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)
For 632 nm, there are minima at:
sin á = 8.283*10^-3 (k=1)
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
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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
Since we have superposition of two maxima b•sinα is the same for two wavelengths,
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
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