![Young’s Double Slit Experiment:
•
S
S• O
P
D
S1
S2
d
y
d/2
d/2
Single Slit Double Slit
Screen
The waves from S1 and S2 reach the point P with
some phase difference and hence path difference
∆ = S2P – S1P
S2P2 – S1P2 = [D2 + {y + (d/2)}2] - [D2 + {y - (d/2)}2]
(S2P – S1P) (S2P + S1P) = 2 yd ∆ (2D) = 2 yd ∆ = yd / D](https://image.slidesharecdn.com/6-230124065431-971ab265/85/6-wave_optics_1-ppt-11-320.jpg)
6.wave_optics_1.ppt
- 1. WAVE OPTICS - I 1. Wavefront 2. Huygens’ Principle 3. Refraction of Light Using Huygens’ Principle Created by Ganesh Ram Saini PGT-Physics JNV Mohangarh , Jaisalmer
- 2. Wavefront: A wavelet is the point of disturbance due to propagation of light. A wavefront is the locus of points (wavelets) having the same phase of oscillations. A line perpendicular to a wavefront is called a ‘ray’. Spherical Wavefront from a point source Cylindrical Wavefront from a linear source Plane Wavefront Pink Dots – Wavelets Blue Envelope– Wavefront Red Line – Ray •
- 3. Huygens’ Construction or Huygens’ Principle of Secondary Wavelets: S New Wavefront (Spherical) New Wave- front (Plane) . . . . . . . . (Wavelets - Red dots on the wavefront) 1. Each point on a wavefront acts as a fresh source of disturbance of light. 2. The new wavefront at any time later is obtained by taking the forward envelope of all the secondary wavelets at that time. Note: Backward wavefront is rejected. Why? . . . . . . . . . •
- 4. Refraction of Light Using Huygen’s Principle i i r r A B C D N N AB – Incident wavefront CD – Refracted wavefront XY – Refracting surface E F G X Y c, μ1 v, μ2 Denser Rarer Let AB is a plane wavefront incident on the interference XY at angle BAC = i .The interference separates a rarer medium from denser medium. 1,2 3 are the crossponding incident rays normal to AB . According to Huygen’s principle, every point on AB is a source of secondary wavelets. Let the secondary wavelet from B strick XY at C in t seconds. BC = c t .............. (i)
- 5. The secondary wavelet from A travel in the denser medium with a velocity v and would cover a distance v t in t seconds .Therefore, with A as centre and radius equal to v t. Draw an arc D. From C draw a tangent plane touching the spherical arc tangentially at D . Therefore, CD is the secondary wavefront after t seconds. 1’ , 2 ‘ , 3 ‘ are the corresponding refracted rays perpendicular to wavefront CD . AD = vt ..........(ii) Similarly, the secondary waveletes starting from anyother point E on the incident wavefront AB , after refraction at F , must reach the point G on CD in the same time in which the secondary wavelets would go from B to C . Therefore , CD is the refracted wavefront . Let r be the angle of refraction i. e . <ACD = r Triangle ACB and triangle ACD are similar triangles . In triangle ACB , sini = BC/AC = ct / AC And in triangle ACD , sinr = AD / AC = vt /AC Sini / sinr = c /v = n n = sini /sinr..........(iii) Whis is known as Snell’s law of refraction .
