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April 29, 2024, 11:43:51 pm

Author Topic: Young's double slit experiment?  (Read 2024 times)  Share 

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Yoda

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Young's double slit experiment?
« on: September 24, 2012, 06:59:13 pm »
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Question 2 AOS2 2011 phys exam 2   

The distance, L , is now increased.
Describe the effect of this change on the spacing of the observed interference pattern.   

My answer tell me if its correct:       Increasing the length will also increase the spacing between the fringes.
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paulsterio

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Re: Young's double slit experiment?
« Reply #1 on: September 24, 2012, 07:11:07 pm »
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Well what you can do is use the Fraunhofer Diffraction equation ;)



Jokes!

More simply,



S is the distance between slits and U is the "amplitude" of the diffracted wave. Thus, increasing the distance between the slits will increase the extent of diffraction.  (Just remember that)

Yoda

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Re: Young's double slit experiment?
« Reply #2 on: September 24, 2012, 07:12:28 pm »
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So is the fringes part wrong?
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Yoda

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Re: Young's double slit experiment?
« Reply #3 on: September 24, 2012, 07:14:34 pm »
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I believe its not talking about the width of the slits, but the distance between the slits and the object its projected on.
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paulsterio

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Re: Young's double slit experiment?
« Reply #4 on: September 24, 2012, 07:17:26 pm »
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Ohh, right, sorry, I hadn't read the question, but yeah, just had a look at it now.

VCAA 2011 AOS2 - Q2

Yep, in that case you're right, it's talking about the distance between the slits and the object that it is projected on.

Increasing the distance will increase the widths between the bands because think of it this way, the further the slits are from the screen, the more "magnified" the pattern will be...etc.

Lasercookie

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Re: Young's double slit experiment?
« Reply #5 on: September 24, 2012, 07:37:09 pm »
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I took a less rigourous approach than Paul and used this less scary looking bit of maths to remember how changes to the experimental set up would affect each other.

fringe spacing where is the wavelength, d is the distance between the slits and L is the distance from the screen.

You can find the reasoning for this on page 414 of Heinemann Physics, in one of those "physics in action" boxes. I don't know the page references for the other textbooks, but I believe it was in there too. It's just a bit of geometry + trig knowledge to derive it.

It's also mentioned here:
http://en.wikipedia.org/wiki/Fraunhofer_diffraction#Diffraction_by_a_double_slit
http://en.wikipedia.org/wiki/Double-slit_experiment#Classical_wave-optics_formulation

This is a diagram similar to the one in Heinemann.
http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/slits.html#c1