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 Light and Optics Light and Optics Physics Help Forum Sep 10th 2014, 01:42 AM #1 Senior Member   Join Date: Jun 2014 Posts: 306 position of first , second and third maximum produced i dont understand the ans attached. can someone show me how do you do this question please? thanks! https://www.flickr.com/photos/110120...6/15010201737/ ans   Sep 10th 2014, 04:36 AM #2 Physics Team   Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,352 You haven't posted the entire question, so it's impossible to explain why the values of x1, x2 and x3 are as shown. But I assume that you have been given a wavelength that originates from S1 and S2 and are looking for positions along the x axis where the waves are in phase. Consequently the length of the hypotenuse must be equal to the length of the base plus a whole number of wavelengths, so that the waves are in phase at P. Using Pythagoras: (x+nL)^2 = 4^2 + x^2 where L is wavelength. Set n = 1 and solve, then n=2, and finally n=3. I think you'll find that the answers are inverted however - I believe x1=7.5m, x2=3m, and x3=1.2m. Last edited by ChipB; Sep 10th 2014 at 08:42 AM.   Sep 10th 2014, 09:08 AM   #3
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Join Date: Jun 2014
Posts: 306
 You haven't posted the entire question, so it's impossible to explain why the values of x1, x2 and x3 are as shown. But I assume that you have been given a wavelength that originates from S1 and S2 and are looking for positions along the x axis where the waves are in phase. Consequently the length of the hypotenuse must be equal to the length of the base plus a whole number of wavelengths, so that the waves are in phase at P. Using Pythagoras: (x+nL)^2 = 4^2 + x^2 where L is wavelength. Set n = 1 and solve, then n=2, and finally n=3. I think you'll find that the answers are inverted however - I believe x1=7.5m, x2=3m, and x3=1.2m.
sorry for that . i upload the wrong photo .hope you can help! thanks! here's the correct one :   Sep 10th 2014, 10:21 AM #4 Physics Team   Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,352 Given that L = 1m, use the technique I described earlier and you have for n=1: (x+1)^2 = 4^2 + x^2 what do you get for x? Then for n=2: (x+2)^2 = 4^2 + x^2 and n=3: (x+3)^2 = 4^2 + x^2  Tags maximum, position, produced Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post kerrymaid Kinematics and Dynamics 6 Feb 1st 2011 10:39 AM kerrymaid Kinematics and Dynamics 0 Jan 26th 2011 01:11 PM ohm Kinematics and Dynamics 4 Aug 18th 2010 06:31 AM christina Kinematics and Dynamics 4 Dec 8th 2009 08:59 AM mola Advanced Optics 1 May 24th 2009 09:34 PM