GREPhysics.NET: GR9277 #20

GR9277 #20

Problem

GREPhysics.NET Official Solution

\prob{20}

In a double-slit interference experiment, d is the distance between the centers of the slits and w is the width of the slit, as shown in the figure above. For incident plane waves, an interference maximum on a distant screen will be "missing" when

$d=\sqrt{2}w$
$d=\sqrt{3}w$
2d=w
2d=3w
3d=2w

Optics $\Rightarrow$ }Missing Fringes

Missing fringes in a double-slit interference experiment results when diffraction minima cancel interference maxima.

From a bit of phasor analysis, one can derive the diffraction factor $\beta/2=\pi w/\lambda \sin\theta$ and the interference factor $\delta/2=\pi d/\lambda \sin\theta$ , where w is the width of the slits and d is the separation (taken from slit centers). The angles belong in the intensity equation given by $I \propto \sin(\beta/2)^2 \cos(\delta/2)^2$ .

Thus, the condition for a double-slit diffraction minimum is given by $\delta/2=m_d\pi=\pi w/\lambda\sin\theta \Rightarrow m_d\lambda = w \sin\theta$ .

Also, the condition for interference maximum is given by $\beta/2=m_i\pi=\pi d/\lambda \sin\theta \Rightarrow m_i\lambda = d\sin\theta$ .

Now, one needs to find the choice that allows for an integer $m_d$ . This immediately eliminates choices (A) and (B). But, this leaves choices (C), (D), and (E). Among the remaining choices, there is only one choice that allows for slits that are smaller than the separation. This is choice (D). Take it.

Alternate Solutions

htr 2019-09-29 15:22:47	If this is a double slit, the distance between two slits is D=d-w. Then for interference maximum,\r\n $(d-w)sin(\\theta)=\\lambdan1$ . \r\n where n1 is any integer. \r\n\r\nNow actually there are two slits so, to have a minimum at some point, all the waves should be cancelled each other (consider derivation of single slit diffraction minimum, in this case the equivalent distance is d). So, \r\n $dsin(\\theta)=\\lambdan2$ .\r\n where n2 is an odd number. \r\n\r\nThen we have, d/(d-w) = n2/n1. \r\n\r\nThe relationship is an integer fraction and also, requires d>w.\r\n\r\nThe choice is D. Reply to this comment
kstephe6 2012-11-05 07:09:44	Just note that d MUST be greater than w because... d= w + (a little bit) since d is from the middle of one slit to the other. So without the "little bit extra" this two slits merge to become one. Then note what many have shown from combining equations from the single and double slit equations... d(some integer) = w(some other integer) combine the two facts and the ONLY valid answer is D *EXTRA INFO BELOW* any number of integers will work here: d2=w, d3=w, d3=2w, etc... the only other restriction is that the integers must be small so as to keep in good agreement with the small angle approximation. In other words, if ETS offer d30=w12 then this would not be a valid answer. Reply to this comment
Kabuto Yakushi 2010-11-12 09:36:44	The problem says that the interference maximum is missing (in other words it is a minimum). Recalling the diffraction and interference formulas for a minimum (dark screen): $(m+\frac{1}{2})\lambda=dsin(\theta)$ equ. 1: interference minimum $(m)\lambda=wsin(\theta)$ equ. 2: diffraction minimum Dividing equation 1 by equation 2 and solving one gets (D). m has to be 1, since 0 would get an undefined result. Reply to this comment
nitin 2006-11-18 22:45:00	One easy way to solve this problem is as follows. The idea is that we need a minimum of the single-slit diffraction to coincide with a maximum of the double-slit interference pattern (which results in a "missing" interference maximum). The angular locations of the minima of the single single-slit diffraction pattern (which modulates or envelopes the double-slit diffraction pattern) are given by: $wsin(\theta)=m_1\lambda$ , where $m_1=1,2,3,...$ The angular locations of the maxima (bright fringes) of the double-slit interference pattern are given by: $dsin(\theta)=m_2\lambda$ , where $m_2=0,1,2,...$ Take the ratio of the 2 equations: $\frac{d}{w}=\frac{m_2}{m_1}$ . Now, it is obvious that $m_2$ will be in general greater than $m_1$ , since we have several interference maxima inside a diffraction maximum. Also, $m_1$ and $m_2$ are integers. The only instance which satisfies these 2 conditions is choice (D). Reply to this comment

Comments

htr
2019-09-29 15:22:47

If this is a double slit, the distance between two slits is D=d-w. Then for interference maximum,\r\n $(d-w)sin(\\theta)=\\lambda*n1$ . \r\n where n1 is any integer. \r\n\r\nNow actually there are two slits so, to have a minimum at some point, all the waves should be cancelled each other (consider derivation of single slit diffraction minimum, in this case the equivalent distance is d). So, \r\n $dsin(\\theta)=\\lambda*n2$ .\r\n where n2 is an odd number. \r\n\r\nThen we have, d/(d-w) = n2/n1. \r\n\r\nThe relationship is an integer fraction and also, requires d>w.\r\n\r\nThe choice is D.

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