Allen Career Institute Pre-Medical Physics PDF

PRE-MEDICAL PHYSICS ENTHUSIAST | LEADER | ACHIEVER STUDY MATERIAL Wave Optics ENGLISH MEDIUM All rights including trademark and copyrights and rights of translation etc. reserved and vested exclusively with ALLEN Career Institute Private Limited. (ALLEN) No part of...

PRE-MEDICAL PHYSICS ENTHUSIAST | LEADER | ACHIEVER STUDY MATERIAL Wave Optics ENGLISH MEDIUM All rights including trademark and copyrights and rights of translation etc. reserved and vested exclusively with ALLEN Career Institute Private Limited. (ALLEN) No part of this work may be copied, reproduced, adapted, abridged or translated, transcribed, transmitted, stored or distributed in any form retrieval system, computer system, photographic or other system or transmitted in any form or by any means whether electronic, magnetic, chemical or manual, mechanical, digital, optical, photocopying, recording or otherwise, or stood in any retrieval system of any nature without the written permission of the Allen Career Institute Private Limited. Any breach will entail legal action and prosecution without further notice. This work is sold/distributed by Allen Career Institute Private Limited subject to the condition and undertaking given by the student that all proprietary rights (under the Trademark Act, 1999 and Copyright Act, 1957) of the work shall be exclusively belong to ALLEN Career Institute Private Limited. Neither the Study Materials and/or Test Series and/or the contents nor any part thereof i.e. work shall be reproduced, modify, re-publish, sub-license, upload on website, broadcast, post, transmit, disseminate, distribute, sell in market, stored in a retrieval system or transmitted in any form or by any means for reproducing or making multiple copies of it. Any person who does any unauthorised act in relation to this work may be liable to criminal prosecution and civil claims for damages. Any violation or infringement of the propriety rights of Allen shall be punishable under Section- 29 & 52 of the Trademark Act, 1999 and under Section- 51, 58 & 63 of the Copyright Act, 1957 and any other Act applicable in India. All disputes are subjected to the exclusive jurisdiction of courts, tribunals and forums at Kota, Rajasthan only. Note:- This publication is meant for educational and learning purposes. All reasonable care and diligence have been taken while editing and printing this publication. ALLEN Career Institute Private Limited shall not hold any responsibility for any error that may have inadvertently crept in. ALLEN Career Institute Private Limited is not responsible for the consequences of any action taken on the basis of this publication. ®  Physics : Wave Optics Pre-Medical CHRISTIAAN HUYGENS (1629 – 1695) DUTCH Physicist, astronomer, mathematician and the founder of the wave theory of light. His book, Treatise on light, makes fascinating reading even today. He brilliantly explained ® the double refraction shown by the mineral calcite in this work in addition to reflection and refraction. He was the first to analyse circular and simple harmonic motion and designed and built improved clocks and telescopes. He discovered the true geometry of Saturn’s rings. THOMAS YOUNG (1773 – 1829) ENGLISH Physicist, physician and Egyptologist. Young worked on a wide variety of scientific problems, ranging from the structure of the eye and the mechanism of vision to the decipherment of the Rosetta stone. He revived the wave theory of light and recognised that interference phenomena provide proof of the wave properties of light. 62  Physics : Wave Optics ® Pre-Medical WAVE OPTICS INTERFERENCE OF LIGHT 1. NATURE OF LIGHT Newton's corpuscular theory of light This theory was given by Newton.  Characteristics of the theory (i) Extremely minute, very light and elastic particles are being constantly emitted by all luminous bodies (light sources) in all directions which are known as corpuscles. (ii) These corpuscles travel with the speed of light.. (iii) When these corpuscles strike the retina of our eye then they produce the sensation of vision. ® (iv) The velocity of these corpuscles in vacuum is 3 × 108 m/s. (v) The different colours of light are due to different size of these corpuscles. (vi) The rest mass of these corpuscles is zero. (vii) The velocity of these corpuscles in an isotropic medium is same in all directions but it changes with the change of medium. (viii) These corpuscles travel in straight lines. (ix) These corpuscles are invisible.  The phenomena explained by this theory (i) Reflection and refraction of light. (ii) Rectilinear propagation of light. (iii) Existence of energy in light.  The phenomena not explained by this theory (i) Interference, diffraction, polarisation, double refraction and total internal reflection. (ii) Velocity of light being greater in rarer medium than that in a denser medium. (iii) Photoelectric effect Huygen's Wave theory of light This theory was enunciated by Huygen in a hypothetical medium known as luminiferrous ether. Ether is that imaginary medium which prevails in all space and is isotropic, perfectly elastic and massless. The velocity of light in a medium is constant but changes with change of medium. This theory is valid for all types of waves. (i) The locus of all ether particles vibrating in same phase is known as wavefront. (ii) Light travels in the medium in the form of wavefront. (iii) When light travels in a medium then the particles of medium start vibrating and consequently a disturbance is created in the medium. (iv) Every point on the wavefront becomes the source of secondary wavelets. It emits secondary wavelets in all directions which travel with the speed of light (v) The tangent plane to these secondary wavelets represents the new position of wave front. 63 ®  Physics : Wave Optics Pre-Medical original A2 A A1 A1 wavefront A Secondar wavefront 1 A2 1 secondar wavelets 2 y 2 S Propagation 3 of light-wave 3 B2 4 B B1 4 B2 B1 The phenomena explained by this theory ® (i) Reflection, refraction, interference, diffraction. (ii) Rectilinear propagation of light. (iii) Velocity of light in rarer medium being greater than that in denser medium. Phenomena not explained by this theory (i) Photoelectric effect WAVEFRONT, VARIOUS TYPES OF WAVEFRONT  Wavefront The locus of all the particles vibrating in the same phase is known as wavefront.  