Qualification > Sciences
ALL CIE PHYSICS DOUBTS HERE !!!
Deadly_king:
--- Quote from: $!$RatJumper$!$ on November 11, 2010, 10:15:49 am ---Alright that makes sense now :) thankx
Ok can you please have a look at S10 P21 Q4b
So, i use y = (D*lambda) / d to find d. Then I did 1 / d to find the answer. I got it right, but then looked at the MS and saw they used d*sin(theta) = n*lambda to find d. Then they did 1 / d to get the answer.
What im confused about is that when are we meant to know what formula to use where?
--- End quote ---
The formula you used can only be used when n = 1 that is only one wavelength is involved. It is normally used to find the fringe width which is the distance between the centres of two consecutive bright fringes or that between two dark fringes.
The formula you used proved to be good here since was actually 1. But if that was not the case you would not have been able to find the correct answer. But it's upto you to choose the method you find easier and more appropriate. The essential thing is that you get the required answer. :D
$!$RatJumper$!$:
--- Quote from: Deadly_king on November 11, 2010, 10:36:33 am ---The formula you used can only be used when n = 1 that is only one wavelength is involved. It is normally used to find the fringe width which is the distance between the centres of two consecutive bright fringes or that between two dark fringes.
The formula you used proved to be good here since was actually 1. But if that was not the case you would not have been able to find the correct answer. But it's upto you to choose the method you find easier and more appropriate. The essential thing is that you get the required answer. :D
--- End quote ---
I see. I just called my friend now and he says that d*sin(theta) = n*lambda can only be used for diffraction gratings and y = (D*lambda) / d
can only be used for double slit experiments. Is this true?
TJ-56:
--- Quote from: $!$RatJumper$!$ on November 11, 2010, 10:41:57 am ---I see. I just called my friend now and he says that d*sin(theta) = n*lambda can only be used for diffraction gratings and y = (D*lambda) / d
can only be used for double slit experiments. Is this true?
--- End quote ---
Yup. That is correct.
$!$RatJumper$!$:
In d*sin(theta) = n*lambda, what does the theta actually stand for?
Its really confusing me because for example,
In W03 Q4(b)(ii), they ask for the angle where the first image is seen. It also says the incident light is at right angles.
But in W06 Q4(b)(i), they ask for the numbers where is it visible so why do we suddenly use a value of 90 for theta in the equation: d*sin(theta) = n*lambda. Wouldnt we need to use the angle like used in W03?
TJ-56:
--- Quote from: $!$RatJumper$!$ on November 11, 2010, 10:53:51 am ---In d*sin(theta) = n*lambda, what does the theta actually stand for?
Its really confusing me because for example,
In W03 Q4(b)(ii), they ask for the angle where the first image is seen. It also says the incident light is at right angles.
But in W06 Q4(b)(i), they ask for the numbers where is it visible so why do we suddenly use a value of 90 for theta in the equation: d*sin(theta) = n*lambda. Wouldnt we need to use the angle like used in W03?
--- End quote ---
To find the maximum number n of the visible image; we use the eequation you stated,
which is n*lambda=d* sin (theta)
Arranged with n the subject: n = [d* sin (theta)]/ lambda
So to find the max value of n, we need the max value of sin theta, that is when theta is 90 degrees.
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