Wave Optics JEE Engineering PDF

Welcome to Wave optics Does Light always exhibit Rectilinear Propagation?  Huygens’ Principle Explanation for bending of light Wave theory of light Chri...

Welcome to Wave optics Does Light always exhibit Rectilinear Propagation?  Huygens’ Principle Explanation for bending of light Wave theory of light Christiaan Huygens Huygens Wave theory of light Wavefront  Wavefront  All points on a particular circle will oscillate with the same phase.  Wavefront is the locus of all points at which the wave disturbance is in the same phase.  Point source creates wave traveling in all directions according to equation. 𝑦 = 𝐴sin(𝜔𝑡 ± 𝑘𝑥)  Points 𝐴, 𝐵, 𝐶 have same phase 𝜔𝑡 ± 𝑘𝑥1.  In 3𝐷, for a point source, the wavefronts are spherically symmetric.  Direction of Propagation of Light Spherical Wavefront The direction of propagation of light wave is perpendicular to the wavefront at any given point.  Properties of Wavefront  Time taken by each ray to propagate from one wavefront to the next wavefront in any medium remains same. 𝑐  Refractive index, 𝑛 = 𝑣 𝑛1 𝑣2 When medium changes, = 𝑛2 𝑣1 ⟹ Speed of light changes. 𝜆  Hence, distance covered by the light in each time interval also changes.  Distance between two consecutive wavefronts = 𝜆. Spherical Wavefront  Spherical wavefront is observed for a point source. Direction of propagation of light: Radially outward and perpendicular to the wavefront at any given point. Planar Wavefront Planar wavefronts are observed when the source is at infinity. Direction of propagation of light: Perpendicular to the plane. Cylindrical Wavefront Cylindrical wavefront is observed for a line source. Direction of propagation of light: Radially outward and perpendicular to the wavefront at any given point.  If the distance between the wavefronts in the medium with refractive S index 𝑛2 is 𝑑2 , then what will be the distance 𝑑1 between the wavefronts in the medium with refractive index 𝑛1 ? We known that, 𝑐 𝑛2 𝑣 𝑛= ⇒ ൗ𝑛1 = 1ൗ𝑣2 𝑣 𝜆 = 𝑣ൗ𝑓 𝜆1 𝑣 𝑛 ⇒ ൗ𝜆 = 1ൗ𝑣2 = 2ൗ𝑛1 2 Here, 𝜆1 = 𝑑1 , and 𝜆2 = 𝑑2 Hence, 𝑑1 𝑛2 = 𝑑2 𝑛1 𝑛2 ⇒ 𝑑1 = 𝑑2 𝑛1  Huygens’ Principle Primary  Every point on the wavefront acts as a Wavefront point source called secondary wave source and generates secondary wavelets.  The common tangent to the secondary wavelets in the forward direction gives Primary the secondary wavefront. Source  Intensity is maximum in forward direction and zero in backward direction. Secondary Secondary Wavelets Wavefront  Shape of Wavefront Object at infinity for a convex lens Object at infinity for a concave lens  Shape of Wavefront Object at infinity for a concave mirror Object at infinity for a prism  Huygens’ Principle - Law of Reflection Reflection of a plane wave by a plane surface Secondary Reflected Incident Wavelet Wavefront 𝐶𝐷 Wavefront 𝐴𝐵 From 𝐴 at time = 𝑡 at 𝑡 = 0 𝐷 𝐵 𝐵𝐶 = 𝑣𝑡 𝑖 𝐷𝐴 = 𝑣𝑡 𝐴 𝐶 𝑡 𝑡=0 Reflecting plane surface  Huygens’ Principle - Law of Reflection Reflection of a plane wave by a plane surface 𝐷 𝐵 𝑟 𝑖 𝐴 𝐶 𝐴 𝐶 From ∆𝐴𝐷𝐶 and ∆𝐶𝐵𝐴: ∠𝐴𝐷𝐶 = ∠𝐴𝐵𝐶 = 90° 𝐴𝐷 = 𝐵𝐶 = 𝑣𝑡 𝐴𝐶 = 𝐴𝐶  Angle of incidence is the angle between the incident wavefront and the reflecting surface. ∴ ∆𝐴𝐷𝐶 ≅ ∆𝐶𝐵𝐴 [𝑅. 𝐻. 𝑆 Congruency] ∠𝐵𝐴𝐶 = 𝑖 → Angle of incidence ⟹ ∠𝑖 = ∠𝑟  Angle of reflection is the angle between the reflected wavefront and the reflecting surface. ∠𝐷𝐶𝐴 = 𝑟 → Angle of reflection  Huygens’ Principle - Law of Refraction Incident 𝑛1 < 𝑛2 Medium 𝑛1 Wavefront 𝐴𝐵 at 𝑡 = 0 𝐵 From ∆𝐴𝐵𝐶: From ∆𝐴𝐷𝐶: 𝑖 𝑣1𝑡 𝐵𝐶 𝑣1 𝑡 𝐴𝐷 𝑣2 𝑡 sin 𝑖 = = sin 𝑟 = = 𝐴𝐶 𝐴𝐶 𝐴𝐶 𝐴𝐶 𝑡=0 𝐴 𝑖 𝐶 𝑟 𝑣2𝑡 sin 𝑖 𝑣1 𝑛2 𝑟 𝑟 = = = Constant sin 𝑟 𝑣2 𝑛1 𝐷 Medium 𝑛2 Secondary Refracted Wavelet Wavefront 𝐶𝐷 From A at 𝑡 = 𝑡  Light waves travel in vacuum, along the 𝑥 − axis. Which of the following may represent the wavefronts? Solution: A 𝑥=𝑐 The direction of propagation → 𝑖Ƹ ⟹ The direction of wavefront → ⊥ 𝑡𝑜 𝑖Ƹ B 𝑦=𝑐 In the given options, the plane ⊥ to 𝑖Ƹ is represented by 𝑥 = 𝑐. C 𝑧=𝑐 D 𝑥+𝑦+𝑧 =𝑐  Does Light always exhibit Rectilinear Propagation? Hole Screen Light behaves like a ray (i.e., exhibits rectilinear propagation) for: Obstacle dimensions ≫𝜆𝑙𝑖𝑔ℎ𝑡 Hole Screen Light behaves like a wave for: Obstacle dimensions ≈ 𝜆𝑙𝑖𝑔ℎ𝑡  OPTICS Study of the behavior and properties of light GEOMETRICAL OPTICS WAVE OPTICS Explains refraction and reflection, Explains reflection and refraction, but not interference, diffraction as well as interference, diffraction and polarization and polarization  Interference Interference is the phenomenon in which two waves superpose to form the resultant wave of lower, higher or same amplitude.  