Types of wavefront The shape of wavefront depends upon the shape of the light source where wavefront originates. On this basis there are three types of wavefronts Comparative study of three types of Wavefront :- S.No. Wavefront Shape Diagram of Variation of Variation of of light shape of amplitude (A) Intensity (I) source wavefront with distance with distance 1 1 1 1. Spherical Point source A∝ or A ∝ I∝ d r r2 O 1 1 1 2. Cylindrical Linear source or slit A∝ or A ∝ I∝ d r r O' 3. Plane Extended large source or point A = constant I = constant source situated at very large distance 64  Physics : Wave Optics ® Pre-Medical CHARACTERISTIC OF WAVEFRONT (a) The phase difference between various particles on the wavefront is zero. (b) These wavefronts travel with the speed of light in all directions in an isotropic medium. (c) A point source of light always gives rise to a spherical wavefront in an isotropic medium. (d) In anisotropic medium it travels with different velocities in different directions. (e) Normal to the wavefront represents a ray of light. (f) It always travels in the forward direction in the medium. 2. INTERFERENCE OF LIGHT When two light waves of same frequency with constant phase difference superimpose over each other, then the resultant amplitude (or intensity) in the region of superimposition is different from the amplitude (or intensity) of individual waves. This modification in intensity in the region of superposition is called interference. ® (a) Constructive interference When resultant intensity is greater than the sum of two individual wave intensities [I > (I1 + I2)], then the interference is said to be constructive. (b) Destructive interference When the resultant intensity is less than the sum of two individual wave intensities [I < (I1 + I2)], then the interference is said to destructive. There is no violation of the law of conservation of energy in interference. Here, the energy from the points of minimum energy is shifted to the points of maximum energy. TYPES OF SOURCES  Coherent sources Two sources are said to be coherent if they emit light waves of the same frequency and start with same phase or have a constant phase difference. They are obtained from a single source. Note : Laser is a source of monochromatic light waves of high degree of coherence.  Incoherent sources Two independent monochromatic sources, emit waves of same frequency. Lamp S1 L1 But the waves are not in phase. So they are incoherent. This is because, atoms screen cannot emit light waves in same phase and these sources are said to be Lamp S2 incoherent sources. By using two independent laser beams it has been possible L2 to record the interference pattern. Independent Source (non-coherent source) METHODS OF OBTAINING COHERENT SOURCE  Division of wave front In this method, the wavefront is divided into two or more parts by reflection or refraction using mirrors, lenses or prisms. Illustration : Young's double slit experiment. P S1 S2 YDS 65 ®  Physics : Wave Optics Pre-Medical  Division of amplitude The amplitude of incoming beam is divided into two or more parts by partial reflection or refraction. These divided parts travel different paths and are finally brought together to produce interference. Illustration : The brilliant colour seen in a thin film of transparent material like soap film, oil film. thin film Condition for sustained interference To obtain the stationary interference pattern, the following conditions must be fulfilled : (a) The two sources should be coherent, i.e., they should vibrate in the same phase or there should be a constant phase difference between them. ® (b) The two sources must emit continuous waves of same wavelength and frequency. (c) The separation between two coherent sources should be small. (d) The distance of the screen from the two sources should be large. (e) For good contrast between maxima and minima, the amplitude of two interfering waves should be as nearly equal as possible and the background should be dark. (f) For a large number of fringes in the field of view, the sources should be narrow and monochromatic. ANALYSIS OF INTERFERENCE OF LIGHT When two light waves having same frequency and equal or nearly equal amplitude are moving in the same direction superimpose then different points have diffeent light intensities. At some point the intensity of light is maximum and at some point it is minimum this phenomenon is known as interference of light. Let two waves having amplitude a1 and a2 and same frequency, and constant phase difference φ superpose. Let their displacement are : y1 = a1sinωt and y2 = a2sin(ωt + φ) y = y1 + y2 = Asin(ωt + θ) Where A = Amplitude of resultant wave θ = New initial phase angle Phasor Diagram By right angle triangle : 2 2 A = (a1 + a2cosφ) + (a2sinφ) 2 A a2 Resultant amplitude A = a12 + a22 + 2a1 a2 cos φ θ φ  a2 sin φ  −1 a1 Phase angle θ = tan    a1 + a2 cos φ  Intensity ∝ (Amplitude)2 ⇒ I ∝ A2 ⇒ I = KA2 so I1 = Ka12 & I2 = Ka22 ∴ I = I1 + I2 + 2 I1 I2 cos φ here, 2 I1 I2 cos φ is known as interference factor.  If the distance of a source from two points A and B is x1 and x2 then Path difference δ = x2 – x1 66  Physics : Wave Optics ® Pre-Medical 2π 2π S A B Phase difference φ= (x2 –x1) ⇒ φ = δ λ λ x1 φ x2 Time difference ∆t = T 2π Phase difference Path difference Time difference φ δ ∆t = = ⇒ = = 2π λ T 2π λ T TYPES OF INTERFERENCE Constructive Interference When both waves are in same phase then phase difference is an even multiple of π ⇒ φ = 2 nπ ; n = 0,1,2.. λ  Path difference is an even multiple of 2 φ δ 2nπ δ λ  =⇒ = ⇒δ = 2n   ⇒δ = nλ (where n = 0,1,2...) 2π λ 2π λ 2 ® T T  When time difference is an even multiple of ∴ ∆t = 2n   2 2  In this condition the resultant amplitude and intensity will be maximum. Amax = (a1 + a2) ⇒ Imax = I1 + I2 + 2 I1 I2 = ( I1 + I2 )2 Destructive Interference When both the waves are in opposite phase. So phase difference is an odd multiple of π.     φ = (2n–1) π ; n = 1, 2... λ λ  When path difference is an odd multiple of , δ = ( 2n –1) , n = 1, 2... 2 2 T T  When time difference is an odd multiple of , ∆ t = (2n–1) , ( n=1,2...) 