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)?
1 answer:
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) × ÷ 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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