Physics 2 - Lecture 4: Interference (Optics) - Huynh Quang Linh

Nội dung text: Physics 2 - Lecture 4: Interference (Optics) - Huynh Quang Linh

1. INTERFERENCE Tran Thi Ngoc Dung – Huynh Quang Linh – Physics A2 HCMUT 2016
2. CONTENTS  Optical path length (OPL),  Malus’s theorem  Conditions for Interference, Constructive Interference, Destructive Interference.  Relationship between Phase Difference and Optical Path Difference.  Change of Phase Due to Reflection  Thin-Film Interference
3. Intensity  Intensity is the wave energy per a unit area, which is perpendicular to the propagation ditection, per unit time.  Intensity is propotional to the squared amplitude of the light wave. I= kA2 [W/m2] We choose the proportional coefficient k =1 I=A2
4. Malus’s theorem Optical path length of light rays between two wave fronts is equal. A 2 L1 n1 A1I n2 IK n2 KB1 L2 n1 A2 H n1HJ n2 JB2 A 1 H n2 IK n2 IJ sin r n i 1 n HJ n IJ sin i i J 1 1 I r n1 sin i n2 sin r n2 K r n2 IK n1HJ B2 L1 L2 B 1
5. Interference of two coherent waves Consider 2 coherent waves arriving at a point M: E A cos(t ) 1 1 1 A E A cos(t )  2 2 2 A2 2 A The resultant wave function is : 1 1 E E1 E2 A1 cos(t 1) A2 cos(t 2) E Acos(t ) 2 2 A A1 A2 2A1A2 cos( 1 2) A sin A sin tan  1 1 2 2 A1 cos 1 A2 cos 2 2 Intensity I A I1 I2 2 I1I2 cos( 1 2 )
6. Relation between Phase difference and Optical Path Difference (OPD) 2 L  ConstructiveInterference : Δ 2mπ ΔL mλ λ Destructive Interference : Δ (2m 1)π ΔL (2m 1) 2
7. Lloyd’s experiment (cont.) Optical Path Difference We Experiment’s expect Results L=(OI+IM)-OM =m M Bright M Dark L=(OI+IM)-OM=(m+1/2) M Dark M Bright Optical Path Difference Experiment’s Results L=(OI+/2 + IM) - OM =m +/2 M Dark = (m+1/2)  L=(OI++/2 IM)-OM=(m+1/2) +/2 M Bright = (m+1) 
8. Thin – film interference  a phenomenon that occurs when incident light waves reflected by the upper and lower boundaries of a thin film interfere with one another.  thin layers of oil on water or the thin surface of a soap bubble. The varied colors observed when white light is incident on such films result from the interference of waves reflected from the two surfaces of the film
9. Thin film of Varied Thickness – Fringes of Equal Thickness - Extended Light source - Thin film of varying thickness, of refractive index n. O The optical path difference of the mắt two waves reflected at the lower H and upper surfaces of the thin i’ i film is : A K M 2 2 n ΔL 2d n sin i  / 2 B OPD depends on incident angles i and thickness d . Angles i and i’ are almost the same. OPD depends on thickness d. Points of equal thichness have the same intensity. Interference fringes are called Fringes of Equal Thickness .
10. Air Wedge The distance between any two consecutive bright fringes or two consecutive dark fringes is called fringe spacing d d  m=2 sin m 1 m m=1 i 2i i m=0 x  i 2 Dark fringe positions O Bright fringe position xd mi m 0,1,2 i x mi m 0,1,2 b 2
11. Glass wedge ( Cont.)  i 2n m=2 n m=1 x m=0 i Bright Fringe Position Dark Fringe Position O x t mi m 0,1,2 i x mi m 0,1,2 s 2
12. Newton’s rings O’ Fringe radius Ro R r R2 (R d)2 2Rd d2 2Rd 1 r 2Rd R m m 1,2,3 Bright Bright 2 r rDark 2RdDark R m m 0,1,2,3 d +/2 O R: curvature radius of the spherical surface of the convex lens.
13. Example 35.4 0.02 10 3 2 10 4(rad) 10 10 2  0.5 10 6 i 1.25mm 2 2 2 10 4 Example 35.5  0.5 10 6 i 0.94mm 2n 2 1.33 2 10 4
14. Example 37.4 Nonreflective Coatings for Solar Cells Solar cells—devices that generate electricity when exposed to sunlight—are often coated with a transparent, thin film ofsilicon monoxide (SiO, n = 1.45) to minimize reflective losses from the surface. Suppose that a silicon solar cell (n = 3.5) is coated with a thin film of silicon monoxide for this purpose (Fig. 37.20). Determine the minimum film thickness that produces the least reflection at a wavelength of 550 nm, near the center of the visible spectrum.  550nm d 94.8nm min 4n' 4 1.45 To finalize the problem, we can investigate the losses in typical solar cells. A typical uncoated solar cell has reflective losses as high as 30%; a SiO coating can reduce this value to about 10%. This significant decrease in reflective losses increases the cell’s efficiency because less reflection means that more sunlight enters the silicon to create charge carriers in the cell. No coating can ever be made perfectly nonreflecting because the required thickness is wavelength dependent and the incident light covers a wide range of wavelengths.