in a double slit experiment the distance between the slits is 5.0 mm and the sli
ID: 3897211 • Letter: I
Question
in a double slit experiment the distance between the slits is 5.0 mm and the slits are 1.0 m from the screen two inteference patterns are seen on the screen one due to light of wavelength 480 nm and the other due to the light iof wavelength 600 nm what is the seperation on the screen betwen the third order m = 3 bright fringes fo the two inteference patterns?
Please explain fully, I'm not able to follow other examples on here that stop at just an equation saying "now just solve" on this one for some reason. Thanks!
Explanation / Answer
he order of the fringe m = 3
the distance between slits and screen L= 1m
the distance between slits d = 5.0mm=5.0*10-3m
?1=480nm, ?2 = 600nm
constructive interference:
dsin? = m? .................(1)
Where sin? = y/L...........(2)
Substituting the equation(2) in equation(1) we get
d(y/L) = m?
solve for y
y = (m?L)/d
Seperation between the third order bright fringes of the two interference patterns is
y2-y1= (m?2L)/d - (m?1L)/d
= (mL/d) x (?2-?1)
= [(3 x 1m) / 5 x 10-3m] x (600 x 10-9 - 480 x 10-9)
= 7.2 x 10-5 m
or 72 ?m
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