Question

A laser with wavelength 480 nm passes through 2 slits of width 6μm. If the 10th dark interference fringe aligns with the first dark fringe from diffraction, what is the slit spacing?

Answer 1

Given data wavelength of the laser lambda $\lambda$ = 480 nm

slit width d = 6 *10^-6 m, for 10 th dark fringe

we know that the condition for dark fringe is

      

$d sin \theta =(m+1/2)\lambda$

$sin \theta =(m+1/2)\lambda/d$   

       sin theta = (10 +0.5)480*10^-9 /(6*10^-6)
       = 0.840
now $\theta$ = arc sin (0.840)
       $\theta$ ​ = 57.140 degrees

now the tenth dark fringe align with first darl fringe then the slitseparation should be

       $d =(m+1/2)\lambda/ sin \theta$
       d = (1+0.5)480*10^-9 /(sin57.140)
       d = 0.85714*10^-6 m

the slit spacing is 0.86 *10^-6 m