Calculating Width of Single Slit Diffraction Pattern

AI Thread Summary
To calculate the width of a single slit in a diffraction pattern, the central bright fringe width of 1.9 cm and the distance to the screen of 1.05 m are used. The equation .5asintheta=.5wavelength is questioned, particularly the inclusion of the factor of .5. The small angle approximation allows for the use of tanθ ≈ sinθ, and the height "h" should be measured from the central maximum. Clarification is provided that "h" is indeed 1.9 cm and that the correct approach involves understanding the diffraction equation properly. Understanding these principles is essential for solving the problem accurately.
phy112
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Homework Statement



The central bright fringe in a single slit diffraction pattern from light of wavelength 412 nm is 1.9 cm wide on a screen that is 1.05 m from the slit.

How wide is the slit?

Homework Equations



i keep trying the equation .5asintheta=.5wavelength
I can't seem to get the right answer but I am sure its easy

The Attempt at a Solution

 
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Express sinθ in terms of pattern position and distance to screen. (Use a small angle approximation.)
 
is it the eq. tantheta=h/D?
 
phy112 said:
is it the eq. tantheta=h/D?
Yes, and for small angles tanθ ≈ sinθ.
 
do you use h=1.9 cm and D=1.05m?
 
phy112 said:
do you use h=1.9 cm and D=1.05m?
Not exactly. Note that in the diffraction equation, "h" is measured from the central maximum.

Also, when you wrote ".5asintheta=.5wavelength" why do you have .5 in there?

Read this: http://hyperphysics.phy-astr.gsu.edu/HBASE/PHYOPT/sinslit.html"
 
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