# WBUT CS Solved Question Papers Polarisation Of Light

# WBUT CS Solved Question Papers Polarisation Of Light

Answer: (c) 1.2. For E-ray and O-ray refractive indices are 1.65 and 1.45 respectively. Then the thickness of the material required for a quarter wave piate for light of wave length 5000A will be [WBUT 2006(DECEMBER>]

a) 0° b) — c) — d)X [WBUT 2010(DECEMBER)]

*Answer: (b) 1.7. In a Nicol-prism the O-ray is totally internally reflected and the E-ray is transmitted. This statement is [WBUT 2010(DECEMBER)]*

a) ^{true} b) false c) partly true d) partly false

*Answer: (c) or (d) 1.8. For a doubly refracting crystal, the refractive indices for the ordinary and extraordinary rays are denoted by ju _{o} and Which of the following relations is valid along the optical axis of the crystal? [WBUT 2011 (JUNE)] *

*a) =//„ b) fl*

_{e}<//„ c) fl_{t}> fi_{o}d) fi_{a}<M_{e}Answer: (a) ’ 1.9. A rotating calcite crystal (a doubly refracting crystal) is placed over an ink dot. On seeing through the crystal, one finds [WBUT 2011(DECEMBER)1

*a) two stationary dots b) two dots moving along a straight line c) one dot rotating about the other d) both dots rotating about a common axise Answer: (c) 1.8. The refractive indices of E-ray and O-ray are respectively 1.65 and 1.45. Th’ 7 the thickness of the material required for a quarter wave plate for liaht of wavelength 500 nm is*_{[WBUT}2012(JUNE)?a) 250 nm b)125nm c) 625 nm d) 740 nm

*Answer: (c)*

2.1. Write short note on Nicol prism as polarizer and analyser.

[WBUT 2005(DECEMBER), 2009(JUNE)]

*A MIS w m Nicol prism is constructed from such a calcite crystal whose length ^{D} is three times its breadth* The end faces of the crystal are cut down so as to reduce the angles at C and E from 71° in the principal section to a more acute angle 68° shown in the figure. The crystal is then cut along the plane AKGJL perpendicular both to k the principal section ACGE and the end faces AC makes an angle of 90° with the and face AC & GE as shown in the figure. The cut surfaces are grounded and polished optically flat and then cemented together by Canada Balsam whose refractive index lies between the reliactivc indices tor the ordinary and the extra ordinary rays for h calcite crystal. p<, = 1.658 ( O-ray r.i for calcite crystal ) fici? — 1-55 ( r.i of canada Balsam ) u_{E} = L486 { E-ray r.i for calcite crystal ) The above values are considered for monochromatic sodium lieht*

If the incident ray makes an angle much smaller than CMS with the surface AiC the ordinary ray will strike the balsam layer at an angle less than the critical angle and hence will be transmitted. If the incident ray makes an angle greater than CMS the extraordinary ray will become more and more parallel to the optic axis AtN and hence its refractive index will become nearly equal to that of calcite for the ordinary ray. This will then also suffer total interna! reflection like ordinary ray. Hence no light will emerge out of the nicol prism. A nicol prism therefore cannot be used for highly convergent or divergent beams. The angle between the extreme rays of the incoming beam is limited to about 28°. Extraordinary ray is not affected and is therefore transmitted through the prism. The value of critical angle for the ordinary ray for calcite to Canada balsam can be calculated as follows. AG —► represents Canada balsam layer. 0_{C} = sin’^{1} [l/|i] 0c — sin^{-1} [1.550/1.658] [v *i = 1,550/1.658] 6_{C} ~ 69° If the incident ray makes an angle much smaller that 0_{C} ordinary ray will be transmitted. Ordinary ray will not be transmitted if the incident angle is greater than this critical angle. A nicors prisms cannot be used for highly convergent or divergent beams. Analysis of plane polarised light ^ POLARIZER Pi ANALYZER P_{2} Fig: a) t n ► o The Nicol’s prism can be used both as a polariser and an analyser. When two nicols are mounted coaxially. The first nicol P, which produces plane polarised light is called the polariser while the second Nicol P_{2} which analyses the incoming light is called the analyser. When the two Nicols are placed with their principal sections parallel to each as shown in figure (i), then the extraordinary ray transmitted freely. If the second prism is gradually rotated, then the intensity of extraordinary ray gradually decreases and when the two nicols are right angle to each other as shown in figure (ii), they are in a crossed position, no light comes from the second prism. This due to fact that when the polarised extra-ordinary ray enters the second nicol prism, it acts as ordinary ray and is totally internally reflected. Therefore the first nicol prism P| produces plane polarised light and second one detects it. 2.2. What is the difference between the extraordinary and the ordinary ray propagating through an uniaxial crystal? [WBUT 2005(DECEMBER)] Answer: Extraordinary Ray is termed as E-ray and ordinary ray is termed as O-ray. In double refraction E-ray is not obeying snell’s law where as O-ray is obeying the same and the consequence of this is E-ray has varying velocity in different direction of the uniaxial crystal where as O-ray has constant velocity throughout the direction of the crystal. According to Huygens principle if we imagine a point source of light in a double refracting medium is the origin of two wave fronts then for the O-ray, for which the velocity of light is the same in all directions the wave front is spherical. For the E-ray * the velocity varies with the direction and the wave front is an ellipsoid of revolution. The velocities of the ordinary and the extraordinary rays are the same along the optic axis.

