1) A double-slit experiment is set up using red light ( = 734 nm). A first order
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Question
1)A double-slit experiment is set up using red light ( = 734 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?
=__________nm
2)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 = 682 nm and the third order bright fringe of light with = 403 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.)
|x| =___________m
Explanation / Answer
here,
A.
theta = arcsin(m*lambda1/d),
where m = 1
theta = arcsin(m2*lambda2/d)
Thus lambda2 = lambda1/m2 ,
where m2 = 0.5, 1.5, etc (until lambda is within the specified range)
m = 1.5 works;
lambda2 = lambda1/1.5 = 489.33 nm
the wavelength of the visible light is 489.33 nm
B.
screen distance , D = 1.8 m
d = 0.08 * 10^-3 m
theta is small enough that we can use the small-angle approximation for deflection y.
y = m * lambda * D/d
for lamda1 = 682 * 10^-9 and m = 2
y1 = 2 * 682 * 10^-9 * 1.8 /( 0.08 * 10^-3)
y1 = 0.030690 m
for lamda2 = 403 * 10^-9 and m = 2
y2 = 2 * 403 * 10^-9 * 1.8 /( 0.08 * 10^-3)
y2 = 0.018135 m
the distance between them = y1 - y2 = 0.012555 m
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