Calculating Wave Length in Monochromatic Light

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Homework Help Overview

The problem involves calculating the wavelength of monochromatic light based on measurements related to a diffraction pattern observed on a screen. The setup includes a distance from the slits to the screen and the position of the dark fringe.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the angle and the position of the dark fringe, with some questioning the application of the m-factor in the wavelength calculation. There is also mention of simplifying assumptions regarding small angles.

Discussion Status

Some participants have provided guidance on the correct interpretation of the m-factor in relation to the dark fringe, suggesting that the original poster's approach may need adjustment. Multiple interpretations of the problem are being explored, particularly regarding the definition of the dark fringe in the context of the equation used.

Contextual Notes

There is a focus on the precision of the answer, with discussions about significant figures based on the given measurements. The original poster's calculations are based on specific assumptions about the geometry of the setup.

Kris1120
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Homework Statement


A screen is illuminated by monochromatic
light.
The distance from the slits to the screen is
7.4 m.
What is the wave length if the distance from the central bright region to the fourth
dark fringe is 2.6 cm.
Answer in units of nm.


Homework Equations


x=L*tan(theta)

sin(theta)=(m+.5) lambda/d


The Attempt at a Solution



theta= inverse tan( 2.6e-2 / 7.4) = .201309

lambda = (.55e-3) sin(.201309) / 4.5 = 429.427 nm
 
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correct, but way too many digits. 430 nm is the proper way to give the answer, since only two digits are given for L and x.

By the way, when the angle is very small (such as is the case here) sin theta = tan theta.

That way, you can save some time and see that x/L=(m lambda)/d

It'll be like that whenever L is much greater than x
 
Kris1120 said:

The Attempt at a Solution



theta= inverse tan( 2.6e-2 / 7.4) = .201309

lambda = (.55e-3) sin(.201309) / 4.5 = 429.427 nm


I don't believe this is correct. The fourth dark fringe does not correspond to m=4 in that equation.
 
Alphysicist is right. Fourth dark fringe is halfway between the 3rd & 4th bright fringe, so the "m-factor" should be 3.5.
 

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