Calculating Width of Single Slit Diffraction Pattern

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Homework Help Overview

The discussion revolves around calculating the width of a single slit based on the characteristics of its diffraction pattern, specifically focusing on the width of the central bright fringe and the wavelength of light used.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between the slit width, the wavelength of light, and the geometry of the diffraction pattern. There are attempts to apply trigonometric relationships and small angle approximations, with questions about the correct variables to use in the equations.

Discussion Status

The discussion is active, with participants providing guidance on how to express certain variables and clarifying the use of specific equations. There is an ongoing exploration of the correct interpretation of the parameters involved in the diffraction equation.

Contextual Notes

Participants note potential confusion regarding the definitions of variables in the context of the diffraction pattern and the specific measurements being used. There is also a mention of a resource for further reading on the topic.

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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