Physics 214 Problem 3 Week 2
Two-slit Interference Experiment
In a two-slit interference demonstration done with a laser, the experimenter must choose the
separation of the slits to give a reasonable interference pattern. She wants the distance between
interference maxima to be 0.25 cm. The laser light from his HeNe laser has a wavelength of
633 nm. The viewing screen should be about 1 meter directly in front of the slits. What must be
the separation between the two slits? Is this slit separation practical? In Lab #1 you will do this
experiment!
(Obviously here we are talking about interference peaks which are separated by a small angle,
since 0.25 cm is small compared to 1 m. Therefore, here you can use the small angle
approximation sin
θ
≈
θ
, where
θ
is in radians.)
Note: We have not drawn a diagram. You should draw your own.
At each interference maximum on the screen the path length difference,
δ
, between the two
sources is an integral number of wavelengths (
δ
= m
λ
). This makes the phase difference
between the two sources an integral multiple of 2π. The position on the screen of the mth
maximum, ym = m
λ
L/d. Thus the separation of adjacent maxima
Δ
y =
λ
L/d. We want Δy =
0.25 cm.
d =
λ
L /
Δ
y = (632 × 10 −9m)(1m) / (0.0025 m) = 0.253 mm
Depending on the available tools, it is quite easy to make slits separated by 0.25 mm. In fact, in
the P214 laboratory we use slits separated by 0.125 - 0.25 mm.
This figure shows the desired interference pattern:
Solution