Solve Bright Fringes Problem: Step-by-Step Guide

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The discussion focuses on calculating the order of interference fringes in a double-slit experiment. The original poster seeks clarification on determining the value of m when identifying nodal and antinodal lines. It is confirmed that dark fringes occur halfway between bright fringes, leading to a calculation of m=4.5 for the fifth dark fringe based on counting from the central antinode. Participants express appreciation for the guidance received in understanding the concepts involved.
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Homework Statement


Original Post:
[post]1998976[/post]
i would like to seek help from anyone of you,could you please help me with these questions?
1)The fourth nodal line on an interference pattern is 8.4 cm from the first antinodal line when the screen is placed 235 cm from the slits. The slits are separated by 0.25 mm.?
the answer for m=2.5 ,can you please help me to check and rectify my mistake if have any of them..is that because of there are 5 bright fringes;then we start count from 1st antinodal line till 4th nodal line,so it has 2.5 steps..am i wright?if not,then how do we know the value of m?is there any way to evaluate m?thanks for your kind help..



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The Attempt at a Solution

 
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The link to your earlier post isn't working. Did you list relevant equations and show an attempt at a solution when you originally posted this problem?
 


heth said:
The link to your earlier post isn't working. Did you list relevant equations and show an attempt at a solution when you originally posted this problem?

hmm,actually i just wanted to know how to figure out the m when i encounter this kind of questions...

eg of a relevant que is:Two slits separated by 0.250 mm produces an interference pattern in which the fifth dark band is located 12.8 cm from the central antinode when the screen is placed a distance of 8.2 meters away.

solution: because central antinode is m=o wright?then counted frm central antinode to 5th dark fringe..so it works out to be 4.5!so frm antinode m=0,go through m=1,2,3,4 when it reaches 4th fringes of bright fringe its m=4.thus que requires 5th dark fringe so the other half contributes which is 4th bright fringes + 1/2 bright fringe ans wll be 4.5,am i wright?if its not the way,could you please show me the exact way to determine m?thank a lot
 


> am i wright.

You certainly are, Wilbur! The dark patches will be halfway between the light patches - so if you imagine/sketch the fringes then count along, you'll get m=4.5 for the 5th dark patch.
 


heth said:
> am i wright.

You certainly are, Wilbur! The dark patches will be halfway between the light patches - so if you imagine/sketch the fringes then count along, you'll get m=4.5 for the 5th dark patch.

oh,ok,i finally understand..thanks for your kind help,is much appreciated..btw,what's wilbur?haha,thanks a lot yea..what's your name?glad to have you to teach me..thanks..
 


Wilbur and Orville Wright - the Wright brothers.
 


heth said:
> am i wright.

You certainly are, Wilbur!

Wilbur and Orville Wright - the Wright brothers.

:smile: :smile: :smile: LOL
 
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