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2 years ago

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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alex41 [277]

Answer:

h=12.41m

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$N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}$

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$M_{I}=0.8*40.0kg*(0.6m} )^{2}\\M_{I}=11.5 kg*m^{2}$

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$K=3317.79J+4147.2J\\K=7464.99J$

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