2 2 In this condition the resultant amplitude and intensity of wave will be minimum. Amin = (a1 – a2) ⇒ Imin = ( I1 − I2 )2 GOLDEN KEY POINTS  Interference follows law of conservation of energy. I max + I min  Average Intensity Iav = = I1 + I2 ∝ a12 + a22 2 I1 w1 a12  Intensity ∝ width of slit ∝ (amplitude)2 ⇒ I ∝ w ∝ a2 ⇒ = = I2 w 2 a22 2 I max  I 1 + I2   a1 + a2  2  a max  2 =  =  = a − a  a  I min  I1 – I2   1 2   min  I max − I min  Fringe visibility V = × 100% when Imin = 0 then fringe visibility is maximum i.e. when both slits I max + I min are of equal width the fringe visibility is the best and equal to 100%. 67 ®  Physics : Wave Optics Pre-Medical Illustrations Illustration 1 π If two waves represented by y1 = 4 sin ωt and y2 = 3 sin (ωt + ) interfere at a point. Find out the amplitude 3 of the resulting wave. Solution π (4)2 + ( 3 ) + 2.(4)(3) cos 2 Resultant amplitude A = a12 + a22 + 2a1 a2 cos φ = ⇒A~6 3 Illustration 2 Two beams of light having intensities I and 4I interferer to produce a fringe pattern on a screen. The phase π difference between the beam is at point A and 2π at point B. Then find out the difference between the 2 ® resultant intensities at A and B. Solution Resultant intensity I = I1 + I2 + 2 I1 I2 cos φ π Resultant intensity at point A is IA = I + 4I + 2 I 4I cos = 5I 2 Resultant intensity at point B, IB = I + 4I + 2 I1 I2 cos 2π = 9I ( cos 2π = 1) ∴IB – IA = 9 I – 5 I ⇒ 4 I Illustration 3 In an interference pattern, the slit widths are in the ratio 1:9. Then find out the ratio of minimum and maximum intensity. Solution Slit width ratio w1 1 I1 w1 a12 1 a1 1 I min (a1 − a2 )2 (a1 − 3a1 )2 4 =  = = = 2 ⇒ = ⇒3a 1 = a2 ∴ = 2 = 2 = =1:4 w2 9 I2 w 2 a 2 9 a2 3 I max (a1 + a2 ) (a1 + 3a1 ) 16 Illustration 4 The intensity variation in the interference pattern obtained with the help of two coherent sources is 60% of the average intensity. Find out the ratio of intensities of two sources. Solution Let Iavg = 100 units 2 I max 160 4  a + a2  4 a1 + a2 2 a1 3 = = ⇒ 1 a − a  =1 ⇒ a − a =1 ⇒ a =1 I min 40 1  1 2  1 2 2 I1 9 = I2 1 Illustration 5 Can two different bulbs, similar in all respect act as coherent sources ? Give reasons for your answer. Solution. No, because the light waves emitted by two independent bulbs will not have stable constant phase difference. 68  Physics : Wave Optics ® Pre-Medical Illustration 6 Two coherent light wave y1=Asin(ωt+π/3) and y2=Asin(ωt+π/4) superimpose at point P if central (zero order) maxima obtained at P then What is difference between x1 and x2 ? y1=Asin(ωt+π/3) x1 P Maxima x2 y2=Asin(ωt+π/4) Solution. π π π  φ δ ∆t  Initial phase difference ∆φ= − =  2π= λ= T  3 4 12   To obtain maxima at P λ π λ ⇒ ∆x= × = 2π 12 24 BEGINNER'S BOX-1 ® 1. Consider interference between two sources of intensities I and 4I. Obtain intensity at a point where phase difference is π π (a) (b) (c) π (d) 4π 2 4 2. Two coherent sources whose intensity ratio is 81 : 1 produce interference fringes. Calculate the ratio of intensity of maxima and minima in the fringe system. 3. YOUNG'S DOUBLE SLIT EXPERIMENT (YDSE) According to Huygen, light is a wave. It is proved experimentally by YDSE. S is a narrow slit illuminated by a monochromatic source of light sends wave fronts in all directions. Slits S1 and S2 become the source of secondary wavelets which are in phase and of same frequency. These waves are superimposed on each other which give rise to interference. Alternate dark and bright bands are obtained on a screen (called interference fringes) placed certain distance from the plane of slit S1 and S2. Central fringe is always bright (due to path length S1O and S2O to centre of screen are equal) and is called central maxima. screen y z=0 S1 S1 x S O d S2 S2 D sodium S2 lamp  In YDSE division of wavefront takes place.  If one of the two slit is closed, the interference pattern disappears. It shows that two coherent sources are required to produce interference pattern.  If white light is used as parent source, then the fringes will be coloured and of unequal width. (i) Central fringe will be white. (ii) The fringe closest on either side of the central white fringe is red and the farthest will appear blue. After a few fringes, no clear fringe pattern is seen. 69 ®  Physics : Wave Optics Pre-Medical CONDITION FOR BRIGHT AND DARK FRINGES Bright Fringe D = distance between slit and screen, d = distance between slit S1 and S2 Bright fringe occurs due to constructive interference. λ  For constructive interference path difference should be even multiple of 2 λ ∴ Path difference δ = PS2 – PS1 = S2 L = (2n) P 2 xn δ xn In ∆PCO tanθ = ; In ∆S1S2L sinθ = D d S1 A dO θ θ δ = nλ for bright fringes δ C xn δ S2 B If θ is small then tanθ ~ sinθ ⇒ = D D d Dλ  The distance of nth bright fringe from the central bright fringe xn = n ® d Dark Fringe Dark fringe occurs due to destructive interference. λ  For destructive interference path difference should be odd multiple of. 2 λ ∴ Path difference δ = (2m –1) 2 (2m − 1)Dλ The distance of the mth dark fringe from the central bright fringe xm = 2d FRINGE WIDTH The distance between two successive second bright bright or dark fringe is known as fringe width. second dark β first bright (n + 1)Dλ nDλ = β x n +1 − x n = − d d (n+1) th first dark β central fringe dark fringe x= 0 n th central bright β Dλ Fringe Width β = dark fringe xn+1 first dark 2 d xn first bright β second dark ANGULAR FRINGE WIDTH S1 fringe α angular fringe β widt width β λ  β λ h Angular fringe width α= , α=  D = d  D d   S2 D 70  Physics : Wave Optics ® Pre-Medical nλ D The distance of n bright fringe from the central bright fringe x n = = nβ th  d λD λD  The distance between n1 and n2 bright fringe x n2 − x n1 = n2 − n1 = (n2 − n1 )β d d (2m − 1)Dλ (2m − 1)β  = The distance of mth dark fringe from central fringe xm = 2d 2 Dλ (2m − 1)Dλ (2m − 1)β  The distance of nth bright fringe from mth dark fringe x n − x m = n − = nβ − d 2d 2  (2m − 1)  x n − x m = n − β  2  GOLDEN KEY POINTS  If the whole apparatus is immersed in a liquid of refractive index µ, then wavelength of light ® λ λ' = since µ >1 so λ' < λ ⇒wavelength will decrease. Hence fringe width (β ∝ λ ) will decrease µ ⇒ fringe width in liquid β' = β/µ. Angular width will also decrease.  With increase in distance between slit and screen D, angular width of maxima does not change, fringe width β increases linearly with D but the intensity of fringes decreases.  If an additional phase difference of π is created in one of the wave then the central fringe becomes dark.  