second Year Physics study material 2015 Slits A and B can be coherent sources because the light waves passing through them are derived from single wavefront that illuminates the slits so they have same frequency and may have constant initial phase difference or no phase difference. The phenomenon of coherence can be explained by means of two vibrating needles. These needles are allowed to oscillate in phase. These needles generate two waves , each of amplitude A at any instant. Phase difference between their displacement does not change with time. consider a point P such that S1P = S2P . At this point , resultant displacement is the sum of individual displacements y1 and y2 of the two waves respectively. y = y1 + y2 But y1 = A sinωt and y2 = A sinωt Therefore y = A sinωt + A sinωt = 2A sinωt Since the intensity is proportional to square of amplitude , the resultant intensity will be given by I = 4I0 Where I0 is intensity produced by individual sources. Consider another point P’ , where S1P’ ≠ S2P’ S2P’ - S1P’ = λ , where λ is wavelength of wave Hence y1 = A sinωt λ is 2π radian)

and y2 = A sin(ωt + 2π) (since phase difference corresponding to

On substituting these values in equation y = y1 + y2 , we get y =2A sinωt . Amplitude of resultant wave is double the amplitude of single wave. Hence Intensity in increased and superposition is constructive. Now consider one more point P’’ , such that S2P’’ - S1P’’ = λ/2 y1 = A sinωt , y2 = A sin(ωt + π) = - A sinωt

Condition for constructive superposition (brightness) can be generalized as

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y =0. , giving zero

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On substituting these values in equation y = y1 + y2 , we get amplitude or destructive superposition.

Second Year Physics study material 2015 S2P’’ - S1P’’ = n λ , where n = 0,1,2…. Condition for destructive superposition (darkness) can be generalized as S2P’’ - S1P’’ = (n + ½) λ , where n = 0,1,2…..

Note : Sources having different frequencies or the sources having same frequency but no stable phase difference are known as incoherent sources. 10. Define Interference of light and Give some examples of Interferenceof light ? A) The phenomenon of redistribution of light energy due to the superposition of light waves from two coherent sources is known as interference of light. When two sources of light having same frequencies and constant phase difference then the redistribution of light energy is not uniform. At certain points the Intensity of light energy is maximum and at certain points Intensity of light energy is minimum. Thus the light energy is redistributed and this phenomenon is called Interference of light. Examples of Interference : 1. When oil spills on water we observe coloured streaks. These are formed due to interference of light reflected from the oil film on water. 2. Soap bubbles appear coloured in sunlight due to interference of light reflected from the soap bubble. 11. What is the relation between phase difference and Path difference of light? A) Phase difference = (2π/wavelength) X Path difference 12. Explain Briefly the Theory of Youngs Double slit Experiment? A) The phenomenon of interference in light was demonstrated by Thomos young. Dark and bright bands in the interference pattern is called interference fringes.

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Light waves emitted from the slits s1 and s2 reach point O on the screen after travelling equal distances. So path difference and hence phase difference between these waves is zero. Therefore they meet at O in phase and hence constructive interference takes place. Thus O is the position of central bright fringe.

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Consider two coherent sources s1 and s2 separated by a small distance d. Let D be the distance between screen and the plane of slits s1 and s2.

Second Year Physics study material 2015 Let the waves emitted by s1 and s2 meet at point P on the screen at a distance y from the central bright fringe. The path difference between these waves at P is given by ∆ x = s2P – s1P

------------------ (1)

From right angled triangle S2BP, S2P = [S2B2 +PB2]1/2 = [D2 + (y +d/2)2]1/2 = D [1 + (y +d/2)2/D2]1/2 -------------------(2) Using Binomial theorem , and neglecting terms containing high powers S2P = D [1 + (y +d/2)2/2D2] = D + (y +d/2)2/2D Similarly from right angled triangle S1AP , S1P = D + (y -d/2)2/2D----------------- (3) substituting 2 and 3 equations in 1 , we get ∆ x =( D + (y +d/2)2/2D) – (D + (y -d/2)2/2D) = 2yd/2D ∆x

= yd/D

Condition for bright fringes If the path difference is an integral multiple of λ , then bright fringe will be formed at P yd/D = mλ

or y = mλD/d

which is the position of mth bright fringe from central bright fringe. Fringe width (β) : The distance between any two successive bright fringes (or successive fringes) is called Fringe width. β = y2 –y1 = 2λD/d - λD/d = λD/d Condition for dark fringes : If the path difference is an odd multiple of λ/2 , then dark fringe will be formed at P yd/D = (m+1/2)λ , where m = 0,1,2…….