2.3. A polarizer and an analyzer are oriented in such a manner such that the maximum intensity is achieved. If the analyzer is rotated through 60° estimate percentage reduction of its maximum intensity. [WBUT 2006(DECEMBER)]

_{0}. Now when analyser is rotated through 60° by using Maul* s law we have l-l

_{0}cos

^{2}0 So, percentage reduction in intensity is (1 – cos

^{2}60) x 100 = 75%. 2.4. Two polarizers are placed at crossed position, (angle between the polarizing planes are 90°)

_{f}a third polarizer with angle 6 with the first one is placed between them. An unpolarizerd light of intensity I is incident on the first one and passes through all three polarizers. Find the intensity of the light that comes out.

i [WBUT 200S(JUNE), 2010(DECEMBER)]

Answer: Optic axis of the l^{sr} polariser is shown in vertical direction and that of the 2^{nd} one in the horizontal direction. Now as per the question optic axis of the third polariser is placed in between two, making an angle 0 with the I^{sl} one. When the incident ray passes through the [^{st} polariser the electric vector vibration amplitude say will be restricted in vertical direction. After passing through the third polariser component of E_{0} i.e., E_{0} cos 0 will be passing parallel to the optic axis of the same. Third polariser optic axis is making an angle 90°- 6 with the 2^{nd} one which is placed horizontal as shown. So a cos(90°- 0) component of E_{0} cos 0 will be transmitted out i.e., finally amplitude of the transmitted light will be E_{0} cos 9cos(90° — o)-E_{Q} cos 0 sin 9

^{2}9 sin

^{2}0 — I

_{Q}cos

^{2}9sin

^{2}9 = Here we assume that the intensity of the incident beam is 7

_{0}.

_{p}) where i

_{p}—*■ angle of polarising. As per the question it is required to determine the polarizing angle in water so we must find out refractive index of water with respect to glass. It will be fi* = 1.5 ; /4= 1.33 ■ 128 2.6. Explain briefly the action of a dichroic Polariod.

1.33

[WBUT 2008(JUNE), 2009(JUNE)j

Answer: Certain doubly refracting crystals have the property of absorbing the O-ray and E-ray unequally. The most outstanding substance is tourmaline. If a beam of unpolarised light is incident on a tourmaline crystal, it is broken into O-ray having vibrations in the plane of incidence and E-ray having vibrations perpendicular to the plane of incidence. On passing through the crystal, the O-ray is absorbed if the crystal is cut of a proper thickness. This phenomenon is known as dichorism. The phenomenon of selective absorption is made use of in the construction of polaroids. These are large sized polarizing films of any cross-section which are used to produce and detect plane polarized light like Nicol prism. Initially, a paste of organic synthetic crystalline material of iodo sulphate of quinine, called herapathite, is prepared. Herapathite is a small needle shaped crystal and possess the property of polarizing the light. These crystals are not stable r>rtd cannot bear even a slight strain. So, the commercial Polaroid is prepared from suspension of this small herapathite crystal in nitrocellulose. Herapathite crystals are arranged side by side with their optic axes all parallel, so that they function as one single crystal in nitro-cellulose and squeezing it out through a fine slit, through which only those crystals pass whose axes are parallel to the length of the slit. The sheet thus produced has millions of tiny crystals with their optic axes parallel to one-auother. The sheet thus prepared is mounted between two thin glass sheets, forming what is called Polaroid. 2.7. State and explain Malus’s taw of polarization. [WBUT 2010(JUNE)] Answer: When a beam of light, polarized by reflection at one plane surface is allowed to fall on the second plane surface at the polarizing angle the intensity of the reflected beam varies with the angle between the planes of the two surfaces. In the Biot’s polariscope it was found that, ihe intensity of the twice reflected beam is maximum when the two planes are parallel and zero when the two planes are at right angles to each other. The same is also true for the twice transmitted beam from the polarizer and analyzer. The law of Malus states that the intensity of the polarized light transmitted through the analyzer varies as the square of the cosine of the angle between the plane of transmission of the analyzer and the plane of the polarizer. In the case of the Biot’s polariscope this angle is between the two reflecting planes. In the figure AP represents the transmission plane of a polariser and AE represents that of analyser. The angle between two planes of analyser and polariser be 0. Let the electric vector of the polarised light emerging out from the plane of polariser is resolved