When wavelength λ1 is used to obtain a fringe n1. At the same point wavelength λ2 is required to obtain a fringe n2 then n1λ1= n2 λ2  When waves from two coherent sources S1 and S2 interfere in space the shape of the fringe is hyperbolic with foci at S1 and S2, if the distance D is very large compared to the fringe width, the fringes will be very nearly straight lines. Illustrations Illustration 7 Laser light of wavelength 630 nm incident on a pair of slits produces an interference pattern in which the fringes are separated by 8.1 mm. A second laser light produces an interference pattern in which the fringes are separated by 7.2 mm. Calculate the wavelength of the second light. Solution λD β λ β 7.2 Fringe separation is given by β = i.e. 2 = 2 ⇒ λ2 = 2 × λ1 = × 630 = 560 nm d β1 λ1 β1 8.1 Illustration 8 A double slit is illuminated by light of wavelength 6000 Å. The slits are 0.1 cm apart and the screen is placed one metre away. Calculate : th (i) The angular position of the 10 maximum in radians and (ii) separation between the two adjacent minima. 71 ®  Physics : Wave Optics Pre-Medical Solution (i) λ = 6000 Å = 6 × 10–7 m, d = 0.1 cm = 1 × 10–3 m, D = 1m, n = 10 nλ 10 × 6 × 10−7 Angular position θn = = = 6 × 10−3 rad. d 10−3 (ii) Separation between two adjacent minima = fringe width β λD 6 × 10−7 × 1 β= = = 6 × 10– 4 m= 0.6 mm d 1 × 10−3 Illustration 9 In Young's double slit experiment the fringes are formed at a distance of 1m from double slits of separation 0.12 mm. Calculate (i) The distance of 3rd dark band from the centre of the screen. ® (ii) The distance of 3rd bright band from the centre of the screen, given λ = 6000Å Solution Dλ (i) x 'm (2m − 1) For mth dark fringe = given, D = 1m = 100 cm, d = 0.12 mm = 0.012 cm 2d (2 × 3 − 1) × 100 × 6 × 10−5 x 'm = = 1.25 cm [ m = 3 and λ = 6 × 10–5 cm] 2 × 0.012 nDλ 3 × 100 × 6 × 10−5 (ii) For nth bright fringe xn = ⇒ xn = = 1.5 cm [n = 3] d 0.012 Illustration 10 In Young's double slit experiment the two slits are illuminated by light of wavelength 5890Å and the distance between the fringes obtained on the screen is 0.2°. The whole apparatus is immersed in water, then find 4 out angular fringe width, (refractive index of water = ). 3 Solution λ α λ λ α λ 0.2 × 3 α air= ⇒ α air = 0.2°⇒ α ∝ λ ⇒ w = w ⇒ λw = air ⇒ αw = air = = 0.15° d α air λ air µ µ.λ 4 Illustration 11 The path difference between two interfering waves at a point on screen is 171.5 times the wavelength. If the path difference is 0.01029 cm. Find the wavelength. Solution 343 = Path difference = 171.5 λ λ = odd multiple of half wavelength. It means dark finge is observed 2 343 0.01029 × 2 = According to question 0.01029 λ ⇒ λ= = 6 × 10−5 cm ⇒ λ = 6000 Å 2 343 72  Physics : Wave Optics ® Pre-Medical Illustration 12 In young's double slit interference experiment, the distance between two sources is 0.1/π mm. The distance of the screen from the source is 25 cm. Wavelength of light used is 5000 Å. Then what is the angular position of the first dark fringe ? Solution β λ λD The angular position θ= = (∴ β = ) The first dark fringe will be at half the fringe width from the D d d mid point of central maximum. Thus the angular position of first dark fringe will be- θ 1  λ  1  5000 × π  180 α= =  =  −3 × 10−10  = 0.45°. 2 2  d  2  0.1 × 10  π BEGINNER'S BOX-2 ® 1. In Young's double slit experiment, the slits are 2mm apart and are illuminated with a mixture of two wavelength, λ = 7500 Å and λ' = 9000 Å. At what minimum distance from the common central bright fringe on a screen 2 m from the slits will a bright fringe from one interference pattern coincide with a bright fringe from the other ? 2. In a Young's double slit experiment the angular width of fringe formed on a distant screen is 0·1 radian. Find the distance between the two slits if wavelength of light used is 6000Å. 3. Two slits in Youngs experiment are 0·02 cm apart. The interference fringes for light of wavelength 6000Å are formed on a screen 80cm away. Calculate the distance of the fifth bright fringe. 4. In Young's double slit experiment, two slits are separated by 3mm distance and illuminated by light of wavelength 480 nm. The screen is at 2m from the plane of the slits. Calculate the separation between the 8th bright fringe and the 3rd dark fringe obtained with respect to central Bright fringe. 5. In a double slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from slits. If the screen is moved by 5 × 10–2 m towards the slits, the change in fringe width is 3× 10–5 m. If the distance between slits is 10–3 m. Calculate the wavelength of light used. 6. In young's experiment for interference of light the slits 0.2 cm apart are illuminated by yellow light (λ = 5896 Å). What would be the fringe width on a screen placed 1m from the plane of slits. What will be the fringe width if the system is immersed in water. (µ w = 4/3) ? 7. The distance between the coherent source is 0.3 mm and the screen is 90 cm from the sources. The second dark band is 0.3 cm away from central bright fringe. Find the wavelength and the distance of the fourth bright fringe from central bright fringe. 8. In Young's double slit experiment, using light of wavelength 400 nm, interference fringes of width 'X' are obtained. The wavelength of light is increased to 600 nm and the separation between the slits is halved. If one wants the observed fringe width on the screen to be the same in the two cases, find the ratio of the distance between the screen and the plane of the slits in the two arrangements. 9. In a Young's double slit experiment, light has a frequency of 6 × 1014 Hz. The distance between the centres of adjacent bright fringes is 0.75 mm. If the screen is 1.5 m away then find the distance between the slits. 10. In a Young's experiment, the width of the fringes obtained with light of wavelength 6000 Å is 2.0 mm. What will be the fringe width, if the entire apparatus is immersed in a liquid of refractive index 1.33 ? 