Double-Slit Experiment problem?

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Homework Help Overview

The discussion revolves around a problem related to the double-slit experiment, specifically focusing on the positioning of bright and dark fringes. The original poster is trying to determine the distance of the first dark fringe from the central maximum, given the distance of the third order bright fringe.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the distance of the first dark fringe based on the provided distance of the third bright fringe but expresses uncertainty due to missing angle and wavelength information. Some participants reiterate the relevant equations for maxima and question how to obtain the necessary angle and wavelength. Others suggest using approximations for small angles.

Discussion Status

The discussion is ongoing, with participants exploring different aspects of the problem. Some guidance has been offered regarding the equations involved and approximations for small angles, but there is no consensus on how to proceed due to the lack of specific values for angle and wavelength.

Contextual Notes

Participants note the absence of specific angle and wavelength information, which are crucial for solving the problem. The original poster also highlights the challenge of working with the provided data.

brinstar
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Homework Statement


In a double-slit experiment, the third order bright fringe is 15 mm from the central fringe. What is the distance of the first (zero-th order) dark fringe from the central maximum?

Homework Equations


(m+.5)(lambda) = dsin(theta) => dark fringe
m(lambda) = dsin(theta) => bright fringe

The Attempt at a Solution


I honestly don't know. Like, there's no angle or wavelength. The most I've come up with is 15 mm divided by 3 (because it's the third) and getting 5 mm. I seriously don't know.
 
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the equation you need is
dsinθ = nλ for maxima...you have this in your post
 
lychette said:
the equation you need is
dsinθ = nλ for maxima...you have this in your post

But where do I get the angle and wavelength from?
 
For angles this small you can use the approximation: sin theta = theta = x / D where x
is the fringe displacement from the central maximum and D is the distance to the screen.
 

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