Superposition Principle “When two or more waves cross at a point, the displacement at that point is equal to the vector sum of the displacements of individual waves” 𝑦 𝑦𝑛𝑒𝑡 𝑦2 𝑦1 𝑥 𝑦Ԧ𝑛𝑒𝑡 = 𝑦Ԧ1 + 𝑦Ԧ2 + 𝑦Ԧ3 … … + 𝑦Ԧ𝑛 𝑂  Resultant of Waves Case 1: When two crest meet Case 2: When crest and trough meet  Phase Difference and Path Difference Path difference (∆𝑥): Difference in the 𝑦1 = 𝐴1 sin 𝜔𝑡 + 𝑘𝑥 path traversed by the two waves. Source 𝑆1 Phase difference (𝛿 ): Difference in the phase angle of the two waves. We know that, 𝑦𝑛𝑒𝑡 = 𝑦1 + 𝑦2 For a path difference 𝜆, phase difference = 2𝜋 𝑃 So, for path difference ∆𝑥, phase difference 2𝜋 = 𝜆 ∆𝑥. Source 𝑆2 2𝜋 𝑦2 = 𝐴2 sin 𝜔𝑡 + 𝑘𝑥 + 𝛿 𝛿= ∆𝑥 𝜆 Combination of Waves 𝑦𝑛𝑒𝑡 = 𝐴1 sin 𝜔𝑡 + 𝑘𝑥 + 𝐴2 sin 𝜔𝑡 + 𝑘𝑥 + 𝛿 𝑦𝑛𝑒𝑡 = 𝐴𝑛𝑒𝑡 sin 𝜔𝑡 + 𝑘𝑥 + 𝛼 𝐴𝑛𝑒𝑡 = 𝐴12 + 𝐴22 + 2𝐴1 𝐴2 cos 𝛿 𝐴2 sin 𝛿 tan 𝛼 = 𝐴1 + 𝐴2 cos 𝛿 If 𝐴1 = 𝐴2 = 𝐴 𝛿 𝐴𝑛𝑒𝑡 = 2𝐴 cos 2  Constructive Interference Interference that produces maximum possible amplitude (or maximum intensity) is called constructive interference. 𝐴𝑛𝑒𝑡 = 𝐴12 + 𝐴22 + 2𝐴1 𝐴2 cos 𝛿 For maximum amplitude: cos 𝛿 = 1 ⇒ 𝛿 = 2𝑛𝜋 𝐴 = 𝐴𝑚𝑎𝑥 = 𝐴1 + 𝐴2 2𝜆 3𝜆 2𝜋 𝑃 𝛿= ∆𝑥 𝜆 2𝜋 2𝑛𝜋 = × (∆𝑥) 𝜆 Path Difference = 3𝜆 − 2𝜆 = 𝜆 Path difference = ∆𝑥 = 𝑛𝜆 Constructive Interference  Constructive Interference Constructive interference occurs when the crest and trough of one wave overlaps with the crest and trough of another wave. 𝑦 𝑦 𝑂 𝑥 𝑦1 = 𝐴 sin(𝜔𝑡 + 𝑘𝑥) + ≡ 𝑥 𝑦 𝑂 𝑦𝑛𝑒𝑡 = 2𝐴 sin(𝜔𝑡 + 𝑘𝑥) 𝑥 𝑦2 = 𝐴 sin(𝜔𝑡 + 𝑘𝑥)  Destructive Interference Interference that produces minimum possible amplitude (or minimum intensity) is called destructive interference. 𝐴𝑛𝑒𝑡 = 𝐴12 + 𝐴22 + 2𝐴1 𝐴2 cos 𝛿 For minimum amplitude: cos 𝛿 = −1 ⇒ 𝛿 = (2𝑛 + 1)𝜋 𝐴 = 𝐴𝑚𝑖𝑛 = 𝐴1 − 𝐴2 7.25𝜆 9.75𝜆 2𝜋 𝛿= ∆𝑥 𝜆 𝑃 2𝜋 (2𝑛 + 1)𝜋 = (∆𝑥) 𝜆 Path Difference = 9.75𝜆 − 7.25𝜆 = 2.5𝜆 2𝑛 + 1 𝜆 Path difference = ∆𝑥 = 2 Destructive Interference  Destructive Interference Destructive interference is resulted when the crest of one wave overlaps with the trough of another wave. 𝑦 𝑦 𝑂 𝑥 𝑦1 = 𝐴 sin(𝜔𝑡 + 𝑘𝑥) + ≡ 𝑂 𝑥 𝑦 𝑦=0 𝑂 𝑥 𝑦2 = 𝐴 sin(𝜔𝑡 + 𝑘𝑥 + 𝜋)  Two-point light sources 𝑆1 and 𝑆2 are separated by a distance of 4.2𝜆. If an observer standing at the centre 𝐶 of the two sources starts moving towards T 𝑆2 , then find the minimum distance travelled by the observer to meet the first maxima. Solution: For maxima at 𝑃: ∆𝑥 = 𝜆 ⇒ 𝑆1 𝑃 − 𝑆2 𝑃 = 𝜆 2.1𝜆 2.1𝜆 𝑥 ⇒ (2.1𝜆 + 𝑥) − (2.1𝜆 − 𝑥) = 𝜆 ⇒ 2𝑥 = 𝜆 𝑆1 𝐶 𝑃 𝑆2 ∆𝑥 = 0 ∆𝑥 = 𝜆 𝑥 = 0.5𝜆  Coherent and Incoherent Sources Coherent Sources Incoherent Sources Same wavelength Different wavelength Same frequency Different frequency Constant phase difference Varying phase difference Coherent Waves Incoherent Waves Note: We will always consider coherent sources in our discussion.  Young’s Double Slit Experiment Particle nature of light Wave nature of light  Young’s Double Slit Experiment According to Huygens’ principle, the sources 𝑆1 and 𝑆2 will behave as independent sources. Light source must be monochromatic. Sources 𝑆1 and 𝑆2 must be coherent. Width of the slit is comparable to the wavelength of light. Waves coming from sources 𝑆1 and 𝑆2 will interference and obtain different 𝐷 interference pattern on the screen.  Young’s Double Slit Experiment 𝑆𝑆1 = 𝑆𝑆2 (∴ No path difference till slits.) ′𝑑′ is distance between slits. 𝑃 ′𝐷′ is distance between screen and slits 𝑦 plane. 𝑑 𝑂 𝜃 𝐵 𝑂𝐵 is the central line. Consider point 𝑃 at a distance 𝑦 from central line 𝑂𝐵. 𝐷 ∠𝑃𝑂𝐵 = 𝜃. 𝑆2 𝑃 > 𝑆1 𝑃 → 𝑆2 𝑃 − 𝑆1 𝑃 = ∆𝑥 ∆𝑥 is the path difference.  Young’s Double Slit Experiment Path difference at any general point 𝑃, ∆𝑥 = 𝑆2𝑃 − 𝑆1𝑃 𝑃 2 2 𝑦 𝑑 𝑑 ∆𝑥 = 𝑦+ + 𝐷2 − 𝑦− + 𝐷2 2 2 𝜃 𝑑/2 𝑑 𝑂 𝐵 𝑑/2 Approximation 1: 𝐷 ≫ 𝑑 Approximation 2: 𝜃 is very small 𝐷 𝑦𝑑 ∆𝑥 ≈ 𝐷 Young’s Double Slit Experiment ∆𝑥 = 𝑆2 𝐴 = 𝑑 sin 𝜃 𝑦𝑑 𝜃 is very small → sin 𝜃 ≈ tan 𝜃 ∆𝑥 ≈ 𝐷 𝑑𝑦 ∆𝑥 ≈ 𝑑 tan 𝜃 = 𝐷  Condition for Constructive Interference For Maxima (Constructive interference): 𝑑𝑦 ∆𝑥 = 𝑛𝜆 ⇒ = 𝑛𝜆 𝐷 𝑃 Where, 𝑛 = 0, ±1, ±2, ±3, … 𝑆1 𝑦 𝑛𝜆𝐷 𝑦= 𝑑 𝑂 𝑑 𝐵 ∆𝑥 𝑆2 𝑛 = 0 corresponds to the central maxima. → 𝑦 = 0 𝐷 𝜆𝐷 𝑛 = ± 1 correspond to the 1𝑠𝑡 maxima. → 𝑦 = ± 𝑑 2𝜆𝐷 𝑛 = ± 2 correspond to the 2𝑛𝑑 maxima. → 𝑦 = ± 𝑑  Condition for Destructive Interference For Minima (Destructive interference): 1 𝑑𝑦 1 ∆𝑥 = 𝑛 + 𝜆 ⇒ = 𝑛+ 𝜆 2 𝐷 2 𝑆1 𝑃 Where, 𝑛 = 0, ±1, ±2, ±3, … 𝑦 𝑑 𝑂 𝐵 1 𝜆𝐷 ∆𝑥 𝑦= 𝑛+ 𝑆2 2 𝑑 𝐷 𝜆𝐷 𝑛 = 0, −1 correspond to the 1𝑠𝑡 minima. → 𝑦 = ± 2𝑑 3𝜆𝐷 𝑛 = 1, −2 correspond to the 2𝑛𝑑 minima. → 𝑦=± 2𝑑  In YDSE, white light is passed through the double slit and interference pattern is observed on a screen 2.5 𝑚 away. The separation between the slits is 0.5 𝑚𝑚. The first violet and red maxima are formed at distances of 2 𝑚𝑚 and 3.5 𝑚𝑚 away from the central white maxima, respectively. The wavelengths of red and violet light, respectively, are: Given: 𝐷 = 2.5 𝑚, 𝑑 = 0.5 𝑚𝑚, 𝑦𝑣𝑖𝑜𝑙𝑒𝑡 = 2 𝑚𝑚, 𝑦𝑟𝑒𝑑 = 3.5 𝑚𝑚 To find: 𝜆𝑣𝑖𝑜𝑙𝑒𝑡 , 𝜆𝑟𝑒𝑑 Solution: Distance of first maxima from the central maxima is given by: 𝐷𝜆 𝑦= 𝑑 For violet light, 𝑦 = 2 𝑚𝑚 For red light, 𝑦 = 3.5 𝑚𝑚 2.5 × 𝜆𝑣𝑖𝑜𝑙𝑒𝑡 2.5 × 𝜆𝑟𝑒𝑑 ⇒ 2 × 10−3 = ⇒ 3.5 × 10−3 = 0.5 × 10−3 0.5 × 10−3 2 ⇒ 𝜆𝑣𝑖𝑜𝑙𝑒𝑡 = × 10−6 𝑚 ⇒ 𝜆𝑟𝑒𝑑 = 0.7 × 10−6 𝑚 5 𝜆𝑣𝑖𝑜𝑙𝑒𝑡 = 400 𝑛𝑚 𝜆𝑟𝑒𝑑 = 700 𝑛𝑚  Resultant Amplitude