73 ®  Physics : Wave Optics Pre-Medical 4. OPTICAL PATH LENGTH Optical path length (OPL) also known as optical length or optical distance is the product of the geometric length of the optical path followed by light and the refractive index of homogeneous medium through which a light ray propagates. time taken by light to travel in the given medium ∆x ∆x × µ µ =t = Light v c In same time distance travelled in vacuum is d = c ×t µ ∆x d = c × ∆x × c d = µ∆x optical path length 5. EFFECT OF THIN FILMS IN YDSE When a glass plate of thickness t and refractive index µ is placed in front of one of the slits in YDSE then the central fringe shifts towards that side in which glass plate is placed because extra path difference is ® introduced by the glass plate. In the path S1P distance travelled by wave in air = (S1 P – t) shifted central t fringe P x S1 d O central S2 path difference (µ–1)t fringe D Distance travelled by wave in the sheet = t Time taken by light to reach up to point P will be same from S1 and S2 S2 P S1 P − t t S P S P + (µ − 1)t = + ⇒ 2 = 1 ⇒ S2 P = S1 P + ( µ –1)t⇒ S2 P – S1 P = ( µ –1)t c c c/µ c c 2π (µ − 1)t Path difference = ( µ –1)t ⇒ Phase difference φ = (µ − 1)t ; Number of fringes displaced = λ λ D(µ − 1)t  xd  Distance of shifted fringe from central fringe x = d  D = (µ − 1)t    β(µ − 1)t Dλ ∴ x= and β = λ d GOLDEN KEY POINTS  If a glass plate of refractive index µ1 and µ2 having same thickness t is placed in the path of rays coming from S1 and S2 then path difference ∆x = (µ1 − µ2 )t β(µ1 − µ2 )t β D  Distance of displaced fringe from central fringe x =  = λ λ d BEGINNER'S BOX-3 1. On placing a thin sheet of mica of thickness 12 × 10–7 m in the path of one of interferring beams in a Young's experiment, it is found that the central bright band shifts a distance equal to the width of a bright fringe. If the wavelength of light used is 6 × 10–7 m. then find refractive index of mica. 2. A central fringe of the interference produced by light of wavelength 6000Å is shifted to the position of 5th bright fringes by introducing a thin glass plate of refractive index 1.5. Calculate the thickness of the plate. 74  Physics : Wave Optics ® Pre-Medical DIFFRACTION OF LIGHT Bending of light rays from sharp edges of an opaque obstacle or aperture and its spreading in the geometrical shadow region is defined as diffraction of light or deviation of light from its rectilinear propogation tendency is defined as diffraction of light. diffraction from obstacle diffraction from aperture ®  Diffraction was discovered by Grimaldi.  Theoretically explained by Fresnel  Diffraction is possible in all type of waves means mechanical or electromagnetic waves shows diffraction. Diffraction depends on two factors : (i) Size of obstacles or aperture (ii) Wavelength of the wave λ a a aperture obstacle Condition of diffraction : Size of obstacle or aperture should be nearly equal to the wavelength of light a i.e. λ~ a ~ 1 λ If size of obstacle is much greater than wavelength of light, then rectilinear motion of light is observed.  It is practically observed when size of aperture or obstacle is greater than 50 λ. The obstacle or aperature does not shows diffraction.  Wavelength of light is of the order 10–7 m. In general obstacle of this wavelength is not present so light rays does not show diffraction and it appears to travel in straight line Sound wave shows more diffraction as compared to light rays because wavelength of sound is large (16 mm to 16m). So it is generally diffracted by the objects of our daily life.  Diffraction of ultrasonic waves is also not observed easily because their wavelength is of the order of about 1 cm. Diffraction of radio waves is very conveniently observed because of its very large wavelength (2.5 m to 250 m). X-rays can be diffracted easily by crystals. It was discovered by Laue. sound sound diffraction of sound from a window 75 ®  Physics : Wave Optics Pre-Medical FRAUNHOFER DIFFRACTION In fraunhofer diffraction both source and screen are effectively at infinite distance from the diffracting device and pattern is the image of source modified by diffraction effects. Diffraction at single slit, double slit are the example of Fraunhofer diffraction. 1. FRAUNHOFER DIFFRACTION DUE TO SINGLE SLIT AB is single slit of width a, Plane wavefront is incident on a slit AB. Secondary wavelets coming from every part of AB reach the axial point O in same phase forming the central maxima. The intensity of central maxima is maximum in this diffraction. where θn represents direction of nth minima, Path difference AB' = a sin θn L1 A B' L2 θn ® S a θn O θn x θn B P D nλ for nth minima a sinθn = nλ ∴ sin θn ≈ θn = (if θn is small) a λ   When path difference between the secondary wavelets coming from A and B is nλ or 2n   or even 2  λ multiple of then minima occurs 2 λ  For minima a sin θn =2n   where n = 1, 2, 3... 2  λ  When path difference between the secondary wavelets coming from A and B is (2n+1) or odd multiple of 2 λ then maxima occurs 2 λ For maxima asin θn = (2n + 1) where n = 1, 2, 3... 2 n = 1 → first maxima and n = 2 → second maxima  Alternate ordered minima and maxima occurs on both sides of central maxima. For nth minima th If distance of n minima from central maxima = xn distance of slit from screen = D , width of slit = a P' xn θn 2nλ nλ O P Path difference δ = a sinθn = ⇒ sin θn = θn 2 a P" xn D In ∆ POP' tan θn = If θn is small⇒ sin θn ≈ tan θn ≈ θn D 76  Physics : Wave Optics ® Pre-Medical nλ D xn nλ xn = ⇒ θ= n = First minima occurs on both sides of central maxima. a D a Dλ x λ For first minima x = and θ= = a D a 2Dλ  Linear width of central maxima wx = 2x ⇒ wx = a 2λ  Angular width of central maxima wθ = 2θ = a SPECIAL CASE Lens L2 is shifted very near to slit AB. In this case distance between slit and screen will be nearly equal to the ® x n nλ nλ f focal length of lens L2 (i.e. D ≈ f ) θn = = ⇒ xn = f a a A L2 P B D f 2λf 2x 2λ wx = and angular width of central maxima wθ = = a f a Fringe width : Distance between two consecutive maxima (bright fringe) or minima (dark fringe) is known as fringe width. Fringe width of central maxima is doubled then the width of other maxima i.e., λD nλD λD β = xn + 1 – xn = (n + 1) – = (other than central maxima) a a a Intensity curve of Fraunhofer's diffraction I0 Intensity I0 I0 I0 I0 22 22 61 61 −2λ/a −λ/a O λ/a 2λ/a Angle θ 77 ®  Physics : Wave Optics Pre-Medical DIFFERENCE BETWEEN INTERFERENCE AND DIFFRACTION (FOR FRAUNHOFER SINGLE SLIT) : Interference Fraunhofer diffraction (1) It is the phenomenon of superposition (1) It is the phenomenon of superposition of two waves coming from two different of two waves coming from two different coherent sources. parts of