HW help (Double Slit, Resolving power etc)

AI Thread Summary
In the discussion, participants analyze a double-slit experiment using red light and explore the conditions for both constructive and destructive interference. They derive the equations for determining the wavelength that would create a dark fringe at the same location as a bright fringe produced by red light. Additionally, they calculate the distance between specific bright fringes for different wavelengths, emphasizing the importance of identifying the correct order of the fringes. The final answers for the calculations are confirmed, with one participant noting the distance as 0.00315 m for part (b). The conversation highlights the application of interference principles in wave optics.
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(a) A double-slit experiment is set up using red light (l = 708 nm). A first order bright fringe is seen at a given location on a screen. What wavelength of visible light (between 380 nm and 750 nm) would produce a dark fringe at the identical location on the screen?
l = nm

All right so i know
d * sin theta = m * Wavelenght

How do i find the Distance and the angle?
 
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The condition for constructive interference is what you have written down (i.e. peak on peak), but destructive interference happens when you have a trough meet a peak, i.e. there is a half wavelength difference so for destructive interference (dark fringe) dsin (\theta) = (m + 0.5)\lambda. You should be able to go from there.
 
(b) A new experiment is created with the screen at a distance of 1.8 m from the slits (with spacing 0.08 mm). What is the distance between the second order bright fringe of light with l = 691 nm and the third order bright fringe of light with l = 414 nm? (Give the absolute value of the smallest possible distance between these two fringes: the distance between bright fringes on the same side of the central bright fringe.)







Ok so
D sin theta = m * Wavelenght

Distance is 2.2?
I am going to find theta
m = 1
Wavelenght is 698 in one part, and then i do the quation again for 413 right?
 
Alt+F4 said:
(b) A new experiment is created with the screen at a distance of 1.8 m from the slits (with spacing 0.08 mm). What is the distance between the second order bright fringe of light with l = 691 nm and the third order bright fringe of light with l = 414 nm? (Give the absolute value of the smallest possible distance between these two fringes: the distance between bright fringes on the same side of the central bright fringe.)







Ok so
D sin theta = m * Wavelenght

Distance is 2.2?
I am going to find theta
m = 1
Wavelenght is 698 in one part, and then i do the quation again for 413 right?
Don't lose track of the statement about the different orders of the finges. What is m for second order, and for third order?
 
OlderDan said:
Don't lose track of the statement about the different orders of the finges. What is m for second order, and for third order?
2 for second order, and 3 for third order
 
got the answer, thanks
 
Last edited:
You don't actually need to find the distance at all. You will have two equations equal to dsin (\theta), and you just have to solve for the 1 unknown (wavelength).
 
Alt+F4 said:
got the answer, thanks

and that would be?
 
.00315 m
...
 
  • #10
Yes that's correct for (b). What answer did you get for (a).
 

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