- 6. Also , the incident ray , normal to the interface XY and refracted ray , all lie in the same plane . This is the second law of refraction Questions Based On Refraction of Light Using Huygens’ Principle 1) Write Snell’s law. 2) Write Huygen’s Law. 3) Explain refraction of light using Huygen’s Principle THANKSYOU
- 7. Interference of Waves: • • Constructive Interference E = E1 + E2 Destructive Interference E = E1 - E2 S1 S2 Bright Band Dark Band Dark Band Bright Band Bright Band Crest Trough Bright Band Dark Band 1st Wave (E1) 2nd Wave (E2) Resultant Wave Reference Line E1 E2 E2 E1 The phenomenon of one wave interfering with another and the resulting redistribution of energy in the space around the two sources of disturbance is called interference of waves. E1 + E2 E1 - E2
- 8. Theory of Interference of Waves: E1 = a sin ωt E2 = b sin (ωt + Φ) The waves are with same speed, wavelength, frequency, time period, nearly equal amplitudes, travelling in the same direction with constant phase difference of Φ. ω is the angular frequency of the waves, a,b are the amplitudes and E1, E2 are the instantaneous values of Electric displacement. Applying superposition principle, the magnitude of the resultant displacement of the waves is E = E1 + E2 E = a sin ωt + b sin (ωt + Φ) E = (a + b cos Φ) sin ωt + b sin Φ cos ωt Putting a + b cos Φ = A cos θ b sin Φ = A sin θ θ Φ a b A b cos Φ b sin Φ A cos θ A sin θ We get E = A sin (ωt + θ) (where E is the resultant displacement, A is the resultant amplitude and θ is the resultant phase difference) A = √ (a2 + b2 + 2ab cos Φ) tan θ = b sin Φ a + b cos Φ
- 9. A = √ (a2 + b2 + 2ab cos Φ) Intensity I is proportional to the square of the amplitude of the wave. So, I α A2 i.e. I α (a2 + b2 + 2ab cos Φ) Condition for Constructive Interference of Waves: For constructive interference, I should be maximum which is possible only if cos Φ = +1. i.e. Φ = 2nπ Corresponding path difference is ∆ = (λ / 2 π) x 2nπ where n = 0, 1, 2, 3, ……. ∆ = n λ Condition for Destructive Interference of Waves: For destructive interference, I should be minimum which is possible only if cos Φ = - 1. i.e. Φ = (2n + 1)π Corresponding path difference is ∆ = (λ / 2 π) x (2n + 1)π where n = 0, 1, 2, 3, ……. ∆ = (2n + 1) λ / 2 Imax α (a + b)2 Iminα (a - b)2
- 10. Comparison of intensities of maxima and minima: Imax α (a + b)2 Imin α (a - b)2 Imax Imin = (a + b)2 (a - b)2 (a/b + 1)2 (a/b - 1)2 = Imax Imin = (r + 1)2 (r - 1)2 where r = a / b (ratio of the amplitudes) Relation between Intensity (I), Amplitude (a) of the wave and Width (w) of the slit: I α a2 a α √w I1 I2 = (a1)2 (a2)2 = w1 w2
- 11. Young’s Double Slit Experiment: • S S• O P D S1 S2 d y d/2 d/2 Single Slit Double Slit Screen The waves from S1 and S2 reach the point P with some phase difference and hence path difference ∆ = S2P – S1P S2P2 – S1P2 = [D2 + {y + (d/2)}2] - [D2 + {y - (d/2)}2] (S2P – S1P) (S2P + S1P) = 2 yd ∆ (2D) = 2 yd ∆ = yd / D
- 12. Positions of Bright Fringes: For a bright fringe at P, ∆ = yd / D = nλ where n = 0, 1, 2, 3, … For n = 0, y0 = 0 For n = 1, y1 = D λ / d For n = 2, y2 = 2 D λ / d …… For n = n, yn = n D λ / d y = n D λ / d Positions of Dark Fringes: For a dark fringe at P, ∆ = yd / D = (2n+1)λ/2 where n = 0, 1, 2, 3, … For n = 0, y0’ = D λ / 2d For n = 1, y1’ = 3D λ / 2d For n = 2, y2’ = 5D λ / 2d ….. For n = n, yn’ = (2n+1)D λ / 2d y = (2n+1) D λ / 2d Expression for Dark Fringe Width: βD = yn – yn-1 = n D λ / d – (n – 1) D λ / d = D λ / d Expression for Bright Fringe Width: βB = yn’ – yn-1’ = (2n+1) D λ / 2d – {2(n-1)+1} D λ / 2d = D λ / d The expressions for fringe width show that the fringes are equally spaced on the screen.
- 13. Distribution of Intensity: Intensity 0 y y Suppose the two interfering waves have same amplitude say ‘a’, then Imax α (a+a)2 i.e. Imax α 4a2 All the bright fringes have this same intensity. Imin = 0 All the dark fringes have zero intensity. Conditions for sustained interference: 1. The two sources producing interference must be coherent. 2. The two interfering wave trains must have the same plane of polarisation. 3. The two sources must be very close to each other and the pattern must be observed at a larger distance to have sufficient width of the fringe. (D λ / d) 4. The sources must be monochromatic. Otherwise, the fringes of different colours will overlap. 5. The two waves must be having same amplitude for better contrast between bright and dark fringes.
- 14. Colours in Thin Films: P Q R S r i t μ It can be proved that the path difference between the light partially reflected from PQ and that from partially transmitted and then reflected from RS is ∆ = 2μt cos r O Since there is a reflection at O, the ray OA suffers an additional phase difference of π and hence the corresponding path difference of λ/2. A B C For the rays OA and BC to interfere constructively (Bright fringe), the path difference must be (n + ½) λ So, 2μt cos r = (n + ½) λ When white light from the sun falls on thin layer of oil spread over water in the rainy season, beautiful rainbow colours are formed due to interference of light. For the rays OA and BC to interfere destructively (Dark fringe), the path difference must be nλ So, 2μt cos r = n λ End of Wave Optics - I