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GR9677 #57
Problem
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Optics$\Rightarrow$}Diffractions

The single slit diffraction formula is $d \sin \theta = \lambda m$, where one has integer m for maxima and half-integers for minima. (Opposite to single-slit interference.)

Given $m=1$, $\theta = 4E-3 rad$, $\lambda = 400E-9 m$, and making the approximation $\sin \theta \approx \theta$ for small angles, one has the following equation for d,

$d\approx \frac{\lambda m}{\theta} = \frac{4E-7}{4E-3}=1E-4$, as in choice (C).

Alternate Solutions
 scottopoly2006-11-03 19:39:18 Use Raleigh's criterion!Reply to this comment
faith
2010-11-11 19:33:52
id like to also point out.

no need to convert radian to degree. if u do so, u have to use sin teta and... its impossible to get the value without memorising the sin table.

unlike me.. i did this with a calculator. hahar..
gman
2010-11-10 18:20:37
Just to clarify...

$d sin(\theta) = m \lambda$ with integer, non-zero $m$ giving successive minima.

$m=1 \Rightarrow$ first minimum.
$\theta = 0 \Rightarrow$ central maximum.

So $d=\frac{400nm}{4E-3} = 1E-4$
 scasplte22011-09-01 18:59:14 This made it so much clearer to me... thank you
azureblue22
2010-11-05 21:33:36

There is also a factor of two hidden in the distance ${d}$, allow me to explain.rnrnFor single slit diffraction the diffraction pattern emerges. because the photons interfere with each other as they pass through the single slit. For a minimum to occur each photon must be $\pi$ radians out of phase with another photon passing through the slit. One way to count the out of phase pairs is to divide the slit in two at $\frac{d}{2}$. Pair the photon at the very bottom of the slit with the photon in the middle of the slit, then think of these two locations as two separate, infinitesimally small slits. Wa-lah! Were back to the double slit pattern! Now move the single slit very far away from the basement wall where you measured the double slit pattern in undergraduate lab. The slit becomes a point and you get familiar results. And look on the bright side, you are never going to see that wall again because you're going off to grad school. Good luck.

P.S. So you see, the formula becomes: $\frac{d}{2}{=sin}\theta\frac{\lambda}{2}$.
azureblue22
2010-11-05 21:32:23
For all you guys who disagree about the factor of two I have your answer. There is also a factor of two hidden in the distance ${d}$, allow me to explain.rnrnFor single slit diffraction the diffraction pattern emerges. because the photons interfere with each other as they pass through the single slit. For a minimum to occur each photon must be $\pi$ radians out of phase with another photon passing through the slit. One way to count the out of phase pairs is to divide the slit in two at $\frac{d}{2}$. Pair the photon at the very bottom of the slit with the photon in the middle of the slit, then think of these two locations as two separate, infinitesimally small slits. Wa-lah! Were back to the double slit pattern! Now move the single slit very far away from the basement wall where you measured the double slit pattern in undergraduate lab. The slit becomes a point and you get familiar results. And look on the bright side, you are never going to see that wall again because you're going off to grad school. Good luck. rnrnP.S. So you see, the formula becomes: $\frac{d}{2}{=sin}\theta\frac{\lambda}{2}$.
mianghazanfar786
2010-07-20 18:17:20
There is a confusion that "one has integer m for maxima and half-integers for minima. (Opposite to single-slit interference.)"

We should use directly for diffraction pattern,
1 for minima and 1/2 for maxima so we use m=1
to solve the question by Bragg's Diffraction law
archard
2010-05-30 13:20:12
If all you knew was that minima occur at half integers, you could make a wild guess by noticing that C and D differ by a factor of a half, and since ETS is always trying to trick you it's probably one of the two, and most likely C because 1 is half of 2.
sullx
2009-11-02 18:39:02
Careful! The value 'm' corresponds to Minimum for single slit diffraction, and maximum for double slit interference.