13. Explain briefly factors depending on fringe width?

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Fringe widths of bright and dark are equal.

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β = y2 –y1 = 3λD/2d - λD/2d = λD/d

Second Year Physics study material 2015 A) Fringe width directly proportional to wavelength of light. Larger will be the fringe width and vice versa. It means fringe width will be greater for red colour than violet colour. Fringe width directly proportional to D , Larger is the distance of the screen from the slits , greater will be the fringe width. Fringe width inversely proportional to d , smaller is the distance between the coherent sources , larger will be the fringe width. 14. List out some important observations regarding interference (general reference) A) If one of the slits is closed , Interference pattern will disappear. If instead of Monochromatic light , if we use white light we will get coloured fringes of various widths . But central spot is white. If two independent sources are used instead of coherent sources , Interference pattern is not observed. 15. What is Diffraction of light? A) The phenomenon of bending of light around the corners of an obstacle or an aperture into the region of geometrical shadow of obstacle is called diffraction of light. Diffraction of light is more pronounced when size of the obstacle/aperture should be the order of wavelength of wave. They are two types of diffraction 1. Fresnel diffraction ( Source or screen both are at finite distance from obstacle or aperture) 2. Fraunhoffer diffraction (Source or screen both are at infinite distance from obstacle or aperture)

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16. Explain theory of diffraction due to a single slit?

Second Year Physics study material 2015 A)

Light which is diverging from a monochromatic source S is made parallel after refraction through convex lens L1 . The refracted light from L1 is propagated in the form of wavefront WW’ . The plane wavefront WW’ in incident on the slit AB of width ‘d’ . According to Huygens principle, each point on the AB acts as a source of secondary disturbance or wavelets. Second lens L2 is used in converging the parallel beam. Now consider a point O on screen, which is placed at a distance D from the slit AB . Since point O is equidistance from A and B , therefore the secondary wavelets from A and B reach the point O in the same phase and the constructive interference takes place at O . In other words point O is the position of central maximum. Now Let the light is diffracted through an angle θ . So the secondary wavelets will also be diffracted through an angle θ . Let these wavelets meet the screen at point P. The point P will be of maximum or minimum intensity depending on the path difference between secondary wavelets (reaching point P) originating from corresponding points of wavefront. To find the path difference between secondary wavelets originating from the corresponding points of A and B of the same plane wavefront , draw AN perpendicular to BB’ . The path difference between these wavelets originating from A to B is BN. From ∆ BAN ,

BN/AB = sin θ

for minima :

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BN = dsinθ = dθ (if θ is small)

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or BN = AB sin θ

Second Year Physics study material 2015 If path difference is equal to one wavelength that is BN = dsinθ = λ , then divide the slit into two equal halves that is AC and CB . Now the path difference between the secondary wavelets originating from A and C is equal to λ/2. So these wavelets will meet at point P out of phase (phase difference π) and hence destructive interference will take place at P . similarly the path difference between the wavelets originating from C and B is λ/2 and hence these will also produce destructive interference at P. Thus Position P will be of minimum intensity. Hence for first minimum = dsinθ1 = λ sin θ1 = λ/d θ1 = λ/d ( if angle is very small) similarly , if BN = 2λ , then the slit AB can be imagined to be divided into four equal halves . Then the path difference between the secondary wavelets originating from the corresponding from of each half = (2λ/4) = λ/2 . Thus these wavelets produce destructive interference and point P is of minimum intensity. Thus for second minimum dsinθ2 = 2λ sinθ2 = 2λ/d θ2 = 2λ/d (for smaller angles) In general , for minima dsinθm = mλ , sinθm = mλ/d θm = mλ/d (for smaller angles) Where θm is the angle giving direction of the mth order minimum and m = 1,2,3---- an integer. For Maxima : If path difference BN = d sinθ is an odd multiple of λ/2 , then the constructive interference takes place at P , Hence point P is position of secondary maxima dsinθm = (m+1/2)λ θm = (m+1/2)λ/d

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Where θm is the angle giving direction of mth order secondary maximum and m = 1,2,3----

Second Year Physics study material 2015 Width of central maximum or principal maximum :

