Physics 133: tutorial week 6 Interference

Transcription

1 Physics 133: tutorial week 6 Interference 1. What is the angle of the sun above the horizon when light reflecte from a lake is completely plane polarize? Take the refractive inex of water as (36 56 ) When the light is completely polarize it is incient at the Brewster angle, for which tan θ B n. Hence θ B arctan Since θ B is the incient angle, the angle of the sun above the horizon is 90 θ B A sugar solution of concentration 1g/cm 3 is foun to rotate the plane of vibration of plane-polarize light passing through it by 6.64 per centimeter of length. The same light passe through a tube of 10cm length containing a sugar solution of unknown concentration unergoes a rotation of its plane of vibration of What is the concentration of the solution? (0.125gcm 3 ) The specific rotation [θ] is constant as we are concerne with the same solution at ifferent concentrations. For the first solution, from θ [θ]cl we have [θ] θ l 1 c cm 1 /gcm 3. For the secon solution therefore, c θ l 1 [θ] gcm Two sources are emitting ientical waves that have a wavelength of m. Determine whether constructive or estructive interference occurs at a point whose istances from the two sources are as follows: (a) 1.32m an 3.08m; (b) 2.67m an 3.33m; (c) 2.20m an 3.75m; () 1.10m an 4.18m. (con.; es.; es.; con.) Destructive interference occurs when the ifference in path length is n wavelengths. (a) m λ Since 4 N, constructive interference occurs (b) 1.5 estructive interference (c) 3.5 estructive interference () 7 constructive interference.

2 4. Re light of wavelength 644nm, from a point source, passes through two parallel an narrow slits which are 1mm apart. Determine the istance between the central bright fringe an the thir bright interference fringe forme on a screen parallel to the plane of the slits an 1.0m away. (1.93mm) For a ouble slit, constructive interference occurs when y λx. We require the istance to the thir bright fringe, which is 3 y. Hence 3 y m. 5. Fringes are prouce by Young s slits with wavelength m. The twentieth bright fringe (consiering the central fringe as of zero orer) is 1.00mm from the axis of the system. Calculate the position of the twelfth bright ringe for light of wavelength m. (0.45mm from the axis) The istance to the 20th bright fringe for light of wavelength λ is 1.0mm, hence 20 y λx x, which gives x For light of wavelength λ , we require 12 y x m. 6. Two slits are 0.158mm apart. A mixture of re light (wavelength 665nm) an yellow green light (wavelength 565 nm) falls on the slits. A flat observation screen is locate 2.24 m away. What is the istance on the screen between the thir orer re fringe an the thir orer yello green fringe? (0.43cm) The istance from the central fringe to the 3r orer green an re fringes are 3 y g 3λ gx an 3 y r 3λ rx respectively. The istance between these maxima is therefore 3 y r 3 y g 3x (λ r λ g ) ( ) m 4.25 mm. 7. A parallel beam of light containing only the wavelengths 490nm an 588nm falls normally on a pair of slits mm apart. At what istance from the central bright fringe on a screen 50 cm away oes a maximum of one wavelength first coincie with a maximum of the other? (4.9mm) Let λ 1 490nm an λ 2 588nm. We require n 1 λ 1 x n 1 y 1 n 2 y 2 or n 2λ 2 x, hence n n Hence the first time the interference fringes of the two colours coincie is for n 1 6 an n 2 5. (The secon time the fringes coincie will be for n 1 12 an n 2 10.) The corresponing istance from the central bright fringe is: n 1 y 1 n 1λ 1 x m 4.9 mm.