of the Wave Electric field of the wave from 𝑆1: 𝐸1 = 𝐸01 sin 𝑘𝑥 − 𝜔𝑡 Electric field of the wave from 𝑆2: 𝐸2 = 𝐸02 sin 𝑘𝑥 − 𝜔𝑡 + 𝛿 𝐸01 , 𝐸02 𝑜𝑟 𝐴01 , 𝐴02 → Electric Field Amplitude. Net electric field after the interference: 𝐸𝑛𝑒𝑡 = 𝐸1 + 𝐸2 = 𝐸0 sin 𝑘𝑥 − 𝜔𝑡 + 𝛿 𝐸02 = 𝐸01 2 2 + 𝐸02 + 2𝐸01 𝐸02 cos 𝛿 𝐸02 sin 𝛿 tan 𝜀 = 𝐸01 + 𝐸02 cos 𝛿  Intensity of Waves 𝐼 = 2𝜋 2 𝑓 2 𝜌𝑣𝐸02 𝑃 𝐸1 𝑆1 𝑦 𝐼 ∝ 𝐸02 𝐸02 = 𝐸01 2 2 + 𝐸02 + 2𝐸01 𝐸02 cos 𝛿 𝑑 𝑂 𝐵 𝐸2 𝑆2 𝐼𝑛𝑒𝑡 = 𝐼1 + 𝐼2 + 2 𝐼1 𝐼2 cos 𝛿 𝐷  Intensity for Identical Waves 𝐼𝑛𝑒𝑡 = 𝐼1 + 𝐼2 + 2 𝐼1 𝐼2 cos 𝛿 For identical slits: 𝐼1 = 𝐼2 = 𝐼 𝑃 𝐼1 𝑆1 𝐼𝑛𝑒𝑡 = 𝐼 + 𝐼 + 2 𝐼 𝐼 cos 𝛿 𝑦 𝐼𝑛𝑒𝑡 = 2𝐼 + 2𝐼 cos 𝛿 = 2𝐼(1 + cos 𝛿) 𝑑 𝑂 𝐵 𝛿 𝐼2 𝑆2 𝐼𝑛𝑒𝑡 = 2𝐼 × 2 cos 2 2 𝛿 𝐷 2 𝐼𝑛𝑒𝑡 = 4𝐼 cos 2  Intensity of Waves For constructive interference: (𝑀𝑎𝑥𝑖𝑚𝑎) 𝐸0𝑚𝑎𝑥 = 𝐸01 + 𝐸02 For destructive interference: (𝑀𝑖𝑛𝑖𝑚𝑎) 𝐸0𝑚𝑖𝑛 = 𝐸01 − 𝐸02 2 2 𝐼𝑚𝑎𝑥 𝐸0𝑚𝑎𝑥 𝐼𝑚𝑎𝑥 𝐸01 + 𝐸02 𝐼 ∝ 𝐸02 ⇒ = = 𝐼𝑚𝑖𝑛 𝐸0𝑚𝑖𝑛 𝐼𝑚𝑖𝑛 𝐸01 − 𝐸02 When cos 𝛿 = 1, 𝛿 = 2𝑛𝜋 (𝑀𝑎𝑥𝑖𝑚𝑎) 2 𝐼𝑛𝑒𝑡 = 𝐼1 + 𝐼2 + 2 𝐼1 𝐼2 ⇒ 𝐼1 + 𝐼2 When cos 𝛿 = −1, 𝛿 = 2𝑛 + 1 𝜋 (𝑀𝑖𝑛𝑖𝑚𝑎) 2 𝐼𝑛𝑒𝑡 = 𝐼1 + 𝐼2 − 2 𝐼1 𝐼2 ⇒ 𝐼1 − 𝐼2 𝐼1 + 𝐼2 2 , constructive interference 𝐼= 𝐼1 − 𝐼2 2 , destructive interference  Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of the maximum intensity to the minimum intensity is 16. The intensity of the waves are in the ratio: 𝐼𝑚𝑎𝑥 Given: = 16 𝐼𝑚𝑖𝑛 𝐼1 To find: 𝐼2 2 𝐼𝑚𝑎𝑥 𝐼1 + 𝐼2 16 Solution: = 16 ⇒ = 𝐼𝑚𝑖𝑛 𝐼1 − 𝐼2 1 ⇒ 𝐼1 + 𝐼2 = 4 𝐼1 − 4 𝐼2 ⇒ 3 𝐼1 = 5 𝐼2 𝐼1 5 ⇒ = 𝐼2 3 2 𝐼1 5 Ratio of intensities: = 𝐼2 3 𝐼1 25 = 𝐼2 9  Intensity Variation 𝑦 Constructive Interference: 2nd order 𝑛𝜆𝐷 𝑦= maxima 𝑑 2𝑠𝑡 order 1𝑠𝑡 order minima 𝜆𝐷 2𝜆𝐷 3𝜆𝐷 𝐼0 maxima 𝑆1 𝑦=± ,± ,±....... 1𝑠𝑡 order 𝑑 𝑑 𝑑 Central minima 𝑑 𝑂 maxima Destructive Interference: 𝐼0 𝑆2 1 𝜆𝐷 𝑦= 𝑛+ 2 𝑑 𝜆𝐷 3𝜆𝐷 5𝜆𝐷 𝑦=± ,± ,±....... 2𝑑 2𝑑 2𝑑 𝐷  Intensity Variation 4𝐼0 Intensity at any point: 𝐼=0 𝐼0 4𝐼0 𝛿 𝑆1 𝐼 = 4𝐼0 cos 2 𝐼=0 2 𝑑 𝑂 4𝐼0 𝛿 = Phase difference between 𝐼=0 𝐼0 the two waves from 𝑆1 and 𝑆2. 𝑆2 4𝐼0 𝐼𝑚𝑎𝑥 = 4𝐼0 , 𝐼𝑚𝑖𝑛 = 0 𝐼=0 4𝐼0 𝐷  Shape of Fringes on Screen If we replace slits by pin holes in YDSE, then we will see a Hyperbolic Fringe pattern. Point sources placed on the perpendicular axis to the screen create concentric circular fringes.  Fringe Width Fringe width is the distance between 𝝀𝑫 two consecutive maxima/minima. 𝑆1 𝒅 3𝜆𝐷 𝜆𝐷 𝛽= − 𝑑 𝑂 2𝑑 2𝑑 𝜆𝐷 𝑆2 𝛽= 𝑑 𝐷  Fringe Width when the setup is inside a Medium 𝜆𝐷 If experimental setup is dipped in liquid Fringe Width 𝛽 = 𝑑 Refractive index 𝑐 𝜇= 𝑣 𝑐 is speed of light in vacuum 𝑣 is speed of light in liquid 𝑓𝜆𝑣𝑎𝑐𝑢𝑢𝑚 𝜆𝑣𝑎𝑐𝑢𝑢𝑚 𝜇= ⇒ 𝜆𝑚𝑒𝑑𝑖𝑢𝑚 = 𝑓𝜆𝑚𝑒𝑑𝑖𝑢𝑚 𝜇 𝜆𝑚𝑒𝑑𝑖𝑢𝑚 𝐷 𝜆𝑣𝑎𝑐𝑢𝑢𝑚 𝐷 Fringe Width inside liquid, 𝛽 = ⇒𝛽= 𝑑 𝜇𝑑  Angular Fringe Width 𝛽 tan 𝜃 ≈ 𝜃 = 𝐷 𝑃 First Maxima 𝜆𝐷 We know 𝛽 = 𝑆1 𝑦=𝛽 𝑑 𝛽 𝜃 Central Maxima After putting 𝛽 value in 𝜃 = 𝑑 𝑂 𝐵 𝐷 𝑆2 𝜆 𝐷 𝜃= 𝑑  Position of Maxima/Minima Constructive Interference: 𝑛𝜆𝐷 𝑦= 𝜇𝑑 𝜆𝐷 2𝜆𝐷 3𝜆𝐷 𝑦=± ,± ,±....... 2𝜆𝐷 𝜇𝑑 𝜇𝑑 𝜇𝑑 𝜇𝑑 Destructive Interference: 3𝜆𝐷 2𝜇𝑑 1 𝜆𝐷 𝑦= 𝑛+ 2 𝜇𝑑 𝜆𝐷 3𝜆𝐷 5𝜆𝐷 𝑦=± ,± ,±....... 2𝜇𝑑 2𝜇𝑑 2𝜇𝑑  Incoherent Light Sources In ultra slow motion, the fringes on the screen flicker. However, the human eye cannot the capture the rapidly changing bright and dark fringes. So, the eyes see a continuous band of light. Net Intensity on the screen is the sum of intensity from two sources. Incoherent Light Sources  In a Young's double slit interference experiment, the fringe pattern is observed on a screen placed at a distance 𝐷 from the slits. The slits are separated by a distance 𝑑 and are illuminated by monochromatic light of wavelength 𝜆. Find the distance from the central point 𝐵 where the intensity falls to half the maximum. Solution: 𝛿 𝐼𝑚𝑎𝑥 cos2 𝐼 = 4𝐼0 𝐼𝑚𝑎𝑥 2 2 𝛿 1 Intensity is half the maximum, 4𝐼0 cos 2 = 4𝐼0 2 2 𝛿 1 𝜋 ⇒ cos = ⇒𝛿= 𝑆1 2 2 2 𝜆𝐷 𝜆𝐷 2𝜋 Phase difference 𝛿 = ∆𝑥 𝑑 4𝑑 2𝑑 𝜆 𝑂 𝐵 2𝜋 𝜋 𝜆 ∆𝑥 = ⇒ ∆𝑥 = 𝜆 2 4 𝑆2 𝑦𝑑 Path difference ∆𝑥 = 𝐷 𝜆 𝑦𝑑 𝜆𝐷 = ⇒ 𝑦= 4 𝐷 4𝑑 𝐷  Two coherent point sources 𝑆1 and 𝑆2 vibrating in phase emit light of wavelength 𝜆. The separation between the sources is 2𝜆. Consider a line T passing through 𝑆2 and perpendicular to the line 𝑆1 𝑆2. What is the smallest distance from 𝑆2 where the intensity is minimum? Given: 𝑑𝑆1 𝑆2 = 2𝜆 To find: 𝑥 2 1 Solution: 2𝜆 + 𝑥2 − 𝑥 = 𝑛 + 𝜆 2 16𝜆 − 2𝑛 + 1 2 𝜆 𝑥= 4 2𝑛 + 1 When 𝑥 > 0, 16𝜆 − 2𝑛 + 1 2 𝜆 > 0 So, 2𝑛 + 1 < 4 3 𝑛< 2 ∴𝑛=1 16𝜆 − 2𝑛 + 1 2 𝜆 16𝜆 − 9𝜆 7𝜆 𝑥= ⇒ 𝑥= 4 2𝑛 + 1 12 12  Figure shows three equidistant slits being illuminated by a monochromatic parallel beam of light. Let 𝐵𝑃0 − 𝐴𝑃0 = 𝜆/3 and 𝐷 ≫ 𝜆. Show that in this case T 𝑑 = 2𝜆𝐷/3. Solution: Optical path difference between 𝐵𝑃0 and 𝐴𝑃0 : ∆𝑥 = 𝑑 sin 𝜃 ≈ 𝑑 tan 𝜃 𝑑 𝑑2 ∆𝑥 = 𝑑 = 2𝐷 2𝐷 𝜆 𝑑2 ∆𝑥 = = 3 2𝐷 2𝜆𝐷 𝑑= 3 Path Difference between the Two Waves  Optical Path length Original wave equation: 𝐸 = 𝐸0 sin 𝑘𝑥0 − 𝜔𝑡 𝐸 𝑥0 𝑥0 + 𝐿 Equation of a wave when it is ahead by a length 𝐿: 𝐿 𝐸 = 𝐸0 sin 𝑘 𝑥0 + 𝐿 − 𝜔𝑡 𝑥 𝐸 = 𝐸0 sin 𝑘𝑥0 − 𝜔𝑡 + 𝑘𝐿 Phase difference ∆𝜙 2𝜋 𝑘= 𝜆 2𝜋 So, ∆𝜙 = 𝐿 𝜆 ∴ Two points on a wave separated by a path 2𝜋 length of 𝐿 will have a phase difference of 𝐿. 