Answer:
The smallest angle is [tex]\theta = 0.00061^o[/tex]
Explanation:
From the question we are told that
The distance of separation is d = 0.9 mm
The wavelength is [tex]\lambda = 550nm[/tex]
General the location of the interference fringes is mathematically represented by
[tex]dsin \theta = m \lambda[/tex]
m is the order of the fringe
Now the smallest angle at which both ray create a bright spot would be when
m= 1
[tex]dsin \theta = \lambda[/tex]
[tex]sin \theta =\frac{\lambda}{d}[/tex]
[tex]\theta = sin ^{-1} (\frac{\lambda}{d} )[/tex]
Substituting values
[tex]\theta = sin ^{-1}(\frac{550*10^{-9}}{0.9*10^{-3}} )[/tex]
[tex]= sin^{-1}(0.0006111)[/tex]
[tex]\theta = 0.00061^o[/tex]