3 8. A layer of transparent plastic (µ 1.61) on glass (µ 1.52) looks ark when illuminate by light of wavelength 589 nm. Fin the two smallest possible nonzero values for the thickness of the layer. (183nm; 366nm) Assuming the materials are illuminate by light travelling through air, there will be a phase change on reflection at the air plastic interface since n air < n plastic, but not at the interface between plastic an glass since n plastic > n glass. Hence there will be estructive interference whenever 2t nλ, where λ is the wavelength of the light in plastic. The smallest non zero values of the thickness of the plastic will be for n 1 an n 2. Therefore t nλ nλ 2 2µ nm Similarly for n 2, t 366nm. 9. A mixture of re light (λ 661nm) an blue light (λ 472nm) shines perpenicularly on a thin layer of petrol (µ 1.40) lying on water (µ 1.33). The petrol layer has a uniform thickness of m. Destructive interference removes one of the colours from the reflecte light. By means of suitable calculations ecie whether the petrol looks re or blue in reflecte light. (blue) There will be a 180 phase change on reflection from the air petrol interface, but not from the petrol water interface since n air < n petrol an n petrol > n water. Destructive interference will therefore occur whenever 2t nλ, where λ is the wavelength of the light in petrol. For re light: n 2t λ 2tµ λ Similarly for blue light n 1.4. There is therefore cancellation of re light but not for blue light, hence the reflecte light appears blue. 10. A mixture of yellow light of wavelength 580nm an blue light of wavelength 450nm is incient normally on an air film mm thick. What is the colour of the reflecte light. (blue) For yellow light: n 2t λ For blue light: n 2t λ The yellow light cancels but not the blue light, hence the reflecte light appears blue.

4 11. White light is incient normally on a film of thickness 900nm an refractive inex 1.5 which is in air. In the reflecte light, which wavelengths (a) are missing an (b) experience constructive interference? (Assume the visible spectrum lies between 400nm an 700nm.) (450nm, 540nm an 675nm are missing; 415nm, 491nm an 600nm are enhance) (a) There is a phase change at the top surface of the film but not at the bottom. Cancellation therefore occurs when 2t nλ, where λ is the wavelength of the light in the film (refractive inex µ 1.5). The visible spectrum is approximately between 400 nm an 700nm (this varies from person to person). For violet light n 2tµ , λ 400 an for re light n 2tµ λ 750 Destructive interference will therefore occur for n 4, 5 an 6, which correspons to light of wavelength 450 nm, 540 nm, an 675 nm. (b) For constructive interference we use the relation n tµ λ which yiels n 6.25 an n 3.36 an hence the same n as for estructive interference. The corresponing wavelengths are 415 nm, 491 nm, an 600 nm. 12. Two flat glass plates are presse together at the top ege an separate at the bottom ege by a strip of tinfoil. The air wege is examine in yellow soium light (wavelength nm) reflecte normally from its two surfaces. If 42 ark fringes are observe, calculate the thickness of the tinfoil. ( cm) Because 42 fringes are observe, n 41 (since n 0, 1, 2,...). There is cancellation for 2t nλ, hence t nλ m.

5 13. Newton s rings are forme with yellow light between the curves surface of a plano convex lens an a flat glass plate in perfect contact. proving any formmula you use, calculate the wavelength of the light, given that the raius of the 10th ark ring from the centre is 0.25cm an the raius of curvature of the lens face is 1.0m. Explain also (a) what you coul expect to observe at the centre of the interference pattern (b) what change in appearance of the rings woul occur if re light were use instea of yellow (other than the change in colour)? (625nm) R R t t r n From the iagram it is clear (from the theorem of pythagoras) that R 2 rn 2 + (R t) 2 rn 2 + R 2 2Rt + t 2. If t r n, this gives 2Rt rn 2. Since the thickness is given by 2t nλ, the wavelength λ r2 n Rn ( ) m (a) At the centre of the ring pattern woul be a ark spot. (b) The iameters of the interference rings woul increase. 14. A piece of curve glass has a raius of curvature of 10.0m an is use to form Newton s rings. Not counting the ark spot at the centre of the pattern, there are one hunre ark fringes, the last one being at the outer ege of the curve piece of glass. The light being use has a wavelength of 654nm. What is the raius of the outermost ark ring in the pattern? (0.0256m) The raius of the nth ring is given by r 2 n nλr m 2, hence r n m. 15. In an experiment on Newton s rings the plano convex lens has a raius of curvature of 3m. Light of wavelength 600nm is use. What is the number of ark rings observe within a raius of 0.8cm? (35) At r 0.8cm: n r2 n λr ( ) However, n must be an integer, hence n 35.