Another Q about double-slit experiment

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SUMMARY

The discussion focuses on calculating the spacing between slits in Young's double-slit experiment using 589nm light and a 2.00m distance from the slits to the screen. The tenth interference minimum is located 7.26mm from the central maximum. The correct formula for determining the spacing is derived from the relationship: distance from central max = (wavelength * distance to screen * order number) / spacing of slits. The calculated spacing between the slits is approximately 162.25 micrometers.

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aquabum619
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Homework Statement


Youngs double-slit experiment is performed with 589nm light and a distance of 2.00m between the slits and the screen. The tenth interference minimum is observed 7.26mm from the central maximum. Determine the spacing on the slits.

Homework Equations


Iavg = (Imax) cos^2 [(pi d sin angle) / lambda]
(it is very frustrating not being able to insert relevant symbols )

The Attempt at a Solution


?
 
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hm you seem to be using the wrong formula, in this case we are not measuring the intensity but using: distance from central max= (wavelength * distance to screen*order number)/spacing of slits. Hoped that helped
 
589nm x 2.00m x 10 / .00726m = spacing of slits

1622589m between slits?
 
aquabum619 said:

Homework Statement


Iavg = (Imax) cos^2 [(pi d sin angle) / lambda]
(it is very frustrating not being able to insert relevant symbols )

Sorry to change the topic, but look here for some info on LaTeX https://www.physicsforums.com/showthread.php?t=8997 if you want to make the equations look right
 

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