𝜆  Optical Path length Optical Path Length in a medium is the corresponding path that light travels in vacuum to undergo the same phase difference. Phase Difference between points 𝐴 and 𝐵 on the wave, travelling in air: 2𝜋 ∆𝜙𝑎𝑖𝑟 = 𝐿 𝜆𝑎 𝑎𝑖𝑟 Phase Difference between points 𝐴 and 𝐵0 on the wave, travelling in medium: 𝐴 𝐵 𝑎𝑖𝑟 2𝜋 2𝜋 𝜆𝑎 ∆𝜙𝑚𝑒𝑑 = 𝐿 = 𝐿 𝜆𝑚𝑒𝑑 = 𝜆𝑚𝑒𝑑 𝑚𝑒𝑑 𝜆𝑎 𝑚𝑒𝑑 𝜇 𝐿𝑎𝑖𝑟 𝜇 2𝜋 ∆𝜙𝑚𝑒𝑑 = 𝜇𝐿 𝜆𝑎 𝑚𝑒𝑑 𝐿𝑚𝑒𝑑 𝐴 𝐵0 If, ∆𝜙𝑎𝑖𝑟 = ∆𝜙𝑚𝑒𝑑 𝜇 2𝜋 2𝜋 𝐿𝑎𝑖𝑟 = 𝜇𝐿 ⇒ 𝐿𝑎𝑖𝑟 = 𝜇𝐿𝑚𝑒𝑑 𝜆𝑎 𝜆𝑎 𝑚𝑒𝑑  Optical Path Difference Optical Path Difference: 𝑂𝑃𝐷 = 𝑂𝑃𝐿𝐼𝐼 − 𝑂𝑃𝐿𝐼 = 𝜇𝑚𝑒𝑑 − 1 𝐿 Phase Difference: 2𝜋 𝛿 = ∆𝜙𝐼𝐼 − ∆𝜙𝐼 = 𝜇𝑚𝑒𝑑 − 1 𝐿 𝜆𝑎 2𝜋 𝛿= 𝑂𝑃𝐷 𝜆𝑎 Optical Path Difference: 𝑂𝑃𝐷 = 𝑂𝑃𝐿𝐼𝐼 − 𝑂𝑃𝐿𝐼 = 𝜇2 − 𝜇1 𝐿 Phase Difference in medium 1 and 2: Optical Path Length in air: 2𝜋 2𝜋 𝑂𝑃𝐿𝐼 = 𝐴𝐵 + 𝜇𝑎𝑖𝑟 𝐿 + 𝐶𝐷 = 𝐴𝐵 + 𝐿 + 𝐶𝐷 ∆𝜙𝐼 = 𝜇1 𝐿 and ∆𝜙𝐼𝐼 = 𝜇 𝐿 𝜆𝑎 𝜆𝑎 2 Phase Difference: Optical Path Length in medium: 2𝜋 𝛿 = ∆𝜙𝐼𝐼 − ∆𝜙𝐼 𝛿= 𝜇2 − 𝜇1 𝐿 𝑂𝑃𝐿𝐼𝐼 = 𝐴𝐵 + 𝜇𝑚𝑒𝑑 𝐿 + 𝐶𝐷 𝜆𝑎 Thin Transparent Film in YDSE Optical Path Travelled by wave 𝐼: 𝜇𝑡 + 𝑆1 𝐵 − 𝑡 Optical Path Travelled by wave 𝐼𝐼: 𝑆2 𝐵 Optical Path Difference: 𝑂𝑃𝐷 = 𝐼 − 𝐼𝐼 = 𝜇𝑡 + 𝑆1 𝐵 − 𝑡 − 𝑆2 𝐵 𝑂𝑃𝐷 = 𝜇𝑡 − 𝑡 = 𝜇−1 𝑡 𝑆1 𝐵 = 𝑆2 𝐵 Introduction of slab causes change in 𝑂𝑃𝐷 by 𝜇 − 1 𝑡. Path length of the ray 𝑆1 𝐵 increases when it encounters the thin film. Now the central maxima does not lie at 𝐵.  Thin Transparent Film in YDSE After inserting thin film, light ray 𝑆1 𝑃 After inserting thin film, Optical Path Difference: has travelled an extra path of 𝜇 − 1 𝑡. light ray 𝑆2 𝑃 has travelled an extra path of 𝜇 − 1 𝑡. ∆𝑥 = 𝐼𝐼 − 𝐼 Total Optical Path Difference at point P: Total Optical Path Difference at ∆𝑥 = 𝑆2 𝑃 − 𝑆1 𝑃 ∆𝑥 = 𝐼𝐼 − 𝐼 point P: ∆𝑥 = 𝐼𝐼 − 𝐼 ∆𝑥 = 𝑆2 𝑃 − 𝑆1 𝑃 + 𝑡 𝜇 − 1 𝑦𝑑 ∆𝑥 = 𝑆2 𝑃 + 𝑡 𝜇 − 1 − 𝑆1 𝑃 ∆𝑥 = 𝐷 𝑦𝑑 𝑦𝑑 ∆𝑥 = −𝑡 𝜇−1 ∆𝑥 = +𝑡 𝜇−1 𝐷 𝐷  Shift in Central Maxima 𝑥 𝑃 At central maxima, 𝑡, 𝜇 𝐼 𝑂𝑃𝐷 = 0 𝑆1 𝑦 𝑥 + Δ𝑥 𝑦𝑑 − 𝜇−1 𝑡 =0 𝜃 𝐼𝐼 𝐷 𝑑 𝑂 𝐵 𝜇 − 1 𝑡𝐷 𝑦= 𝑆2 𝑑 𝜇 − 1 𝑡𝐷 Shift in central maxima = 𝑑 𝐷 Screen  Number of Fringes Shifted Shift in central maxima: Shift in central maxima: 𝜇 − 1 𝑡𝐷 𝑑 𝜇 − 1 𝑡𝐷 𝑆= 𝑠ℎ𝑖𝑓𝑡 𝑖𝑛 𝑐𝑒𝑛𝑡𝑟𝑎𝑙 𝑚𝑎𝑥𝑖𝑚𝑎 𝑑 Number of Fringes shifted = 𝑓𝑟𝑖𝑛𝑔𝑒 𝑤𝑖𝑑𝑡ℎ 𝑦 𝜇 − 1 𝑡𝐷 𝑑 Number of fringes shifted: 𝑛= = × 𝛽 𝑑 𝜆𝐷 𝜇−1 𝑡 𝜇−1 𝑡 𝑛= 𝑛= 𝜆 𝜆  The Young’s double slit experiment is done in a medium of refractive index 4/3. A light of 600 𝑛𝑚 wavelength is falling on the slits having 0.45 𝑚𝑚 separation. The lower slit 𝑆2 is covered by a thin glass sheet of thickness T 10.2 𝜇𝑚 and refractive index 1.5. The interference pattern is observed on a screen placed 1.5 𝑚 from the slits. Find the location of the central maximum on the 𝑦 −axis. A 7.5 𝑚𝑚 𝜇 = 4/3 B 10.2𝑚𝑚 𝑆1 𝐼 𝑂 C 4.25 𝑚𝑚 𝑆2 𝐼𝐼 𝑃0 D 1.25 𝑚𝑚 Screen  4 Given: 𝜇𝑚𝑒𝑑 = , 𝑑 = 0.45 𝑚𝑚, 𝐷 = 1.5 𝑚, 𝜆 = 600 𝑛𝑚, 𝑡 = 10.2 𝜇𝑚, 𝜇𝑡 = 1.5 3 To find: Position of central maxima Solution: 𝑂𝑃𝐷 = 𝐼 − 𝐼𝐼 𝜇 = 4/3 𝑂𝑃𝐷 = 𝑆1 𝑃0 𝜇𝑚𝑒𝑑 − 𝜇𝑔 𝑡 + 𝑆2 𝑃0 𝜇𝑚𝑒𝑑 − 𝜇𝑚𝑒𝑑 𝑡 𝑆1 𝑂𝑃𝐷 = 𝑆1 𝑃0 − 𝑆2 𝑃0 𝜇𝑚𝑒𝑑 − 𝑡 𝜇𝑔 − 𝜇𝑚𝑒𝑑 𝐼 𝑦𝑑 𝑂 𝑂𝑃𝐷 = 𝜇𝑚𝑒𝑑 − 𝑡 𝜇𝑔 − 𝜇𝑚𝑒𝑑 𝑦 𝐷 𝑆2 𝐼𝐼 At central maxima 𝑂𝑃𝐷 = 0 𝑃0 𝑦𝑑 𝜇𝑚𝑒𝑑 − 𝑡 𝜇𝑔 − 𝜇𝑚𝑒𝑑 = 0 Screen 𝐷 𝑡𝐷 𝑦= 𝜇 − 𝜇𝑚𝑒𝑑 𝑑𝜇𝑚𝑒𝑑 𝑔 Substituting all the values we get, 𝑦 = 4.25 𝑚𝑚  In YDSE, find the thickness of a glass slab (𝜇 = 1.5) which should be kept in front of upper slit 𝑆1 so that the central maxima is formed at a place where T 5𝑡ℎ bright fringe was lying earlier (before inserting slab). 𝜆 = 5000 𝐴ሶ Given: 𝜆 = 5000 𝐴,ሶ 𝜇 = 1.5, To find: 𝑡 5𝐷𝜆 𝑛𝐷𝜆 Solution: 𝑦5𝐵 = 𝑦𝑛𝐵 = 𝑑 𝑑 𝜇 − 1 𝑡𝐷 Shift in central maxima: 𝑦= 𝑑 𝜇 − 1 𝑡𝐷 5𝐷𝜆 = 𝑑 𝑑 5𝜆 5000 × 10−10 𝑡= =5× 𝜇−1 1.5 − 1 𝑡 = 5 𝜇𝑚  Two transparent slabs, having equal thickness but different refractive indices 𝜇1 and 𝜇2 𝜇1 > 𝜇2 , are pasted side by side to form a composite slab. This slab is placed just after the double slits in a Young’s experiment so that the light from T one slit goes through one material and light through other slit goes through other material. What should be the minimum thickness of the slab so that there is a minimum at point 𝑃0 which is equidistant from the slits? Given: Wavelength = 𝜆, Refractive index = 𝜇1 & 𝜇2 To find: 𝑡1 = 𝑡2 = 𝑡 =? Solution: Path difference due to both the slabs at 𝑃0 : 𝑡 𝜇1 − 1 𝑡 − 𝜇2 − 1 𝑡 = 𝜇1 − 𝜇2 𝑡 𝑆1 𝜇1 For minima at 𝑃0 , we know, 𝑑 𝑂 1 𝜇2 𝑃0 Δ𝑥 = 𝑛 + 𝜆 2 𝑆2 1 ⇒ 𝜇1 − 𝜇2 𝑡 = 𝑛 + 𝜆 2 𝐷 For 𝑡 to be minimum, put 𝑛 = 0, Screen 𝜆 𝑡= 2 𝜇1 − 𝜇2  Phase change in Reflection & Refraction Air 𝜋 𝑖 Reflected from Phase change No phase Denser Medium Yes change 𝜇 𝑟 Rarer Medium No No phase change Air  A narrow slit 𝑆 transmitting light of wavelength 𝜆 is placed at a distance 𝑑 above a large plane mirror. The light coming directly from the slit and that coming after reflection interfere at a screen placed at a distance 𝐷 from the T slit. (𝑎) What will be the intensity at a point just above 𝑂. Given: Wavelength = 𝜆, Distance of screen = 𝐷, Distance of source from the mirror = 𝑑 To find: Intensity just above 𝑂 Solution: 𝑆𝑃 ≈ 𝑆𝑄𝑃 → path difference = 0 (when 𝑃 is just above 𝑂) δ=𝜋 (due to reflection of light from a denser medium) If phase difference is odd integral multiple of 𝜋, destructive interference will take place. 