Answer:
0.66 degrees
Explanation:
The computation of the angle of the third dark interference is shown below:
The condition of the minima is
Path difference = (2n +1) × [tex]\lambda[/tex]÷ 2
For third minima, n = 2
Now
xd ÷ D = (2 × 2 + 1) × [tex]\lambda[/tex]÷ 2
d tan Q_3 = 5[tex]\lambda[/tex] ÷ 2
tan Q_3 = 5[tex]\lambda[/tex] ÷ 2d
Q_3 = tan^-1 × (5[tex]\lambda[/tex] ÷2d)
= tan^-1 × (5 × 580 × 10^-9) ÷ (2 × 0.000125)
= 0.66 degrees