Central maximum occurs at O on screen, where angle θ is zero. that is path difference is zero so all points in the slit being in same phase lead to maximum brightness. Let f be the focal length of lens L2 (placed very close to slit) and the distance of first minimum on either side of the central maximum be x tan θ = x/f Since the Lens L2 is very close to the slit , so f = D , tan θ = x/D since θ is very small , so tan θ ≈ sin θ , sin θ = x/D

---------(1)

Also, for first minimum , d sinθ = λ or sinθ = λ/d -----------(2) from equations 1 and 2 x/D = λ/d

or x = λD/d

This is the distance of first minimum on either side from the center of central maximum. width of central maximum is given by 2x = 2λD/d 17) What are the factors effecting width of central maxima ? A) It is directly proportional to wavelength of light used. so it is small for violet colour and maximum for red colour.

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if width of slit (d) is small , width of central maximum is large and viceversa

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Width of central maximum is directly proportional to distance D between plane of slit and screen, so width will increase with increase in D

Second Year Physics study material 2015 18) What is the difference between Interference and diffraction of light? A) Interference It is the result of superposition of waves from two wavefronts originating from the same source Interference fringes are all of same width All Maxima are bright and all minima are perfectly black Over a large area on screen interference is observed

Diffraction It is due to superposition of wave from different points of the same wavefront Fringes are not of the same width Maxima are of decreasing intensity , minima are not perfectly dark Over a small region on screen diffraction is observed.

19) Define Resolving power of an optical instrument? A) The ability of an optical instrument to produce distinctly separate images of two objects located very closely to each other is called its Resolving power. Resolving power can also be defined as the reciprocal of limit of resolution. smaller is the limit of resolution , larger is its resolving power. 20) Explain briefly Resolving power of an Microscope? A) Let PQ be object of size d placed in front of objective lens AB of the microscope. The central maxima of the images of P and Q are P’ and Q’ respectively. As the size of object PQ decreased, d decreases , central maximum P’ moves closer to the central maximum Q’. According to Rayleigh criterion , two points P and Q are resolved if central maximum P’ coincides with the minimum N near to Q’. The corresponding d = dmin gives the resolving limit of microscope. Since N is the first minimum around Q’ so , (QA + AN) – (QB + BN) = 1.22λ or (QA – QB) + (AN – BN) = 1.22 λ -----------------(1) Also P’ coincides with N , so

using equation 2 in equation 1

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(AN – BN) = (PB – PA) --------------------- (2)

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PA + AN = PB + BN

Second Year Physics study material 2015 (QA – QB) + (PB – PA) = 1.22 λ ----------------- (3) if β is the semi angle subtended by the objective AB of the object PQ , then (QA – QB) + (PB – PA) =2 dmin sin β Hence equation 3 becomes 2 dmin sin β= 1.22 λ dmin = 1.22 λ /2 sin β Generally the high resolving microscope used in laboratories has liquid between the object and objective of microscope. If n is the refractive index of the liquid , then the wavelength in liquid becomes λ/n and the resolving limit of microscope is given by

dmin = 1.22 λ /2 nsin β where nsin β is called numerical aperture of the objective lens of the microscope. Resolving power of the microscope is defined as the reciprocal of the resolving limit of the microscope, therefore Resolving power of microscope = 2nsin β/1.22 λ 21) Explain briefly Resolving power of Astronomical telescope? A) The minimum angle subtended at the objective lens of the telescope by two distinct objects (Lying close to each other) whose images are just resolved is called limit of resolution of the telescope. The reciprocal of limit of resolution is called resolving power of telescope.

In ∆ABC , sin dθ = AC/AB

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The path difference of between BP2 and AP2 is given by , ∆x =AC

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Let dθ be minimum angle subtended at the objective lens L of the telescope by two distant objects O and O’. The images of these objects are formed at P1 and P2 respectively. According to Rayleigh criterion, the two images P1 and P2 will be just resolved if the central maximum of the image P1 of the object O coincides with the first minima of the image P2 of object O’ or vice -versa

Second Year Physics study material 2015 ∆x = AB sin dθ

if D (= AB) be the diameter of the objective lens L of the telescope then ∆x = D sin dθ The angular separation of the center of the central maximum and the first minima for a circular aperture ( of a lens) can be obtained ∆x = λ or

D sin dθ = λ

since dθ is very small angle , so sin dθ = dθ D(dθ) = λ dθ = λ/D But for a circular aperture , dθ = 1.22 λ/D Which is the limit of resolution of the telescope

it depends on diameter of the aperture of objective lens and wavelength of light used.

