Single slit diffraction - distance between 1st&2nd order dark fringes

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Homework Help Overview

The problem involves single slit diffraction, specifically calculating the distance between the first and second-order dark fringes on a screen when monochromatic light passes through a slit. The parameters include the wavelength of light, the width of the slit, and the distance to the screen.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formula used for calculating fringe positions and question the meaning of the variables involved. There is an exploration of the conditions under which the approximation for small angles is valid, and the relationship between angle and fringe distance is examined.

Discussion Status

Some participants have provided clarifications regarding the variables in the formula and the conditions for using approximations. There is an ongoing exploration of the relationship between angle and distance, with no explicit consensus reached on the correct approach to the problem.

Contextual Notes

Participants are working under the constraints of understanding single slit diffraction and the implications of using different formulas based on angle approximations. There is uncertainty regarding the handling of dark fringes specifically.

louza8
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Homework Statement


Monochromatic light with wavelength 580nm passes through a single slit 2.60um wide and 2.10m from a screen.

Find the distance between the first- and second-order dark fringes on the screen.

Homework Equations



y=(n*lamda*L)/d

The Attempt at a Solution



y1=580nm*2.1m/2.1um=0.468m
y2=580nm*2.1m*2/2.1um=0.937m

y2-y1=0.468m

I am not sure how to handle the dark fringes in single slit diffraction, so I am not sure where I am going wrong.

Thanks for people's help in advance.
 
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Do you know what the letters mean in the formula y=(n*lambda*L)/d and when is it a good approximation?

ehild
 
y is the distance between the fringes hitting the screen, n is the order, lambda is the wavelength, L is the distance to the screen, d is the width of the slit?

i don't know when it is a good approximation or what you mean by that.
 
Yo know a formula which contain the angle of deviation from the original direction of the light ray, hitting the slit. It is the exact one. ehild
 
n*lamda = d*sin(theta) ?

so would i use the angle found theta, and the known length L in a triangle to find the difference in y?
 
Yes. Calculate theta for n=1 and for n=2. The relation between L and y is y/L=tan(theta). If theta was very small, not more than a few degrees, you could approximate sin(theta) with tan(theta) and the equation in your first post would be valid. It is not the case now.

ehild
 
thanks ehild, helpful as always
 

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