Answer:
0.152 m
Explanation:
The condition for constructive interference is
d sinθ= mλ  (m= 0,1,2,3...)
the slit width
d= 1/N
d= 10^(-3)/520
= 1.92×10^(-6)
The angular spread for the red light is
[tex]\theta_r= sin^{-1}(\frac{m\lambda}{d})[/tex]
[tex]\theta_r= sin^{-1}(\frac{656\times10^{-9}}{1.92\times10^{-6}})[/tex]
= 19.97°
The angular spread of blue light is
[tex]\theta_b= sin^{-1}(\frac{m\lambda}{d})[/tex]
[tex]\theta_b= sin^{-1}(\frac{456\times10^{-9}}{1.92\times10^{-6}})[/tex]
=14.66°
The distance between the first order red fringe and blue fringe is,
y= y_r- y_b = Ltanθ_r -Ltanθ_b
=1.50( tan19.97°-tan14.66°) = 0.152 m
