Phy 122 - Assignment 12:
A. 1.
2.
B. (a) Grating equation: m
= d sin
m = 3 (3rd order)
(3) (5.00 x 10-7 m) = d sin 32°
d = 1.5 x 10-6 = 2.83 x 10-6 m = 2.83 x 10-4 cm
sin 32°
Slits per cm = 1/(cm per slit) = 1/(2.83 x 10-4) = 3.53 x 103 ans.
(b) From the grating equation: sin
m d
Try larger and larger values of m until sin passes 1. The sine function can’t
ever be more than 1, so that shows you the largest possible m.
sin
sin
(5)( 5.00 x 10-7 m)/(2.83 x 10-6 m) = .883 ok
(6)( 5.00 x 10-7 m)/(2.83 x 10-6 m) = 1.06 nope.
ans: 5th order
C. 1. Oil’s n must be less than water’s. (and more than
air’s.) If both rays reflect off a higher n, both undergo the
same 180° phase shift. That way, interference is
constructive when the thickness is nearly zero.
2.
D. 1. One ray reflects off a higher n (the soap film), the other off a lower n (the
air under the film). This gives one ray a 180° phase shift and the other ray
none. So, with a path difference of about zero, they are 180° out of phase.
2.
E. As mentioned in the notes, (m + ½) = d sin θ is derived from the
assumption that the screen or observer is “far from the slits.” Since this
observer’s distance is similar to the distance between the sources, that approach
won’t work.
A third minimum would be where the path difference is 2.5λ = (2.5)(1.715 m)
= 4.29 m. But the closest he can get is to lean his head against the bottom
speaker, where he is 0 m form one and 4.00 m from the other. So, the largest
possible path difference is 4.00 m. There is no third minimum. That makes the
answers:
a) Two.
b) 8.90 m & 1.82 m
F.