Go Back   Physics Help Forum > High School and Pre-University Physics Help > Light and Optics

Light and Optics Light and Optics Physics Help Forum

Reply
 
LinkBack Thread Tools Display Modes
Old 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
Attached Thumbnails
position of first , second and third maximum produced-img_20140910_163204-1-.jpg  
ling233 is offline   Reply With Quote
Old Sep 10th 2014, 04:36 AM   #2
Physics Team
 
ChipB's Avatar
 
Join Date: Jun 2010
Location: Morristown, NJ USA
Posts: 2,344
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.
ChipB is offline   Reply With Quote
Old Sep 10th 2014, 09:08 AM   #3
Senior Member
 
Join Date: Jun 2014
Posts: 306
Originally Posted by ChipB View Post
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 :
Attached Thumbnails
position of first , second and third maximum produced-img_20140910_163240.jpg  
ling233 is offline   Reply With Quote
Old Sep 10th 2014, 10:21 AM   #4
Physics Team
 
ChipB's Avatar
 
Join Date: Jun 2010
Location: Morristown, NJ USA
Posts: 2,344
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
ChipB is offline   Reply With Quote
Reply

  Physics Help Forum > High School and Pre-University Physics Help > Light and Optics

Tags
maximum, position, produced



Thread Tools
Display Modes


Similar Physics Forum Discussions
Thread Thread Starter Forum Replies Last Post
Calculate Object Position kerrymaid Kinematics and Dynamics 6 Feb 1st 2011 10:39 AM
Position/Motion Vector Getting a future position kerrymaid Kinematics and Dynamics 0 Jan 26th 2011 01:11 PM
position time ohm Kinematics and Dynamics 4 Aug 18th 2010 06:31 AM
position of cart ! christina Kinematics and Dynamics 4 Dec 8th 2009 08:59 AM
What double slit maximum is closest to the 2nd single slit secondary maximum mola Advanced Optics 1 May 24th 2009 09:34 PM


Facebook Twitter Google+ RSS Feed