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  GR0177 #70
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Optics}Interference

The single-slit interference equation for bright fringes is given by m\lambda \propto d \sin \theta \approx d \theta (for small angles), where d is the width of the slit and m is an integer.

Since c=\lambda \nu, one can relate the above equation to frequency to get m c/f \propto d \theta. Increasing frequency would decrease the angle. Thus, the fringes would get closer together. Increasing the frequency by a factor of 2 would decrease the separation by 2, as in choice (B).
See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
carle257
2010-04-01 23:04:36		Recall also that as the wavelength becomes small compared to the slit, the spread of the fringes will become smaller. If you remember that the separation is linear in \lambda then you can deduce answer (B) which is half of the original separation. Just another way of looking at it without remembering the exact diffraction equations.Alternate Solution - Unverified
Comments
Ddrew
2012-08-08 04:27:12		Recall the formula for the distance between fringles \Delta x=\frac{\lambda D}{d}, where D is a distance between a screen and a source of light; d stands for a distance between slits.
From here, once easly could rewrite for frequency \Delta x=\frac{c D}{d \nu}.		NEC
D-Rock
2010-09-18 17:58:36		You know from a vague memory of diffraction formulas that this will be linear (I mean, \sin \theta = \theta here), and if your wavelength's tiny enough, they won't even see the slits, so limiting case gives that the angles will go to zero, hence it must B.NEC
carle257
2010-04-01 23:04:36		Recall also that as the wavelength becomes small compared to the slit, the spread of the fringes will become smaller. If you remember that the separation is linear in \lambda then you can deduce answer (B) which is half of the original separation. Just another way of looking at it without remembering the exact diffraction equations.Alternate Solution - Unverified
tau1777
2008-11-03 19:35:02		i can't really tell if this is the same solution as what's been posted. so here goes: constructive interference for double slit is given by dsin\theta = n\lambda. then i made a small angle approximation (really for fun, just assuming that it could be done, and since we have no info about angles) so we get d\theta= n\lambda. then remembering that sin\theta=\delta y/R where R is the distance from the slits to the screen, and \del y is a height on the screen from y=0 where maximum interference peak is. then solving for solving for theta, and setting the equations equal to one another we get Ld\theta\lambda n = \del y. then we know that if frequency gets doubled the wavelength should be halved and then the distances \delta y on the screen will also be halved.
chemicalsoul
2009-10-30 00:19:40		 I did this way too.
NEC
phys2718
2008-09-28 10:28:17		This is not a single slit interference problem, it is a double slit problem - and d is the distance between the very narrow slits, not the width of a slit. The interference maxima equation is however correct ( but for the DOUBLE slit problem).NEC

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i can't really tell if this is the same solution as what's been posted. so here goes: constructive interference for double slit is given by dsin\theta = n\lambda. then i made a small angle approximation (really for fun, just assuming that it could be done, and since we have no info about angles) so we get d\theta= n\lambda. then remembering that sin\theta=\delta y/R where R is the distance from the slits to the screen, and \del y is a height on the screen from y=0 where maximum interference peak is. then solving for solving for theta, and setting the equations equal to one another we get Ld\theta\lambda n = \del y. then we know that if frequency gets doubled the wavelength should be halved and then the distances \delta y on the screen will also be halved.

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