the same wave front. (2) In interference pattern, all bright lines (2) All bright lines are not equally bright are equally bright and equally spaced. and equally wide. Brightness and width goes on decreasing with the angle of diffraction. (3) All dark lines are totally dark (3) Dark lines are perfectly dark. Their contrast with bright lines and width goes ® on decreasing with angle of diffraction. (4) In interference bands are large in number (4) In diffraction bands are few in number. GOLDEN KEY POINTS  The width of central maxima ∝ λ, that is, more for red colour and less for blue. i.e., wx ∝ λ as λblue < λred ⇒ wblue < wred  For obtaining the fraunhofer diffraction, focal length of second lens (L2) is used. wx ∝ λ ∝ f ∝ 1/a width will be more for narrow slit  By decreasing linear width of slit, the width of central maxima increase. Illustrations Illustration 13 –5 Light of wavelength 6000Å is incident normally on a slit of width 24 × 10 cm. Find out the angular position of second minimum from central maximum ? Solution a sinθ = 2λ given λ = 6 × 10–7 m, a = 24 × 10–5 × 10–2 m 2λ 2 × 6 × 10−7 1 sinθ = = = a 24 × 10−7 2 ∴ θ = 30° 78  Physics : Wave Optics ® Pre-Medical Illustration 14 Light of wavelength 6328Å is incident normally on a slit of width 0.2 mm. Calculate the angular width of central maximum on a screen distance 9 m ? Solution given λ = 6.328 × 10–7 m, a = 0.2 × 10–3 m 2λ 2 × 6.328 × 10−7 6.328 × 10−3 × 180 wθ = = radian = = 0.36° a 2 × 10−4 3.14 Illustration 15 Light of wavelength 5000Å is incident on a slit of width 0.1 mm. Find out the width of the central bright line on a screen distance 2m from the slit ? Solution 2 fλ 2 × 2 × 5 × 10−7 wx = = = 20 mm ® a 10−4 Illustration 16 The Fraunhofer diffraction pattern of a single slit is formed at the focal plane of a lens of focal length 1m. The width of the slit is 0.3 mm. If the third minimum is formed at a distance of 5 mm from the central maximum then calculate the wavelength of light. Solution nfλ ax n 3 × 10−4 × 5 × 10−3 xn = ⇒ λ= = = 5000 Å [ n = 3] a fn 3 ×1 Illustration 17 Find the half angular width of the central bright maximum in the Fraunhofer diffraction pattern of a slit of width 12 × 10–5 cm when the slit is illuminated by monochromatic light of wavelength 6000 Å. Solution λ  sin θ = θ = half angular width of the central maximum. a λ 6 × 10−5 a = 12 × 10 cm, λ = 6000 Å = 6 × 10 cm ∴ = 0.50 ⇒ θ = 30° –5 –5 sin θ= = a 12 × 10−5 Illustration 18 Light of wavelength 6000 Å is incident on a slit of width 0.30 mm. The screen is placed 2 m from the slit. Find (a) the position of the first dark fringe and (b) the width of the central bright fringe. Solution The first fringe is on either side of the central bright fringe. here n = ±1, D = 2 m, λ = 6000 Å = 6 × 10–7 m x ax ∴ sin θ = ⇒a = 0.30 mm = 3 × 10–4 m ⇒ a sin θ = nλ ⇒ = nλ D D nλ D 1 × 6 × 10−7 × 2  −3 (a) x= ⇒ x =±  −4  =±4 × 10 m a  3 × 10  The positive and negative signs corresponds to the dark fringes on either side of the central bright fringe. (b) The width of the central bright fringe y = 2x = 2 × 4 × 10–3 = 8 × 10–3 m = 8 mm 79 ®  Physics : Wave Optics Pre-Medical Illustration 19 A Slit of width a is illuminated by monochromatic light of wavelength 650nm at normal incidence. Calculate the value of a when - (a) the first minimum falls at an angle of diffraction of 30° (b) the first maximum falls at an angle of diffraction of 30°. Solution λ λ 650 ×10−9 650 ×10−9 (a) for first minimum sinθ1 ∴a= = = = 1.3 × 10–6 m a sin θ1 sin 30° 0.5 3λ 3λ 3 × 650 × 10 –9 (b) For first maximum sinθ1 = ∴a= = = 1.95 × 10–6 m 2a 2sin θ 2 × 0.5 Illustration 20 Red light of wavelength 6500Å from a distant source falls on a slit 0.50 mm wide. What is the distance between the first two dark bands on each side of the central bright of the diffraction pattern observed on a ® screen placed 1.8 m. from the slit. Solution Given λ = 6500Å = 65 × 10–8 m, a = 0.5 mm = 0.5 × 10–3 m., D = 1.8 m. Required distance between first two dark bands will be equal to width of central maxima. 2λD 2 × 6500 × 10−10 × 1.8 Wx = = = 468 × 10–5 m = 4.68 mm a 0.5 × 10−3 BEGINNER'S BOX-4 1. A slit of width 0·15 cm is illuminated by light of wavelength 5 × 10–5 cm and a diffraction pattern is obtained on a screen 2·1m away. Calculate the width of central maxima. 2. The light of wavelength 600nm is incident normally on a slit of width 3mm. Calculate the angular width of central maximum on a screen kept 3m away from the slit. 3. Red light of wavelength 6500Å from a distant source falls on a slit 0.50 mm wide. What is the distance between the first two dark bands on each side of the central bright of the diffraction pattern observed on a screen placed 1.8 m from the slit. 4. In a single slit diffraction experiment first minimum for λ1 = 660 nm coincides with first maxima for wavelength λ2. Calculate λ2. 2. FRESNEL DISTANCE & VALIDITY OF RAY OPTICS It is a distance upto which ray optics can be used. Beyond this distance spreading due to diffraction will be so large that we will not get clear image. It is a distance in which beam width becomes double due to diffraction. θ 2a 2λZ z = 2a a a a2 Z= λ This relation shows that for distance much smaller than Z, the spreading due to diffraction is smaller compared to the size of the beam. It becomes comparable when the distance is approximately Z. For distance much greater than Z, the spreading due to diffraction dominates over. Ray optics is not valid in this reion. Relation also shows that straight line propagation is valid in the limit of λ tending to zero. 80  Physics : Wave Optics ® Pre-Medical Illustration 21 Upto what distance we should apply ray optics (Fresnel distance) if beam width is 2mm and wave length of light 500 nm. Solution a2 4 × 10−6 = z = λ 5 × 10−7 z = 8m. 3. DIFFRACTION DUE TO DOUBLE SLIT : In the double slit experiment, we must note that the pattern on the screen is actually a superposition of single slit diffraction from each slit or hole, and the double slit interference pattern as shown in figure. It show a diffraction peak in which there appear several fringes of smaller width due to double – slit ® interference. S1 2λf a λf = a d I d a S2 f Resultant pattern a → width of slit Number of interference fringes inside diffraction pattern (central maxima) : 2λf / a 2d N= = λf a d It is not depending upon distance