into two components as shown. To derive the law, let us suppose that the angle between the two planes of transmission is 0 at any instant. The light vector, AP = a, in the plane polarised light emerging from the polariser may be resolved into two components, as shown in the figure, AE = a cos 0 Along the plane of transmission ‘ AO = a sin 0 perpendicular to the plane of transmission Perpendicular component is eliminated in the analyser while the parallel component is freely transmitted through it. Therefore, the intensity I of light that emerges from the analyser is given by. I = a^{2}cos^{2} 9 ■ I = l_{o}cos^{2}0 Where To is the intensity of the plane polarised light incident on the analyser. 2.8. a) Calculate the minimum thickness of a calcite plate which would convert plane polarized light into circularly polarized light. ju_{e} =1.486 and //_{0} =1.658 and /U5890A. b) What is Brewster’s angle? [WBUT 2005(DECEMBER)_{?} 2009{JUNE), 2010(JUNE)] Answer: The minimum thickness of a calcite plate r is calculated from , . ^{X} 5890×1 O’^{8} ^ a) t = ————- ——- – =————— – = 0.856×10 cm 4 {H’-Mo} 4{1.658-1,486} – b) The angle of incidence for which the reflected beam is polarised is called Brewster’s law.//; =1.5; ^ = 1.33-^//; =-p~ = 1.128 .‘.i_{p} = tan’^{1} (1.128) = 48.437° Which is the required value of the Brewster’s angle. 2.9. How can you get an ellipfically polarized light from an linearly polarized light by using an optical device? A plane polarized light is incident on a piece of quartz cut parallel to the axis. Find the least thickness for which the ordinary and the extraordinary rays combine to form plane polarized light given that //_{0} =1.5442 and JU_{e} -1.5533, /l = 5xl0^{5}cm.

[WBUT 2010 (DECEMBER)]

Answer: When linearly polarized light is allowed to fall normally on another quarter wave plate such that the direction of vibration of the plane polarized light makes an angle 0 other than 0°, 45° , 90° with the direction of the optic axis of the quarter wave plate. The quarter wave plate breaks the incident plane wave into E and O waves of unequal amplitudes and introduces a phase difference of 7t/2 between them. These two

^{5xl}–—-—- =

^{5xl}P_ = 0.00274 cm ■ 2 [1.5533-1.5442] 0.0182 2.10. a) What is Polaroid? Give instances of its practice application. b) A quartz plate with thickness of 0.1436 mm is used as phase retardation plate. For what wavelengths in the visible region (450 – 800 nm) will it act as a quarter wave plate (//

_{o}= 1.5443,//, = 1.55333). [WBUT 2011 (JUNE)] Answer: A Polaroid is artificially made a thin and large transparent sheet or film of crystalline polarizing material capable of producing and analysing of plane polarized beams of light of large cross-section.

- Polaroids are mostly used as polarising sunglasses*

*2, *Polaroids are used in windows of railways, aeroplane etc in which there is a fixed outer and rotatable inner polaroid to control the amount of light entering it from outside* . b) Wavelength of the light can be calculated from the thickness of the quartz plate (2rc + ])/t _{w} 4(ju-ju_{Q})t t = —7——— – —> X = ———— —. Where n = 0_{T}1,2,3……….. From different values of n ^{4}iMe”Mo) (^{2}« + 0 wavelength can be calculated and. For n — 3 we get ^ = 740nm and n = 4 we get X — 576 nm which is in the desired visible region. 2.11. a) Explain Brewster’s law. [WBUT 2012(JUNE)J b) Show that when light is incident on a transparent substance at the polarizing angle, the reflected and the refracted rays are perpendicular to each other. Answer: Refer to Question No. 3.3(a). 3.1. a) What is meant by polarisation of light? b) State Brewster’s law and hence prove that the angle between the reflected and the refracted rays is 90°. c) The refracted index of glass plate is 1.6. Calculate the polarization angle and the corresponding angle of the refracted ray. [WBUT 2005(DECEMBER)]

_{p}is the angle of polarisation \x be the refractive index of the medium concerned. Let unpolarised light is incident at an angle equal to the polarizing angle on the separation of two medium: MM’ BC is reflected as per the law of reflection and other part is refracted to the 2

^{nd}medium as BD.

