Question

Light waves with two different wavelengths, 632 nm and 474 nm, pass simultaneously through a single slit whose width is 7.63 x 10-5 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?

Answer 1

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 = k1/k2, 632/474 = 4/3.
Therefore k1=4, k2=3.
Now
sinα = k1•λ1/b =4•632•10^-9/ 7.63•10^-5 =0.033.
As the angle is very small tanα = sinα = 0.033.
tan α= x/L,
x =L• tanα =1.3•0.033 = 0.043 m = 4/3 cm

Answer 2

My mistake.
It should be
k1•λ1= k2•λ2,
λ1/λ2 = k2/k1, 632/474 = 4/3.
k1 =3 k2 = 4
sinα = k1•λ1/b =3•632•10^-9/ 7.63•10^-5 =0.0248
x=L• tanα =L• sinα =1.3•0.0248 =0.0323 m = 3.23 cm