Posted by Jim on Friday, March 30, 2012 at 1:23pm.
Lets use the condition of diffraction minimum for one split of the width b
bsinα =k1λ1
bsinα =k2λ2,
Since we have superposition of two maxima
bsinα is the same for two wavelengths,
k1λ1= k2λ2,
λ1/λ2 = k1/k2, 632/474 = 4/3.
Therefore k1=4, k2=3.
Now
sinα = k1λ1/b =463210^-9/ 7.6310^-5 =0.033.
As the angle is very small tanα = sinα = 0.033.
tan α= x/L,
x =L tanα =1.30.033 = 0.043 m = 4/3 cm
My mistake.
It should be
k1λ1= k2λ2,
λ1/λ2 = k2/k1, 632/474 = 4/3.
k1 =3 k2 = 4
sinα = k1λ1/b =363210^-9/ 7.6310^-5 =0.0248
x=L tanα =L sinα =1.30.0248 =0.0323 m = 3.23 cm
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