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= D/ 1.22 λ

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Resolving power of a telescope

Second Year Physics study material 2015 22) Write a short note on Validity of Ray optics ? A) Fresnal distance is defined as the distance of the screen from the slit or aperture when the spreading of light due to diffraction from the center of screen is equal to size (width) of the slit . It is represented by ZF . ZF = d2/λ Fresnal distance depends on Square of size of slit and wavelength of light 23) What is polarization of light ? Explain briefly experimental polarization of light ? A) The phenomenon of restricting the vibrations of light vector or electric field vector in a particular direction in a plane perpendicular to direction of propogation of light is called polarization of light. Tourmaline crystal is used to polarize the light and hence it is called Polarizer. Ordinary light is represented as

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Polarized light can be represented in two ways . One way where vibrations parallel to plane of paper and another way where vibrations perpendicular to paper.

Second Year Physics study material 2015 Consider two polaroids , Polarizer P and Analyzer A . Let both polaroids P and A are placed in such a way that their pass axes or polarizing directions are parallel to each other. When unpolarized light incident on Polaroid P , it is found tha intensity of light transmitted by the Polaroid is reduced to Half tohe original value of intensity of incident light. When ordinary light is incident on Polaroid P , electric vector vibrating parallel to the pass axis of Polaroid P passes through it , while the electric vector vibrating perpendicular to the pass axis of Polaroid is blocked. so the light transmitted through the Polaroid P has only electric field vector parallel to the axis. Now place the two polaroids P and A such that their axes are perpendicular to each other. when ordinary light is incident on the Polaroid P , it is found that the intensity of light transmtted thorugh the Polaroid is zero. 24) State and Explain Malus law ? A) This law states that “ Intensity of the polarized light transmitted through the analyzer varies as square of the cosine of the angle between the plane of transmission of the analyzer and the plane of the polarizer. Let E be the amplitude of the light transmitted through the analyzer and θ be the angle between the planes of polarizer and analyzer. Resolve E into two components 1) E cosθ along OP (parallel to plane of transmission of analyzer) 2. E sinθ along OV (perpendicular to plane of transmission of analyzer) only E cosθ component is transmitted through the analyzer. we know Intensity α (Amplitude)2 Intensity of transmitted light through the analyzer is given by I α (E cosθ)2 I = kE2 cos2θ

A) A Plane polarized light can be created by means of two methods

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25) State and Explain methods for creating Plane polarization light?

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I = Io cos2θ which is known as Malus law

Second Year Physics study material 2015 1. Polarization of light by scattering 2. Polarization of light by reflection Polarization of light by Scattering : Let a beam of unpolarized light be incident on a small dust particles or air molecules (scatterer) along X axis. Since light is transverse in nature , therefore all the possible directions of vibrations of electric vector E in the unpolarized light lie in Y-Z plane . Hence the electrons associated with the scatterer will be set with vibrations along Y and Z axis. A vibrating electrons behaves as an oscillating dipole and emits radiations in all directions expects its own line of vibration. when an observer observe the scatterer along Y axis then the light reaching the observer comes from the electrons vibrating along Z axis. This light is linearly polarized. Polarization of light by reflection : When ordinary light or unpolarized light falls obliquely on a glass slab , it is partially reflected and partly refracted. Both the reflected and refracted beams of light are partly polarized. The incident ordinary light has two components , one component of incident light lies in the plane parallel to the surface of the glass slab and this component is reflected. The other component lies in the plane perpendicular to the glass surface is refracted. Both the reflected and refracted components of light are partly polarized. At a particular angle of incidence , the reflected light is completely plane polarized. The angle of incidence at which the reflected light is completely plane polarized is known as polarization angle . This is also known as Brewster angle. when angle of incidence ordinary light is equal to polarization angle , then reflected and refracted beams of light are perpendicular to each other.

26) What is Brewster law ? A) The refractive index of refractive medium (n) is numerically equal to the tangle of angle of polarization.

from figure

∟BOY + ∟COY = 90

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Experimentally reflected component and refracted component of light are mutually perpendicular then ∟i = ∟iB

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n = tan iB

Second Year Physics study material 2015 (90 –r) + (90-r’) = 90 (90-IB) + (90-r’) = 90 r’ = 90- IB using n = sin i/sin r’ = sin iB/sin (90 –θP) = sin iB/cos θP n = tan iB

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Brewster law is used to calculate polarizing angle for any two media in contact when their refractive indices are already known.

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