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Ipatiy [6.2K]
2 years ago
6

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)?

Physics
1 answer:
Dmitry_Shevchenko [17]2 years ago
5 0

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) × \lambda÷ 2

For third minima, n = 2

Now

xd ÷ D = (2 × 2 + 1) × \lambda÷ 2

d tan Q_3 = 5\lambda ÷ 2

tan Q_3 = 5\lambda ÷ 2d

Q_3 = tan^-1 × (5\lambda ÷2d)

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

= 0.66 degrees

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<h2>When two object P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of Q. The mass of P is half that of Q. The ratio of the specific heat capacity of P to Q​</h2>

Explanation:

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