𝑆 𝐼𝑛𝑒𝑡 = 0 𝑑 𝑃 𝑄 𝑂 𝐷 Screen  A narrow slit 𝑆 transmitting light of wavelength 𝜆 is placed at a distance 𝑑 above a large plane mirror. The light coming directly from the slit and that coming after reflection interfere at a screen placed at a distance 𝐷 from the T slit. (𝑏) At what distance from 𝑂 does the first maximum occur? Given: Wavelength = 𝜆, Distance of screen = 𝐷, Distance of source from the mirror = 𝑑 To find: 𝑦1𝑠𝑡 𝑚𝑎𝑥𝑖𝑚𝑎 2𝑑𝑦 Solution: ∆𝑥 = 2𝑑 sin 𝜃 = 2𝑑 tan 𝜃 = 𝐷 𝑃 𝑆1 2𝑑𝑦 𝜆 ∆𝑥𝑛𝑒𝑡 = + = 𝑛𝜆 = 1 𝜆 𝑦 𝐷 2 𝑑 𝜃 𝑄 2𝑑𝑦 𝜆 ⇒ = 𝑂 𝐷 2 𝐷 𝜆𝐷 𝑑 Screen 𝑦= 𝑆2 4𝑑  A narrow slit 𝑆 transmitting light of wavelength 𝜆 is placed at a distance 𝑑 above a large plane mirror. The light coming directly from the slit and that coming after reflection interfere at a screen placed at a distance 𝐷 from the T slit. (𝑐) If 𝑑 = 1 𝑚𝑚, 𝐷 = 1 𝑚 and 𝜆 = 700 𝑛𝑚, then find the fringe width. Given: 𝜆 = 700 𝑛𝑚, 𝐷 = 1 𝑚, 𝑑 = 1 𝑚𝑚 To find: Fringe width = 𝛽 𝜆𝐷 𝜆𝐷 𝑃 𝑆1 Solution: 𝛽 =2× = 4𝑑 2𝑑 𝜆𝐷 𝑑 700 × 10−9 × 1 −4 𝑚 𝑄 4𝑑 𝛽= = 3.5 × 10 𝜃 2 × 10−3 𝑂 𝐷 𝛽 = 0.35 𝑚𝑚 𝑑 𝜆𝐷 𝑆2 4𝑑  A narrow slit 𝑆 transmitting light of wavelength 𝜆 is placed at a distance 𝑑 above a large plane mirror. The light coming directly from the slit and that coming after reflection interfere at a screen placed at a distance 𝐷 from the slit. T (𝑑) If the mirror reflects only 64 % of the light energy falling on it, what will be the ratio of the maximum to the minimum intensity in the interference pattern observed on the screen? Given: Wavelength = 𝜆, Distance of screen = 𝐷, Distance of source from the mirror = 𝑑 𝐼2 = 0.64 𝐼0 𝐼𝑚𝑎𝑥 To find: 𝐼𝑚𝑖𝑛 𝐼1 = 𝐼0 𝑃 2 2 𝑆1 Solution: 𝐼𝑚𝑎𝑥 = 𝐼1 + 𝐼2 , 𝐼𝑚𝑖𝑛 = 𝐼1 − 𝐼2 2 2 𝐼𝑚𝑎𝑥 𝐼1 + 𝐼2 𝐼0 + 0.64 𝐼0 𝑄 𝐼2 = 0.64 𝐼0 = 2 ⇒ 2 𝜃 𝐼𝑚𝑖𝑛 𝐼1 − 𝐼2 𝐼0 − 0.64 𝐼0 𝑂 2 2 𝐷 1 + 0.8 1.8 = = 1 − 0.8 2 0.2 2 𝑆2 𝐼𝑚𝑎𝑥 = 81 𝐼𝑚𝑖𝑛  Condition for Thin Film Interference 𝜋 The film thickness should be comparable to 𝑖 air the wavelength of light 𝑑 ≈ 𝜆. The angle of incidence should be small 𝑖 ≈ 0. 𝑟 𝑑 𝜇 The incident light should be white (non - monochromatic). air  Thin Film Interference Interference of light wave being reflected off Air two surfaces that are at a distance 𝜋 comparable to its wavelength is known as thin 𝑖 film ”Thin film interference”. Inside the film, when a particular colour’s path No phase difference is even integral multiple of the change wavelength, it undergoes constructive 𝜇 𝑟 interference, so these colours appear bright. No phase change When a particular colour’s wavelength is odd integral multiple of the path difference inside the film, it undergoes destructive interference, so these colours don’t appear at all. Air  Interference due to Thin Film from Transmitted Light angles are very small 𝑟 𝑑 air 𝑖 𝑑 cos 𝑟 𝑑 𝜇 𝑟 𝑟 Path difference of 1𝑠𝑡 and 𝑑 air =2 𝜇 2𝑛𝑑 2𝑛𝑑 transmitted light wave cos 𝑟 1𝑠𝑡 When angle is very small, 𝑟 ≈ 0 cos 𝑟 ≈ 1 Path difference ∆𝑥 = 2𝜇𝑑  Interference due to Thin Film from Transmitted Light angles are very small For constructive interference: air 𝑖 𝛿 = 2𝑛𝜋 Or 2𝜇𝑑 = 𝑛𝜆 𝑑 𝜇 𝑟 For destructive interference: air 1 𝛿 = 2𝑛 + 1 𝜋 Or 2𝜇𝑑 = 𝑛 + 𝜆 2  Interference due to Thin Film from Reflected Light angles are very small 𝑑 𝑑 𝑟 cos 𝑟 1𝑠𝑡 2𝑛𝑑 air 𝑖 Path difference of 1𝑠𝑡 and 𝑑 𝑑 𝜇 𝑟 =2 𝜇 ≈ 2𝜇𝑑 2𝑛𝑑 reflected light wave cos 𝑟 Because phase change of 𝜋 after 𝜆 reflection of 1 , path difference = − 𝑠𝑡 air 2 𝜆 Total path difference ∆𝑥 = 2𝜇𝑑 − 2  Interference due to Thin Film from Reflected Light angles are very small For constructive interference: 𝜆 2𝜇𝑑 − = 𝑛𝜆 2 1 𝜋 2𝜇𝑑 = 𝑛 + 𝜆 2 air 𝑖 For destructive interference: 𝑟 𝑑 𝜇 𝜆 1 2𝜇𝑑 − = 𝑛+ 𝜆 ⇒ 2𝜇𝑑 = 𝑛 + 1 𝜆 2 2 air 2𝜇𝑑 = 𝑛𝜆 𝑛 is an integer.  Interference due to Thin Film For constructive interference: Colours will be strongly reflected/transmitted. Destructive interference: Colours will be poorly reflected/transmitted. This gives coloured appearance of the film.  A soap film of refractive index 1.33 is illuminated by the light of wavelength 400 𝑛𝑚 at an angle of 45°. If there is complete destructive interference then, T find the thickness of the film. Given: 𝜇 = 1.33, 𝑖 = 45°, 𝜆 = 400 𝑛𝑚 To find: 𝑡 sin 𝑖 sin 45 Solution: 𝜇= ⇒ 1.33 = sin 𝑟 sin 𝑟 45° soap film 3 sin 𝑟 = ⇒ cos 𝑟 = 0.85 cos 𝑟 = 1 − sin2 𝑟 4 2 For Destructive interference: 2𝜇𝑡 = 𝑛𝜆 cos 𝑟 2 1.33 𝑡 ⇒ = 1 400 × 10−9 0.85 ⇒ 𝑡 = 1.27 × 10−7 𝑚  Diffraction Bending of a wave or its deviation from its original direction of propagation while passing through a small obstruction is known as diffraction. Original direction Every point on the wavefront makes a secondary wave according to Huygens Principle. Condition for bending: size of obstacle ≈ 𝜆 Diffraction is explained by wave nature of light.