580 nm light shines on a double slit with d=0.000125 m. What is the angle of the third dark interference minimum (m=3)? (Remember, nano means 10^-9.)
(Unit=deg)

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) × ÷ 2

For third minima, n = 2

Now

xd ÷ D = (2 × 2 + 1) × ÷ 2

d tan Q_3 = 5 ÷ 2

tan Q_3 = 5 ÷ 2d

Q_3 = tan^-1 × (5 ÷2d)

= tan^-1 × (5 × 580 × 10^-9) ÷ (2 × 0.000125)

= 0.66 degrees

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