**Unformatted text preview: **Physics 7 C Quiz 2
First Name Last Name ID Section You are performing a double slit experiment very similar to the one from DL 4 by shining a laser on two narrow slits
spaced 7.5 X 10—5 meters apart. However, by placing a piece of crystal in one of the slits, you are able to make it so that
the rays of light that travel through the two slits are 71' out of phase with each other (that is to say, A¢0 = 7r). If you
observe that on a screen placed 4 meters from the two slits that the distance between the bright spot closest to the center
of the interference pattern and the center of the pattern is 1.5 cm, what is the wavelength of the laser? (You may use the
approximation that sin 0 m tan0 for small 0.) From DL 4, we recall that the path length difference for the double slit was given by A1: = dsin 0, where 0 is the angle made relative to the horizontal and d = 7.5 x 10—5 m is the slit separation distance. By trigonometry, we know tan0 = %, hence using our approximation sin0 m tan0 we have that sin0 z “ii. We now observe that for a point of constructive interference (like the bright spot on the wall we are considering)
A<I> = 2n7r (3) Here, since the waves through the two slits have the same frequency, we need only consider contributions to A<I> from the
path lengths difference and the ﬁxed phase constant difference A¢0, which is given as 71'. Hence Ax 2Am 2dsin0
Aq)=2n7T—27TT+7F=>)\—2n_1— 271—1 (4) If we now plug in our known values for d and sin0 and use n = 1 for the bright spot closest to the center of the pattern, we get . 1
,\ z 2(7.5 x 10-5 m) ($) e 5.63 x 10—7 m = 563 nm (5) Fun facts: 1) = Af, f = %, y(:1:,t) = Asin (27% :27r§ +¢0)+y0, D—Do = 10log10 (é), Ioc A2, A2: = dsin0 = nA ...

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- Fall '08
- MAHMUD
- Physics