_{A}

^{N c}

B A i 1 \ | |

1 \ “ | |

I | |

a | £VD |

Plane Polarised M’ |

_{p}– Comparing (i) and (ii)

_{p}= (tt/ 2) – r

_{P}+ r = tz/2 As i

_{p}+ r = k/2 , /_CBD is also equal to ti/2. Therefore, the reflected and the refracted rays are at right angles to each other. c) Using Brewster’s law, = tan (i

_{p}) where i

_{p}is polarising angle.

! .732 = tan i_{p} ^ i_{p} = 60°

So polarising angle = 60° Now at polarising angle, reflected & refracted ray are 90° apart. /. reflected angle = i_{p} = 60° /. angle of refraction r = ( 180°- i_{p} – 90° ) = ( 90° – i_{p} ) = 30° 3.2. a) At what angle should light be incident on a glass plate of refractive index // — 1.5697 to get a plane polarized light by reflection? ■ b) An analyzing nicol examines two adjacent plane polarized beams A and B whose planes of polarization are mutually perpendicular. In one position of the analyzer the beam B shows zero intensity. From this position, a rotation of 30° shows the two beams as of equal intensity. Deduce the ratio of two intensities Ia/I_{b} of the two beams. [WBUT 2011 (JUNE)] Answer: a) Using Brewster’s formula // = tan(i_{p}) we have 1.5697 = tan(i_{p}) We have i_{p} = tan^{-1} (/y) —> i_{p} = tan^{-1} (L5697) = 57.5° b) Horizontal and vertical bold lines represent the direction of vibration of electric field vectors of beam A and B. The dashed lines represent the optic axis of the nicol prism. Now if the nicol is rotated through 30° as it is shown. So the components of the field vectors of beam A and beam B will be along the direction of the optic axis of the nicol , So by the condition we have 1_{A} cos^{2} 30° — I_{B} cos^{2} 60″ ~ = the required intensity ratio. / _{B} 3 3.3. a) Using Brewster’s law show that light incident on a transparent substance at polarizing angle gives reflected and refracted rays at right angles to each other. b) Describe how circularly polarized light and unpolarized light may be distinguished by experiment. c) Unpoiarized light is incident on two polarizing sheets placed on top of the other. Each sheet reduces the intensity of unpolarized light by 50%. What must ^e the angle between the characteristic directions of sheets if the intensity gf the transmitted light is I of the intensity of the incident beam?