​  Diffraction of Light Waves Fraunhofer Diffraction Fresnel Diffraction Source and screen are at infinite Source and screen are at finite distance from diffraction element. distance from diffraction element. High intense interference pattern Diffraction was first demonstrated by is observed on screen. this experiment. 𝑆𝑂 𝑆𝑂  Fraunhofer Diffraction – Path Difference 𝜃 𝑏 𝑃 2 𝜃 𝑏 𝑃𝑂 𝑏 sin 𝜃 2 Path difference between the waves, 𝑏 Screen ∆x = sin 𝜃 2  Fraunhofer Diffraction – first Dark Fringe The condition for 1𝑠𝑡 dark fringe formed at point 𝑃: 𝑏 𝜆 sin 𝜃 = 𝑃 2 2 𝜃 𝑏 𝑃𝑂 𝑏 sin 𝜃 = 𝜆 The condition for 𝑛𝑡ℎ dark fringe: 𝑏 sin 𝜃 = 𝑛𝜆 Screen Central Maxima is at 𝑃0 , where 𝜃 = 0  Fraunhofer Diffraction – Intensity at general point Electric Field Amplitude at 𝑃: 𝐸0 sin 𝛽 𝐸′ = 𝛽 = phase difference 𝑃 𝛽 𝜃 2𝜋 𝑏 𝑏 𝑃𝑂 Where 𝛽 = 𝜆 2 sin 𝜃 When 𝜃 = 0, 𝛽 = 0, 𝐸 ′ = 𝐸0 Intensity ∝ 𝐸 2 Intensity at a general point: Screen 𝐼𝑂 sin2 𝛽 𝐼= 𝛽2 Where 𝐼0 is the intensity at Central Maxima  Fraunhofer Diffraction – Graph of Intensity 𝐼 𝐼𝑂 sin2 𝛽 𝜋𝑏 sin 𝜃 𝐼= 𝛽= 𝛽2 𝜆 𝐼𝑜 sin 𝜃 = 0 ; Central Maxima 𝑛𝜆 sin 𝜃 = ; Minima 𝐼𝑂 𝑏 𝜆 2𝜆 3𝜆 22 𝐼𝑂 sin 𝜃 = ± , ± , ± … … … … 𝑏 𝑏 𝑏 62.5 Maximum intensity is distributed 𝜆 𝜆 between − 𝑏 and 𝑏. 2𝜆 𝜆 0 𝜆 2𝜆 sin 𝜃 − − 𝜆 𝑏 𝑏 𝑏 𝑏 If 𝑏 is large, 𝑏 → 0, then Single fringe is observed and no diffraction is seen.  Fraunhofer Diffraction – Graph of Intensity 𝑏 Fraunhofer Diffraction ⟶ Intensity YDSE ⟶ Intensity is same for all decreases away from the centre. fringes. 𝐼0 → Intensity at the central bright fringe 𝐼0 → Intensity from a single slit. 𝐼𝑂 sin2 𝛽 𝜙 𝐼= 𝐼 = 4𝐼𝑂 cos 2 𝛽2 2  Difference between Interference and Diffraction Interference Diffraction It is the phenomenon of superposition of It is the phenomenon of superposition of two two waves coming from two different waves coming from two different parts of the coherent sources. same wavefront. All bright fringes are not equally bright and In interference pattern, all bright fringes are equally wide. Brightness and width equally bright and equally spaced. decreases with the angle of diffraction. All dark fringes are perfectly dark, but their All dark fringes are perfectly dark. contrast with bright fringes and width decreases with angle of diffraction. In interference, bright and dark fringes are In diffraction, bright and dark fringes are large in number for a given field of view. fewer for a given field of view.  A beam of light of wavelength 600 𝑛𝑚 from a distant source falls on a single slit 1 𝑚𝑚 wide and a resulting diffraction pattern is observed on a screen 2 𝑚 away. The distance between the first dark fringes on either side of central bright fringe is 𝜆 Solution: First maxima is formed at distance 𝑏 away on both side of central maxima. sin 𝜃 ≈ 𝜃 (𝜃 is small) 𝜆 𝜃= 𝑏 𝛼 𝜃 𝑦 1𝑚𝑚 𝜃 2𝜆 𝛼 = 2𝜃 = 𝑏 𝑦 2𝜆 2𝜆𝐷 𝛼= = ⟹ 𝑦= 𝐷 𝑏 𝑏 𝐷 =2𝑚 2 × 600 × 10−9 ×2 𝑦= = 24 × 10−4 𝑚 10−3 𝑦 = 2.4 𝑚𝑚  Single Slit Diffraction – 𝟏𝒔𝒕 Secondary Maxima Condition for 1𝑠𝑡 secondary maxima (bright): 𝐴 3𝜆 𝑏 sin 𝜃 = 2 The angle of diffraction (𝜃1B ) for 1st maximum (bright) is: 𝑏 3𝜆 𝜃1𝐵 = 2𝑏 𝑁 The position of 1st maximum (bright) from the centre of the screen is: 𝐵 3𝜆 2 3𝐷𝜆 𝑦1𝐵 = 2𝑏  Single Slit Diffraction – 𝒏𝒕𝒉 Secondary Maxima The angle of diffraction 𝜃𝑛 for 𝑛𝑡ℎ maxima (bright) is: Second secondary maxima 2𝑛 + 1 𝜆 𝜃𝑛𝐵 = First secondary maxima 2𝑏 Central maxima The position of 𝑛𝑡ℎ maxima (bright) from the centre of the screen is: First secondary maxima 2𝑛 + 1 𝐷𝜆 𝑦𝑛𝐵 = 2𝑏 Second secondary maxima Screen Where, 𝑛 = 1, 2, 3, 4 … … …  Single Slit Diffraction Angular position for minima Angular position for maxima 5𝜆 2𝜆 + + 2𝑏 𝑏 3𝜆 𝜆 + 2𝑏 + 𝑏 0 𝜆 − 𝑏 3𝜆 2𝜆 − 2𝑏 − 𝑏 5𝜆 − 2𝑏 𝜆 2𝜆 𝑛𝜆 3𝜆 5𝜆 (2𝑛 + 1)𝜆 sin 𝜃 = ± , ± , … … , ± sin 𝜃 = ± ,± ,……,± 𝑏 𝑏 𝑏 2𝑏 2𝑏 2𝑏  In a diffraction pattern due to a single slit of width 𝑎, the first minima is observed at an angle 30° when the light of wavelength 5000 𝐴ሶ is incident T on the slit. The first secondary maxima is observed at an angle of: Given: 𝜃1𝐷 = 30°; 𝜆 = 5000 𝐴ሶ To find: Angular position of first secondary maximum (𝜃1𝐵 ) Solution: Angular position of first minima is given by, 𝜆 1 𝜆 sin 𝜃1𝐷 = ⟹ = 𝑎 2 𝑎 Angular position of first secondary maxima is given by, 3𝜆 3 sin 𝜃1𝐵 = ⟹ sin 𝜃1𝐵 = 2𝑎 4 3 𝜃1𝐵 = sin−1 4  Fraunhofer Diffraction for Hole When a monochromatic light is incident on the hole we see concentric circular bright and dark spots on the screen. The size of hole should be comparable to the wavelength of incident light. 𝜃 𝑅 𝑏 Central bright spot contains the most energy. Brightness of the rings decreases as we move away from the centre. For the first dark ring, 𝐷 1.22𝜆 1.22𝜆𝐷 sin 𝜃 = 𝑅= 𝑏 𝑏  A convex lens of diameter 8.0 𝑐𝑚 is used to focus a parallel beam of light of wavelength 6200 Å. If the light be focused at a distance of 20 𝑐𝑚 from the lens, what would be the radius of the central bright spot? Given: 𝐷 = 20 𝑐𝑚, 𝑏 = 8 𝑐𝑚 and 𝜆 = 6200 Å To Find: 𝑅 Solution: Radius of central bright spot, 𝜃 𝑅 1.22𝜆𝐷 𝑅= 𝑏 6200 Å 1.22 × 6200 × 10−10 × 20 × 10−2 𝑏 = 8 𝑐𝑚 ⇒𝑅= 8 × 10−2 20 𝑐𝑚 𝑅 = 1.89 × 10−6 𝑚 Binary Star  Limit of Resolution Unresolved: Diffraction discs from both sources overlap. 