OSU PHYSICS 1251 - Example Problem: Interference Part 2
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Example Problem Interference Part 2 Physics 1251 TA Brian Clark 11 23 2015 1 Two speakers are coherent sources of sound that emit sound waves in phase Point P is a distance d1 2 45 m from one speaker and a distance d2 1 78 m from the other speaker The speed of sound is 343 m s Find three di erent frequencies that will experience destructive interference at P For destructive destructive interference at P we want the path length di erence between each of the speakers and P to be equal to some odd half integer multiple of wavelength d d2 d1 m 1 2 m 0 1 2 Additionally we know that v f so d m 1 2 v f Solving for f we nd v 2 d m 0 f0 256 Hz m 1 f1 768 Hz m 2 f2 1280 Hz 3v 2 d 5v 2 d 2 Light of wavelength 600 nm passes through a pair of slits each slit is 50 m wide and the slits are separated by 170 m a How many bright fringes occur on a screen 1 3 m wide that is 1 5 m past the slits We want to count the total number of bright fringes in the double slit pattern out to the angle where the end of the screen is The screen is 1 5 m away from the slits and the edge of the screen is located at a distance of 1 3 2 0 65 m from its center so arctan 23 4 0 65 1 5 Next we can use the equation representing double slit interference maxima to calculate the order m of the last bright fringe on the screen dsin m m m 112 7 dsin This means there are 112 fringes to the right and to the left of the central bright maximum at m 0 So in total we have b How many bright fringes occur in the whole pattern N 112 112 1 225 fringes The last bright fringe in the entire pattern occurs when the angle of interference is 90 Thus we set sin 1 in the double slit interference maximum equation and nd the order of the last interference maximum d 1 m m 283 3 170 m 600 m Meaning there are 283 full fringes on each side of the di raction pattern and then the one in the middle So the total in the pattern is N 283 283 1 567 fringes 3 Two colors of light pass through a double slit at the same time The second interfer ence minimum of color A is at the same angle as the third maximum of color B What is the ratio of A B The equations describing interference maxima and minima for a double slit are Since the second minimum of color A and the third maxima of color B are located at the same angle we can write d sin m maxima d sin m 1 2 minima d sin A d sin 3 B 3 2 A 3 B 2 A B d sin m minima d sin m 1 2 maxima d sin B d sin 2 A A 2 B 1 75 A B 7 4 7 2 3 2 7 2 4 Two colors of light pass through a single slit at the same time The second di raction minimum of color A is at the same angle as the third maximum of color B What is the ratio of A B The equations describing interference maxima and minima for a single slit are Again the second minimum of color A is at the same angle as the third maximum of color B 5 A 0 500 micrometers thick lm of n 1 25 soap is suspended in mid air a For what wavelengths of visible light is the re ected light a maximum For what wavelengths of visible light is the re ected light a minimum We are going to employ the equations for constructive and destructive thin lm interference in the case where n3 air n2 soap n1 air Then we have 2t m 1 2 maxima 2t m minima n n For maxima the only m with a in the visible spectrum 400 700 nm is m 2 2 500 nm For minima m 2 and m 3 both work with m 2 2 625 nm and m 3 3 416 nm b If instead of being suspended in mid air the soap lm was applied to a piece of glass with say nglass 1 5 what would those wavelengths be If the lm as applied to glass then now n3 glass n2 soap n1 air so the condi tions for destructive and constructive interference change and our answers above just swap roles Maxima are now found at 625 nm and 416 nm and minima at 500 nm

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# OSU PHYSICS 1251 - Example Problem: Interference Part 2

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