Double Slit Wave Interference Question

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SUMMARY

The discussion centers on calculating the position of the second order dark bands in a double slit interference experiment using blue light with a wavelength of 482 nm. The slits are separated by 0.15 mm, and the screen is positioned 2.00 m away. The calculations performed using the equations sin θ = (m + 0.5) λ/d and y = L tan θ yield a distance of 16 mm from the central axis for the second order dark band, which contradicts the textbook's stated value of 38 mm. The confusion arises from the interpretation of the order of the dark fringes.

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Homework Statement



Blue light (λ = 482 nm) is directed through parallel slits separated by 0.15 mm. A fringe pattern appears on a ascreen 2.00 m away. How far from the central axis on either side are the second order dark bands?

λ = 4.82 x 10-7 m
d = 1.5 x10-4 m
L = 2.00 m
m = 2
y= ?

Homework Equations



sin θ = (m + 0.5) λ/d
y = L tan θ

The Attempt at a Solution



I used m = 2 (second order dark bands) and used the first equation which gave me the angle formed between the central axis and the second order dark band, which came out to be 0.46 degrees. I then used the second formula, which is the formula for the distance between the central bright fringe and any point on the screen, which gave me 0.016 m or 16 mm. The textbook says 38 mm, but I'm sure I've done it correctly
 
Last edited:
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Your calc checks out.
I took m = 0 as being the first order dark fringe and m = 1 as the second order, which works out to 9.6 mm.
I don't see any way to get 38!
 

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