Physics
posted by Jim .
Light waves with two different wavelengths, 632 nm and 474 nm, pass simultaneously through a single slit whose width is 7.63 x 105 m and strike a screen 1.30 m from the slit. Two diffraction patterns are formed on the screen. What is the distance (in cm) between the common center of the diffraction patterns and the first occurrence of the spot where a dark fringe from one pattern falls on top of a dark fringe from the other pattern?

The condition for an intensity minimum is b sin á = k ë
where b is the slit width.
For 474 nm, there are minima at
sin á = 6.21*10^3 (k=1)
1.242*10^3 (k=2)
1.864*10^2 (k=3)
2.484*10^2 (k=4)
For 632 nm, there are minima at:
sin á = 8.283*10^3 (k=1)
1.657*10^3 (k=2)
2.485*10^2 (k=3)
Clearly, the k=4 minimum of ë = 474 nm radiation coincides with the k = 3 minimum of 632 nm radiation. The distance from the central maximum is
d = 1.30 m*sin á = 3.23*10^2 m
= 3.23 cm 
For further reading, try
http://www.walterfendt.de/ph14e/singleslit.htm
The a' symbol in my answer above was meant to be the diffraction angle alpha, and the e with two dots was supposed to be the wavelength, lambda 
Let’s use the condition of diffraction minimum for one split of the width b
b•sinα =k1•λ1
b•sinα =k2•λ2,
Since we have superposition of two maxima b•sinα is the same for two wavelengths,
k1•λ1= k2•λ2,
λ1/λ2 = k2/k1,
632/474 = 4/3.
Therefore, k1=3, k2=4.
Now sinα = k1•λ1/b =3•632•10^9/ 7.63•10^5 =0.0248.
As the angle is very small tanα = sinα,
x=L• tanα =L• sinα =1.3•0.0248 =0.0323 m = 3.23 cm