Yosun: "The single slit diffraction formula is d \sin \theta = \lambda m, where one has integer m for maxima and half-integers for minima. (Opposite to single-slit interference.) "
 niux2009-11-05 14:15:02 Gee. Seems Yosun using different semantics?. I agree with RootMeanSquare explanation, and for single slit, m should tell the number of minima you are dealing with.
tan
2009-10-23 23:09:13
Central maximum occurs at $\theta$ = 0
First minimum occurs at d sin $\theta$ = $\lambda$ /2
Answer should be 5 x $\10^-5$.
What's wrong?
boundforthefloor
2006-11-24 23:43:14
Wolfram explains this well giving minima at $Sin(\theta) = m\lambda/d$ and the primary maxima at 0.

see:

http://scienceworld.wolfram.com/physics/FraunhoferDiffractionSingleSlit.html
 physicsisgod2008-10-29 22:12:18 Yea, you can think of it like the center max is actually at $\frac{d}{2}$, and then the optical path difference between a ray diffracted at an angle $\theta$ at the center ($\frac{d}{2}$), and one at 0, is $\frac {\lambda}{2}$ if there's a minimum (destructive inteference), and that has to be equal to $\frac{dsin\theta}{2}$.
RootMeanSquare
2006-11-24 14:31:54
Okay, I'll never remember that I don't have to log in before posting...

Anyway: It seems here is a little confusion about the factor 2. $d sin \theta$ gives you the optical retardation between waves originating a distance d apart within the slit. For a full minimum, you need to find a pair of beams that cancel each other out for every point within the slit, i.e. you actually need a optical retardation of $\lambda/2$ for point $d/2$ apart. Hence you end up with $\lambda /\alpha$ as the width of the slit.
 mhas0352007-03-25 21:54:27 Thanks! Some clarity!
RootMeanSquare
2006-11-24 14:25:50
scottopoly
2006-11-03 19:39:18
Use Raleigh's criterion!
 physicsisgod2008-10-29 22:04:35 For those of you who are not as enlightened as our fortunate friend scottopoly, Raleigh's Criterion is a formula for deriving the distance between two point sources necessary to resolve them, given their angular separation, $\alpha_{s}$, and the wavelength of emitted light, $\lambda$. The formula is $\alpha_{s} = \frac{1.22 \lambda}{d}$ But Raleigh's Criterion only applies for a circular slit, thus the factor of 1.22. If you just use the regular old equation for the distance between the center maximum and the first minimum given a single slit diffraction, $\lambda m = d sin\theta$, and the small angle approximation, you'll derive the same formula without the 1.22 factor Source: http://www.ux1.eiu.edu/~cfadd/3050/Ch20WO/OpRes.html
 apr20102010-04-06 14:20:00 Very nice, also makes me remembering the raileigh criterion!
buddy.epson
2006-10-13 16:01:42
Re: Previous User Comment

For single slit diffraction, m corresponds to the center of the dark band (minima). For multiple slits and x-ray diffraction, it corresponds to the center of the bright bands (maxima). The posted solution is correct.
 pablojm2006-10-28 14:39:50 In fact, nahmad is right and the problem is incorrectly worded. You can check in any optics book that the formula for minima is d(theta)=m(lambda), where m is a half-integer. Plugging in m=0 and the given values for (lambda) and (theta), you get answer B. The mistake in the original solution is that you stated that m should be a half-integer for minima, and then you plugged in m=1. Cheers, Pablo
 f4hy2009-04-03 17:37:10 The problem states angle between the first minimum and the central max. Not the angle between two mins.
2006-03-31 22:40:22
I don't understand. The problems gives the angle between the first minimum and the central max. Doesn't that correspond to m=1/2 which leads to d = 5 x 10^-5 which is B?
 suyq2010-11-07 19:59:33 I think B is the right answer!
 nimesh.nic2011-11-01 01:20:14 I think this is the right ans ...please someone explain .....

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$