浙江大学大学物理甲下cha.ppt

Exercise Lesson Wave optics 1. Interference Phase difference and optical path difference (1) Double-slit interference (2) Interference from thin film If i =0, Example 1: Two slips interference, S1S2=0.7mm, D=100cm, =500nm. Now the single slip is shifted and SS1- SS2= /2, an optical medium plates with thickness of l and index of refraction of 1.5 placed at the back of one slits. The point o is 4th minimum fringe. Find (1) l, (2) the distance the central maximum to the point o. Solution: D r1 r2 x Example 2: There is an oil film (n2=1.20) on flat grass plate (n1=1.52). A light ray (=600nm) illuminates normally and is observed from above by reflected light. Find the thickness of the film at 5th bright fringe from the edge of oil film. Solution: At the edge of film, d=0 (m=0), is bright. Thus the 5th bright fringe is m=4. Example 3: As shown, =560nm, center point is dark, there are 20 dark rings in the outer region. Find the thickness of the film at center. n=1.4 n=1.5 Solution: If the 1st minimum correspond to m=0, the center point is m=20. If the 1st minimum correspond to m=1, the center point is m=21. Example 4 (E41-40): 2. Diffraction (1) Single-slit diffraction (2) Grating diffraction (3) Resolving power Airy spot: Grating: Principle maxima (4) missing order The angle is the angle between the incident x-ray and the plane. (5) Braggs law Example 5: (E43-12) Assume that light is incident on a grating at an angle as shown. Show that the condition for a diffraction maximum is d(sin sin) =m. If d=2m, =590nm, =30, d=3a, how many fringes can be observed? k=-1, 0, 1, 2, 4, 5, these 6 fringes are observed. Example 6: Assume that the limits of the visible spectrum are arbitrarily chosen as 400 and 760nm. Calculate the number of rulings per mm of a grating that will spread the first-order spectrum through an angle range of 20.0. Solution: Example 7: E43-22 Solution: (a) condition (1) is satisfied, (b) condition (3) is satisfied, d/a=3, a=800nm. (c) dsin=m2, let 90, m=4. All orders are 0, 1, 2. The 3 is missing order, the 4 is not observed actually. Example 8: To design a grating, when it is illuminated at normal incidence by white light, the light with =600nm is viewed in second order at 30, and the doublet of = 0.05 nm is barely resolved. Meanwhile there are no other principle maxima. Solution: When =400nm, Example 9: E43-11 Solution: If the second order spectra overlaps the third order, it is because the 700nm second order line is at a larger angle than the 400nm third order line. No matter what the value of d, the above formula come into existence. 3. Polarization (1) Law of Malus: I= Im cos2 (2) Brewsters law (3) Double refraction: no and ne Quarter-wave plate and circular polarization* Polarizing sheet Quick quiz Two ideal polarizing sheets are stacked so that none of the incident unpolarized light is transmitted. A third polarizing sheet is slipped between the first two sheets at an angle 45 to the bottom sheet. The fraction of light transmitted through the entire stack is (1) still zero. (2) 1/8. (3) 1/4. (4) 1/2. Example 10: Draw the reflection ray and refraction ray with their polarizing states. Example 11: P44-5 When the polarized light passes through the quarter- wave plate two times, its polarization plane is 90 rotation and no light will pass through the polarizing sheet. The coin can be visible. Side A must be the polarizing sheet, and that sheet must be at 45with the optical axis. Example 12: A light ray is composed of a linear polarized light and a unpolarized light. When it passes through a polarizing sheet, the intensity of transmitted wave is varied with the orientation of the sheet, and Imax=5Imim.