Physics Help Forum Quicksilver lamp

 Dec 5th 2018, 11:32 AM #1 Junior Member   Join Date: Dec 2018 Posts: 2 Quicksilver lamp The light from a quicksilver lamp passes a filter were all wavelengths except 546,1 nm absorbs. The light that passes the filter is made parallel with the help of a lens and is sent through a slit. On a screen 7.00 m from the slit the bending pattern is studied, se figure (the lights intensity distribution on a screen after bending in a slit). Use the figure to determine the slit width. I think that I need to somehow determine an angle from the figure and then use it in b*sin(x) = m*λ to solve for b. Thanks
Dec 6th 2018, 05:11 AM   #2
Senior Member

Join Date: Oct 2017
Location: Glasgow
Posts: 277
 Originally Posted by Legion I think that I need to somehow determine an angle from the figure and then use it in b*sin(x) = m*λ to solve for b.
Close, but the formula is this:

$\displaystyle \frac{s d}{D} = \lambda$

where:

s = slit width
d = distance between peak of diffraction pattern and the first minimum
D = distance between slit and screen
$\displaystyle \lambda$ = wavelength of light

A quick google showed this website, which explains things in more detail

SINGLE SLIT DIFFRACTION PATTERN OF LIGHT

 Tags lamp, quicksilver

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