of screen. a → 0 (when a is very small, diffraction pattern becomes very flat) n→ ∞ This will be the case of pure interference pattern. Illustration 22 In double slit experiment if width of each slit is 0.1mm and separation between the slits is 1mm. How many interference fringes we will get inside the central maxima diffraction pattern. Solution 2λf / a 2d 2 × 10−3 n= = = = 20 λf / d a 0.1 × 10−3 81 ®  Physics : Wave Optics Pre-Medical POLARISATION Experiments on interference and diffraction have shown that light is a form of wave motion. These effects do not tell us about the type of wave motion i.e., whether the light waves are longitudinal or transverse. The phenomenon of polarization has helped to establish beyond doubt that light waves are transverse waves. UNPOLARISED LIGHT An ordinary beam of light consists of a large number of waves emitted by the atoms of the light source. Each  atom produces a wave with its own orientation of electric vector E so all direction of vibration of are equally probable. Z Y Y X ® unpolarised light unpolarised light propagating along. X-axis The resultant electromagnetic wave is a superposition of waves produced by the individual atomic sources and it is called unpolarised light. In ordinary or unpolarised light, the vibrations of the electric vector occur symmetrically in all possible directions in a plane perpendicular to the direction of propagation of light. POLARISATION The phenomenon of restricting the vibration of light (electric vector) in a particular direction perpendicular to the direction of propagation of wave is called polarisation of light. In polarised light, the vibration of the electric vector occur in a plane perpendicular to the direction of propagation of light and are confined to a single direction in the plane (do not occur symmetrically in all possible directions). After polarisation the vibrations become asymmetrical about the direction of propagation of light. POLARISER Tourmaline crystal : When light is passed through a tourmaline crystal cut parallel to its optic axis, the vibrations of the light carrying out of the tourmaline crystal are confined only to one direction in a plane perpendicular to the direction of propagation of light. The emergent light from the crystal is said to be plane polarised light. Nicol Prism : A nicol prism is an optical device which can be used for the production and detection of plane polarised light. It was invented by William Nicol in 1828. Polaroid : A polaroid is a thin commercial sheet in the form of circular disc which makes use of the property of selective absorption to produce an intense beam of plane polarised light. PLANE OF POLARISATION AND PLANE OF VIBRATION : The plane in which vibrations of light vector and the direction of propogation lie is known as plane of vibration A plane normal to the plane of vibration and in which no vibration takes place is known as plane of polarisation A plane of vibration B E plane polarised light O' unpolarised G light F plane of D C polarisation 82  Physics : Wave Optics ® Pre-Medical EXPERIMENTAL DEMONSTRATION OF POLARISATION OF LIGHT Take two tourmaline crystals cut parallel to their crystallographic axis (optic axis). A P O Ordinary Polariser light First hold the crystal A normally to the path of a beam of colour light. The emergent beam will be slightly coloured. Rotate the crystal A about PO. No change in the intensity or the colour of the emergent beam of light. Take another crystal B and hold it in the path of the emergent beam of so that its axis is parallel to the axis of the crystal A. The beam of light passes through both the crystals and outcoming light appears coloured. A B P O ® Ordinary Polariser light Now, rotate the crystal B about the axis PO. It will be seen that the intensity of the emergent beam decreases and when the axes of both the crystals are at right angles to each other no light comes out of the crystal B. A B P O Ordinary Polariser light If the crystal B is further rotated light reappears and intensity becomes maximum again when their axis are parallel. This occurs after a further rotation of B through 90°. This experiment confirms that the light waves are transverse in nature. The vibrations in light waves are perpendicular to the direction of propogation of the the wave. First crystal A polarises the light so it is called polariser. Second crystal B, analyses the light whether it is polarised or not, so it is called analyser. 1. METHODS OF OBTAINING PLANE POLARISED LIGHT  Polarisation by reflection : The simplest method to produce plane polarised light is by reflection. This method was discovered by Malus in 1808. When a beam of ordinary light is reflected from a surface, the reflected light is partially polarised. (The degree of polarisation of the polarised light in the reflected beam is greatest when it is incident at an angle called polarising angle or Brewster's angle). Unpolarised Partially polarised light light µ1 µ2 83 ®  Physics : Wave Optics Pre-Medical  Polarising angle : Polarising angle is that angle of incidence at which the reflected light is completely plane polarised light.  Brewster's Law : When unpolarised light strikes at polarising angle θP on an interface separating two medium then the reflected light is completely polarised. Also reflected and refracted rays are normal to each other. This is known as Brewster's law. In case of polarisation by reflection : (i) For i = θP refracted light is partially polarised. (ii) For i = θP reflected and refracted rays are perpendicular to each other. (iii) For i < θP or i > θP both reflected and refracted light are partially polarised. θP R D ® θP µ 1 < µ2 D R θP > 45° θP < 45° µ 1 > µ2 µ1sin θP = µ2sin r θP µ1 θP µ1sin θP = µ2sin(90–θP) µ1sin θP = µ2cos θP µ2 µ tan θP = 2 µ1 r=90–θP Illustration 23. A ray of light from a denser medium strikes a rarer medium at an angle of incidence i. If the reflected and refracted rays are mutually perpendicular to each other then what is the value of critical angle ? Solution The situation in accordance with the given problem is shown in figure. Applying Snell's law at the boundary at C, µ D sin r ' µD sin i = µR sin r' ⇒ = µ =... (i) µR sin i But according to given problem, r' + 90° + r = 180° r' + r = 90° i.e., r' = 90° – r or r' = 90° – i [as ∠r = ∠i] rarer o sin(90 − i) cos