𝑆1 𝑆2 Just resolved: The periphery of the diffraction 𝑆1 disc of one object touches the centre of the diffraction disc of the other object. 𝑆2 Well resolved: The diffraction discs formed by 𝑆1 two objects are well separated from each other. 𝑆2  Rayleigh criterion The Rayleigh criterion specifies the minimum separation between two light sources that may be resolved into distinct objects. Unresolved Just resolved Well resolved (Rayleigh criterion)  Limit of Resolution of a Telescope Telescope lens 𝑆𝑡𝑎𝑟1 ∆𝜃 𝑅 𝑆𝑡𝑎𝑟2 Just resolved Clear image is formed when the diffraction discs from two sources is just resolved. Distance between diffraction disc > 𝑅 ⇒ Well resolved Distance between diffraction disc < 𝑅 ⇒ Unresolved  Resolving Power of Telescope Angular limit of resolution of telescope: ∆𝜃 𝑅 𝑏 1.22𝜆 ∆𝜃 = 𝑏 ∆𝜃 = Limit of resolution 𝑓 Angle subtended by the first dark fringe > ∆𝜃 ⇒ Well resolved Angle subtended by the first dark fringe < ∆𝜃 ⇒ Unresolved  Radius of the Central Bright Spot Radius of central bright region is: 𝑅 = 𝑓∆𝜃 ∆𝜃 𝑅 1.22𝜆𝑓 𝑏 𝑅= 𝑏 Resolving power of a telescope: 1 𝑅. 𝑃. = ∆𝜃 𝑓 𝑏 𝑅. 𝑃. = 1.22𝜆 Bigger lens ⇒ larger 𝑏 ⇒ smaller Limit of Resolution ∆𝜃 ⇒ higher Resolving Power 𝑅. 𝑃.  Disadvantages of using Lens Chromatic aberration 1. Difficult and expensive to build large lenses. 2.Providing mechanical support to large lenses require large and complex machinery. 3. Chromatic aberration ( light rays passing through a lens focus at different points, depending on their wavelength). Fact: The largest lens objective in use has diameter of 40 inch (~1.02 𝑚). It is at the Yerkes Observatory in Wisconsin, USA.  Advantages of using Mirror as the Objective in Telescope 1. No chromatic aberration 2. Parabolic mirror used to counter spherical aberration. 3. Large mirrors can be Eyepiece supported from the back. Objective mirror Fact: The viewer sits near the focal point of the mirror, in a small cage.  Calculate the limit of resolution of a telescope objective having a diameter of 200 𝑐𝑚, if it has to detect light of wavelength T 500 𝑛𝑚 coming from a star. Given: 𝑏 = 200 𝑐𝑚, 𝜆 = 500 𝑛𝑚 JEE Main 2019 To find: Limit of resolution of telescope Solution: Limit of resolution of telescope is given by: 1.22𝜆 ∆𝜃 = 𝑏 1.22 × 500 × 10−9 ∆𝜃 = 200 × 10−2 305 × 10–9 𝑟𝑎𝑑𝑖𝑎𝑛  Limit of Resolution of a Microscope Lens Formula: 1 1 1 − = 𝑣 𝑢 𝑓 𝑣 𝑣 𝑣 𝑣 1− = ⇒ 1−𝑚 = 𝑚= 𝑢 𝑓 𝑓 𝑢 𝑣 𝑚 =1− 𝑓 𝑣 𝑣 𝑚≈− 𝑣≫𝑓 ⇒ ≫1 𝑓 𝑓 Clear images can be seen in the microscope if the diffraction discs are just resolved. 1.22𝜆 We know that: 𝑑=𝑣× 𝑏𝑚 Angular limit of resolution of microscope: 1.22𝜆 1.22𝜆𝑓 𝑑= × −𝑚𝑓 = 𝑏𝑚 𝑏 1.22𝜆 sin 𝜃 ≈ 𝜃 = 𝑏 𝑏 tan 𝛽 = ⇒ 𝑏 = 2𝑓 tan 𝛽 2𝑓 1.22𝜆 𝑅 = 𝑣𝜃 = 𝑣 × 𝛽 is small, 𝑏 = 2𝑓 sin 𝛽 𝑏 𝑅 𝑅 1.22𝜆 1.22𝜆𝑓 1.22𝜆 Magnification of convex lens: 𝑚= ⇒𝑑= 𝑑=𝑣× 1.22𝜆𝑓 𝑑𝑚𝑖𝑛 = 𝑏𝑚 𝑑𝑚𝑖𝑛 = = 2 sin 𝛽 𝑑 𝑚 𝑏 2𝑓 sin 𝛽  Microscope immersed in Oil Resolving power of a microscope: 1 2𝜇 sin 𝛽 𝑅. 𝑃. = ⇒ 𝑅. 𝑃. = 𝑑𝑚𝑖𝑛 1.22𝜆 Resolving power of the microscope increases when it is Note: The product 𝜇sin𝛽 is called the numerical aperture and immersed in a medium. is sometimesmarked on the objective. 𝜆 When the setup is immersed in oil, 𝜆𝑚𝑒𝑑 changes to 𝜇 1.22𝜆𝑚𝑒𝑑 1.22𝜆 𝑑𝑚𝑖𝑛 = ⟶ 𝑑𝑚𝑖𝑛 = 2 sin 𝛽 2𝜇 sin 𝛽  Validity of Ray Optics Consider a single slit of width ‘𝑎’.Diffraction pattern will be observed. There will be central maxima due to diffraction. Angular size of central maximum, 𝜆 𝜃= 𝑎 𝜆𝑧 The width of diffracted beam after it has travelled by 𝑧, 𝑦 =𝑧×𝜃 = 𝑎 𝑎2 If 𝑦 ≈ 𝑎 𝑧≈ = 𝑧𝐹 𝑧𝐹 is Fresnel distance. 𝜆 If 𝑧 < 𝑧𝐹 , the ray optics is valid. If 𝑧 > 𝑧𝐹 , spreading due to diffraction dominates.  For what distance is ray optics a good approximation when a plane light wave is incident on a circular aperture of width 2 𝑚𝑚 T having wavelength 600 𝑛𝑚? Given: 𝑎 = 2 𝑚𝑚; 𝜆 = 600 𝑛𝑚 Solution: Ray optics is a good approximation up to Fresnel distance only. 𝑎2 𝑧≈ = 𝑧𝐹 𝜆 2 × 10−3 2 𝑧𝐹 = = 6.7 𝑚 600 × 10−9 If 𝑑 < 6.7 𝑚 ⟶ Ray optics holds. If 𝑑 > 6.7 𝑚 ⟶ Spreading due to diffraction dominates.  Scattering When a parallel beam of light passes through a medium, a part of it appears in directions other than the incident direction. This phenomenon is called scattering of light. Light is an EM wave, it oscillates the charged particles in a medium because of its oscillating electric field. Incident light Oscillating charged particles emit EM waves. If the oscillating electric field of incident light has frequency 𝑓, the frequency of scattered A particle wave will also have frequency 𝑓.  