How Does Monochromatic Light Create a Bright Fringe in a Double Slit Experiment?

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Monochromatic light creates a bright fringe in a double slit experiment due to constructive interference, which occurs when the path difference between light waves from the two slits is an integer multiple of the wavelength. The relevant equation for maxima is λ/d = x/L, where λ represents the wavelength, d is the distance between the slits, x is the distance of the bright fringe, and L is the distance from the slits to the screen. The correct path difference for a bright fringe is 1 wavelength (E). The confusion arises from the assumption that specific values for x and L are necessary, but the general principles of interference apply regardless of these distances. Understanding these concepts clarifies the relationship between slit separation and fringe formation.
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Homework Statement


When monochromatic light passes through a double slit, a bright fringe is formed due to a path difference of
A) 1/4 "wavelength symbol"
B) 1/2 "wavelength symbol"
C) 3/4 "wavelength symbol"
D) 7/8 "wavelength symbol"
E) 1 "wavelength symbol"

please help by starting me off.
gracias :)
 
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You need to make at least some effort. What have you studied in class on this topic?
 
Start with the definitions(what do monochromatic and slit imply?) and equations to solve this problem. Also, the wavelength symbol is the greek symbol lambda.
 
for maxima (bright fringe)
lambda / d = x/L
where lambda is wavelength, d is distance between slits, x is distance of bright fringe, L is distance of screen from slits.

lamda = xd/L
answer is E) 1 lambda
 
for bright fringe, path diff is n*lambda
for dark band, it is (2n - 1)*lambda/2
 
cristo said:
You need to make at least some effort. What have you studied in class on this topic?

this is a question that was in my REA SAT II physics book. we never studied this in my physics class a year ago. and there's no mention of it in my REA book. or other study books i have.
 
spideyunlimit said:
for maxima (bright fringe)
lambda / d = x/L
where lambda is wavelength, d is distance between slits, x is distance of bright fringe, L is distance of screen from slits.

lamda = xd/L
answer is E) 1 lambda

that's what the book says the answer is.
but, I'm still really confused.
so the fact it's a double slit doesn't matter??
and that we don't know x and L?
 
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