Ok, the following problem I've tried several times to solve, although I suspect that somewhere along the line I'm just approaching what appears to be a simple problem in the wrong way: ------------------------------- A two-point source operates at a frequency of 1.0 Hz to produce an interference patter in a ripple tank. The sources are 2.5 cm apart and the wavelength of the waves is 1.2 cm. Calculate the angles at which the nodal lines in the pattern are located far from the sources. (Assume the angles are measured from the central line of the pattern.) -------------------------------- So here's what is given: d = 2.5 cm f = 1.0 Hz lambda = 1.2 cm Some formulas/expressions to work with: 1. v = f(lambda) 'v' is the speed of the wavelength 2. (d)sin(theta) = (n-1/2)lambda Where 'theta' is the angle of the nth nodal line, 'lambda' is the wavelength, and 'd' is the distance from the sources. 3. lambda = (d)sin(theta)/(n-1/2) This is just a variation of (1). 4. theta + alpha = 90 degrees *EDIT* 5. theta prime + alpha = 90 degrees theta prime = theta sin(theta prime) = x/L, sin(theta) = (n-1/2)(lambda/d) 6. Since sin(theta prime) = sin(theta) x/L = (n-1/2)(lambda/d), so therefore lambda = (x/L)[d/(n-1/2)] 'x' is the perpendicular distance from the right bisector to the point P on the nodal line 'L' is the distance from the midpoint between the two sources to the point P Although I think it's safe to say that the 'x' and 'L' variables are irrelevant in this case because they aren't given, and are not in the formula used to solve for angles. I suppose that the same goes for the 'n' nodal number, since it's not given, I need to derive an equation which only involves the variables given, as well as sin(theta), so that I can solve for theta, and, in addition, alpha. If there any helpful suggestions on how to approach this problem, it would be much appreciated.