i µR r' C So equation (i) reduces to = µ = = cot i... (ii) A 90° B sin i sin i denser i r 1 µD But by definition, θC =sin −1 and from equation (ii) µ = cot i i=r µ  1  So θC =sin −1   = sin (tan i). –1  cot i  84  Physics : Wave Optics ® Pre-Medical By Refraction In this method, a pile of glass plates is formed by taking 20 to 30 microscope slides and light is made to be incident at polarising angle 57°. According Brewster's law, the reflected light will be plane polarised with vibrations perpendicular to the plane of incidence and the transmitted light will be partially polarised. Since in one reflection about 15% of the light with vibration perpendicular to plane of paper is reflected therefore after passing through a number of plates emerging light will become plane polarised with vibrations in the plane of paper. reflected light 57° S refracted light By Dichroism ® Some crystals such as tourmaline and sheets of iodosulphate of quinone have the property of strongly absorbing the light with vibrations perpendicular of a specific direction (called transmission axis) and transmitting the light with vibration parallel to it. This selective absorption of light is called dichroism. So if unpolarised light passes through proper thickness of these crystals, the transmitted light will plane polarised with vibrations parallel to transmission axis. Polaroids work on this principle. Polaroid optic axis is perpendicular to the plane of tourmaline S paper crystal Ordinary transmission light axis By scattering : When light is incident on small particles of dust, air molecules etc. s(having smaller size as compared to the wavelength of light), it is absorbed by the electrons and is re-radiated in all directions. The phenomenon is called as scattering. Light scattered in a direction at right angles to the incident light is always plane- polarised. y Unpolarised light polarised polarised light light z x Unpolarised light By Double Refraction : It was found that in certain crystals such as calcite, quartz and tourmaline, etc., incident unpolarised light splits in into two light beams of equal intensities with perpendicular polarisations. One of the rays behaves as ordinary light and is called O-ray (ordinary ray) while the other does not obey laws of refraction and is called 85 ®  Physics : Wave Optics Pre-Medical E-ray (extra-ordinary ray). This is why when an object is seen through these crystal is rotated, one image (due to E-ray) rotates around the other (due to O-ray). Canada balsam layer E– ray E- ray Unpolarized Unpolarize O- ray light light Calcite O–ray Blackened surface By using the phenomenon of double refraction and isolating one ray from the other we can obtain plane polarised light which actually happens in a Nicol-prism. Nicol-prism is made up of calcite crystal and in it E-ray is isolated from O-ray through total internal reflection of O-ray at canada balsam layer and then absorbing it at the blackened surface as shown in figure. Law of Malus ®  When a completely plane polarised light beam is incident on analyser, then intensity of emergent light varies as the square of cosine of the angle between the planes of transmission axis of the analyser and I ∝ cos θ ⇒ I = I0 cos θ 2 2 the polarizer. plane of polariser A A θ θ θ (i) If θ = 0° then I = I0 maximum value (Parallel arrangement) (ii) If θ = 90° then I = 0 minimum value (Crossed arrangement)  If plane polarised light of intensity I0(= KA2) is incident on a polaroid and its vibrations of amplitude A make angle θ with transmission axis, then the component of vibrations parallel to transmission axis will be Acosθ while perpendicular to it will be Asinθ. Polaroid will pass only those vibrations which are parallel to transmission axis i.e. Acos θ,  I0 ∝ A2 I = K(Acosθ) = KA cos θ 2 2 2 So the intensity of emergent light  If an unpolarised light is converted into plane polarised light its intensity becomes half.  If light of intensity I1, emerging from one polaroid called polariser is incident on a second polaroid (called analyser) the intensity of light emerging from the second polaroid is I2 = I1 cos θ θ = angle between the transmission axis of the two polaroids. 2 GOLDEN KEY POINTS  Our eyes are nearly insensitive to polarisation of light, but this is not universally true for animals.  If the angle of incidence is 0° or 90° the reflected beam is unpolarised.  A Nicol prism is made by cutting a calcite crystal in a certain way. 86  Physics : Wave Optics ® Pre-Medical Illustrations Illustration 24 For a given medium, the polarising angle is 60°. What will be the critical angle for this medium? Solution Here ip = 60° Thus, µ = tan ip = tan 60° = 3 If ic is the critical angle for the medium. 1 1 1 µ= or sin i c= = or sin ic = 0.5774 or ic = 35°16' sin i c µ 3 Illustration 25 A ray of light is incident on the surface of glass plate of refracive index 1.5 at polarising angle. What is the –1 angle of refraction? (tan (1.5) = 56° 19') Solution ® As µ = tan ip, tan ip = 1.5 or ip = tan–1 (1.5) or ip = 56°19' As r + ip = 90°, r = 90° – ip = 33°41' BEGINNER'S BOX-5 1. Refractive index of water is 1.33. Calculate the angle of polarisation for light reflected from the surface of a lake. (tan–1 1.33 = 53°.4') 2. A ray of light strikes a glass plate at an angle of 60°. If the reflected and the refracted rays are perpendicular to each other, find the index of refraction of glass. 3. A parallel beam of light is incident at an angle of 60° on a plane glass surface and the reflected beam is completely polarised. (a) What is the angle of refraction in glass? (b) What is the refraction index of glass? 4. Light reflected from the surface of a glass plate of refractive index 1.732 is linearly polarised. Calculate the angle of refraction in glass. 5. Critical angle for a certain wavelength of light in case of glass is 40°. Find the polarising angle and angle of 2 refraction in glass corresponding to this. (sin 40° = ) 3 ANSWER'S BEGINNER'S BOX-1 BEGINNER'S BOX-3 25 1. 1·5 2. 6 × 10–6 m 1. (a) 5I (b) 7.8I, (c) I (d) 9I 2. 16 BEGINNER'S BOX-4 BEGINNER'S BOX-2 1. 1·4 mm 2. 0.023° 1. 4.5 mm 2. 6µm 3. 4.68 mm 4. 440 nm. 3. 1·2 cm 4. 1.76 × 10–3 m BEGINNER'S BOX-5 5. 6 × 10–7 m 1. 53°4' 2. 1.732 6. 0.2948 mm, 0.2211 mm 3. (a) 30° (b) 1.732 4. 30° 7. 8 × 10–3 m, λ = 0.66 × 10–6 m 5. 57.3°, 32.7° 3 8. 9. 10–3 m. 10. 1.5 mm 1 87