d) A quarter wave plate is fabricated with smallest possible thickness for wavelength of 589.3mm. What phase retardation will be obtained with this plate with light having wavelength of 435.8mm. You may neglect the variations of refractive indices with wavelength for this problem. [WBUT 2011 (DECEMBER)] Answer: a) According to Brewster’s Law the tangent of the angle of polarisation is numerically equal to the refractive index of the medium. Mathematically Where i_{p} is the angle of polarisation jl be the refractive index of the medium concerned. Interpretation of Brewster’s law: Let unpolarised light is incident at an angle equal to the polarizing angle on the separation of two medium MM\ BC is reflected as per the law of reflection and other part is refracted to the 2nd medium as BD. For completely unpolarised light two components (one perpendicular electric vector to the plane of incidence known as S component represented as dots and other parallel electric vector to the plane of incidence known as p components represented as X sign) of electric vector are of equal magnitude on the average. The reflected ray has a predominance of S components and at polarising angle reflected beam contains fully S components so as it can be treated as completely polarised light* On the other hand on refracted ray contains both S component and p conponents. So the refracted beam is termed as partially polarised, sin i From Snell’s law JJ = —7– -. .(2) sin r sini From Brewster’s law we have, jl — tan i — cos; Comparing (2) and (3) we have cos i = sin r = cos . . 71 . , k i = —~ r or 1 + r — — ^{p} 2 ^{p} 2 As i_{p} + r = 71 !2 , Z. CBD is also equal to 7l!2. Therefore, the reflected and the refracted rays are at right angles to each other. b) Both circularly and unpolarised light is seen through a Nicol prism. On rotation of the Nicol prism about the ray as an axis we observe no variation of intensity in both the cases.lt is because in the case of unpolarised light one vibration or the other is always parallel to the principal section of the Nicol and hence we get uniform intensity. For circularly polarized light as because of the two equal amplitudes of vibrations present in it, one of the components passes through the Nicol prism and the other is refused transmission. The intensity of the transmitted beam is, therefore, not altered as the Nicol is rotated. Let us allow to fall light normally on quarter-wave plate and then examine it through the Nicol prism. If the intensity varies between maximum and zero, it ts circularly polarized light. It is because the circularly vibration on entering the quarter wave plate is broken up into two linear vibrations of the same amplitude mutually perpendicular to each other. The quarter wave plate further introduces a phase change of 7Z12. On examination through the rotating nicol, maximum intensity is observed when principal section of the Nicol is parallel to the vibration of the plane polarized light. Light will be cut off completely when the principal section of the Nicol is perpendicular to the direction of the resultant polarized vibration. The unpolarised light after passing through a quarter-wave plate remains unpolarised and hence no change in intensity will be observed when passing through a Nicol prism which is rotated gradually. c) Through the l^{sl} polarizing light intensity will be reduced to half. So new intensity of light after passing through 1^{st} polarizing sheet will be I^{1} = — Let the characteristic directions of two polarising sheet be 0* When the polarised light through the 1^{st} polariser falls on the 2^{nd} on it will follow Mai us law . So after passing the 2^{nd} polariser new intensity will be /” = /’ cos^{2} e = ^2-cos^{2} 0 . 2 But according to the problem — — — cos^{2} 9 —» 9 — cos ^ J— — 35*2° 3 2 V 3 Which is the required angle, d) Using the minimum thickness formula of an quarter wave plate we have A _ 5893 _ 1473.25 ^{—} Mo) 4 ~ Mo) i 2 k From the relation Phase difference = -—x path difference *A*‘ So path difference is now (jl_{E} – ji_{()})t = 1473.25 2 7t So, Phase difference —– ——– x 1473.25 = 0.67^: 4358 3.4. a) State Malus law. A beam of polarized light makes an angle of 60° with the axis of the Polaroid sheet. How much intensity of light is transmitted through the sheet? b) What is retardation plate? How can you distinguish between circularly polarized light and unpolarized light with the help of a quarter wave plate and a Nicol prism? c) The critical angle of glass with respect to air is 41°, Find the refractive index of the medium and the angle of refraction of the light incident on the glass plate at the polarizing angle. [WBUT 2012(JUNE)] Answer: a) The law of Malus states that the intensity of the polarized light transmitted through the analyzer varies as the square of the cosine of the angle between the plane of transmission of the analyzer and the plane of the polarizer. According to Main’s Law, / = /_{0}cos^{2}# Where /_{0} is the original intensity I_{x} is the intensity of the transmitted light through analyser & 0 is the angle between the axes of the polaroid sheet. I transmitted through the sheet. b) A retardation plate is an optically transparent material which resolves a beam of polarized light into two orthogonal components; retards the phase of one component relative to the other; then recombines the components into a single beam with new polarization characteristics. Distinction between circularly and Unpolarised light: Observation; Both circularly and unpolarised light is seen through a Nicol prism. On rotation of the Nicol prism about the ray as an axis we observe no variation of intensity in both the cases.lt is because in the case of unpolarised light one vibration or the other is always parallel to the principal section of the Nicol and hence we get uniform intensity. For circularly polarized light as because of the two equal amplitudes of vibrations present in it, one of the components passes through the Nicol prism and the other is refused transmission. The intensity of the transmitted beam is, therefore, not altered as the Nicol is rotated. Remedy: Let us allow to fall light normally on quarter-wave plate and then examine it through the Nicol prism. Inference: If the intensity varies between maximum and zero, it is circularly polarized light. It is because the circularly vibration on entering the quarter wave plate is broken up into twp linear vibrations of the same amplitude mutually perpendicular to each other. The quarter wave plate further introduces a phase change of 7t 12. On examination through the rotating nicol, maximum intensity is observed when principal section of the Nicol is parallel to the vibration of the plane polarized light. Light wilt be cut off completely when the principal section of the Nicol is perpendicular to the direction of the resultant polarized vibration. The unpolarised light after passing through a quarter-wave plate remains unpolarised and hence no change in intensity will be observed when passing through a Nicol prism which is rotated gradually. c) If Q_{c} be the critical angle, then ji = ——— = ——————————– ^ jj = 1.524 sin $_{c} sin 41 Now by using Brewster’s law, fl = tan(i_{p}) Hence tan(i^) = 1.524 So i_{p} =56.73 is the angle of incidence . Since at polarising angle sum of angle of incidence and refraction is 90° i.e., (i_{p} +r =90°). So angle of refraction will be r = (90 “56.73) = 43.27″