Chapter 27. Interference

Problem 5.

lambda = [d sin(theta) ]/ m = 660 nm

Problems 11 and 12: Not covered in class.

Problem 19.

a. m=1, first dark fringe

sin(theta) = m lambda / W ==> theta = 0.21

b. if W = 1.8 microns, lambda = 675 nm ==> theta = 22

Problem 27. Solution done in class

lambda / W = y/L = 225 W / (18000 W) = 0.013

Problem 44. Not covered in class.

Problem 50. Done in class.

third dark fringe (m=2) ==> sin(theta)= (2+.5) lambda/d

fourth dark fringe (m=4) ==> sin(theta)= 4 lambda'/d

then

lambda' = 2.5 lambda / 4 = 403 nm