Rayleigh’s Law of Scattering Intensity of scattering depends on 𝜆 − 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑉 Scatters most 1. Wavelength of light 2. Size of particles causing scattering 𝐼 𝜆 increases 𝐵 When size of particles < 𝜆 and Intensity 𝐺 of scattering Intensity of scattered wave ∝ 1 𝜆4 decreases 𝑌 𝑂 𝜆 − 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑅 Scatters least Unpolarized Light The light having electric field oscillations in all directions in the plane perpendicular to the direction of propagation. Examples of source of unpolarized light: Candle, Bulb, Sun etc. Note: Light wave is coming out of the screen, and arrows show direction of oscillation of electric field.  Plane Polarized Light Plane Polarized light – When electric field at a point always remains parallel to a fixed direction as time passes. Plane of polarization – Plane containing electric field and direction of propagation. 𝑥 − 𝑎𝑥𝑖𝑠 → Direction of propagation. 𝑦 − 𝑎𝑥𝑖𝑠 → Oscillation of electric field. 𝑥𝑦 𝑝𝑙𝑎𝑛𝑒 → Plane of polarization. When the polarizer is placed in the path of unpolarized light, the direction of oscillation of the electric field becomes parallel to the transmission axis. Note: If any unpolarized light of intensity 𝐼0 is incident on a 𝐼 polarizer, we get a polarized light of intensity 0. 2  Law of Malus Electric field amplitude of the wave after crossing the polarizer: 𝜃 𝐸 = 𝐸0 cos 𝜃 𝐸0 𝐸0 cos 𝜃 Intensity of the wave after crossing the polarizer: 2𝐼0 𝐼0 𝐼 𝐼 = 𝐼0 cos 2 𝜃 𝐼 ∝ 𝐸2 Unpolarized Polarized 𝐼 = Intensity of transmitted light light light 𝐼0 = Intensity of incident light  Law of Malus Case 1 → when 𝜃 = 0° : Case 2 → when 𝜃 = 90° : 𝐼 = 𝐼0 cos 2 0° = 𝐼0 𝐼 = 𝐼0 cos2 90° = 0 𝐼0 𝐼0 𝐼0  Two ‘crossed’ polaroids 𝐴 and 𝐵 are placed in the path of a light beam. In between these, a third polaroid 𝐶 is placed whose polarization axis makes an angle 𝜃 with the polarization axis of the polaroid 𝐴. If the T intensity of light emerging from the polaroid 𝐴 is 𝐼𝑜 , then the intensity of light emerging from polaroid 𝐵 will be Solution: Intensity of light emerging from polaroid 𝐶: 𝐼𝐶 = 𝐼0 cos2 𝜃 {Applying the law of Malus.} Intensity of light emerging from polaroid 𝐵: 𝐼𝐵 = 𝐼𝐶 cos2 90° − 𝜃 𝐼𝐵 = 𝐼0 cos2 𝜃. cos2 90° − 𝜃 𝐼0 𝐼𝐵 = 𝐼0 cos2 𝜃. sin2 𝜃 = 2 sin 𝜃 cos 𝜃 2 4 𝐼0 2 𝐼𝐵 = sin (2𝜃) 4  Unpolarized light with amplitude 𝐴0 passes through two polarizers. The first one has an angle of 30° clockwise to vertical and second one has an angle of T 15° counter-clockwise to the vertical. What is the amplitude of the light emitted from the second polarizer? Given: 𝐴 = 𝐴0 To find: 𝐴2 Solution: 𝐴0 We know that: 𝐼 ∝ 𝐴2. 𝐴1 = 2 ∴𝐴∝ 𝐼 𝐴1 𝐼1 Again, = 𝐴1 𝐼1 𝐴2 𝐼2 ⇒ = 𝐴2 𝐼2 𝐼2 = 𝐼1 cos2 𝜃 𝜃 = 30° − −15° = 45° 𝐼0 𝐼1 = 2 𝐼0 𝐼0 1 2 𝐼0 𝐼2 = cos2 45° = = 𝐴1 𝐼1 𝐼0 1 2 2 2 4 = = = 𝐴0 𝐼0 2𝐼0 2 𝐴2 𝐼2 𝐼0 1 𝐴0 = = = 𝐴2 = 𝐴0 𝐴0 𝐼0 4𝐼0 4 2 𝐴1 = 2  Polarization by Reflection Oscillations in the incidence plane 𝑖 Oscillations ⊥ to the incidence plane Medium(𝜇) The reflected light has more vibrations perpendicular to plane of incidence. The refracted light has more vibration parallel to the plane of incidence. The percentage of polarization in reflected light changes as we change the angle of incidence 𝑖.  Brewster’s Law For a particular angle of incidence 𝑖𝐵 , the reflected light becomes completely plane polarized. 𝑖𝐵 The required condition for this purpose is: 𝑖𝐵 + 𝑟 = 90° Medium(𝜇) 𝑟 Brewster’s Law tan 𝑖𝐵 = 𝜇 𝑖𝐵 = Brewster angle/polarising angle If the light ray travels from one medium to another with refractive 𝜇1 and 𝜇2 respectively, then Brewster’s law becomes, 𝜇2 tan 𝑖𝐵 = 𝜇1  Relation between Critical Angle and Brewster Angle Brewster’s Law tan 𝑖𝐵 = 𝜇 𝑖𝐵 = Brewster angle/polarising angle 𝑖𝐵 Critical Angle: Medium(𝜇) 1 𝑟 sin 𝜃𝐶 = 𝜇 1 ∴ tan 𝑖𝐵 = sin 𝜃𝐶 1 1 𝑖𝐵 = tan−1 And 𝜃𝐶 = sin−1 sin 𝜃𝐶 tan 𝑖𝐵  The polarizing angle of diamond is 67°. The critical angle of diamond is T nearest to: [Given tan 67° = 2.36 ] Given: 𝑖𝐵 = 67° To find: Critical angle 1 Solution: Critical angle is given by: 𝜃𝐶 = sin−1 tan 𝑖𝐵 1 𝜃𝐶 = sin−1 tan 67° 1 𝜃𝐶 = sin−1 2.36 1 1 1 1 < ⇒ sin−1 < sin−1 ⇒ 𝜃𝐶 < 30° 2.36 2 2.36 2 ∴ Out of the four option only 22° is less than 30°  Scattering by Polarization Observer If an unpolarized light gets scattered from air molecule, light perpendicular to original ray is plane polarized. Air molecule If we draw a plane perpendicular to the incident light, then from every viewpoint Incident Light on the plane we can see the plane polarized light. Observer  A ray of light, travelling in air, is incident on a glass slab with angle of incidence 60°. It is found that the reflected ray is plane polarized. The velocity T of light in the glass is: Given: 𝑖𝑝 = 60° To find: Velocity of light in glass (𝑣𝑔 ) Plane polarized Solution: As reflected light is plane 60° polarized, Air 𝑖𝑝 = 60∘ 90° According to Brewster’s law, Glass (𝜇) 𝜇 = tan 𝑖𝑝 = tan 60∘ = 3 𝑣𝑎 𝑣𝑎 As, 𝜇 = ⇒ 𝑣𝑔 = 𝑣𝑔 𝜇 3 × 108 𝑣𝑔 = = 3 × 108 𝑚/𝑠 3