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3241004551 [841]
3 years ago
6

A diffraction grating with 750 slits per mm is illuminated by light which gives a first-order diffraction angle of 34.0°. What i

s the wavelength of the light?
Physics
1 answer:
lesya [120]3 years ago
6 0
<h2>Answer: 745.59 nm</h2>

Explanation:

The diffraction angles \theta_{n} when we have a slit divided into n parts are obtained by the following equation:

dsin\theta_{n}=n\lambda (1)

Where:

d is the width of the slit

\lambda is the wavelength of the light  

n is an integer different from zero

Now, the first-order diffraction angle is given when n=1, hence equation (1) becomes:

dsin\theta_{1}=\lambda (2)

We know:

\theta_{1}=34\°

In addition we are told the diffraction grating has 750 slits per mm, this means:

d=\frac{1mm}{750}

Solving (2) with the known values we will find \lambda:

\lambda=(\frac{1mm}{750})sin(34\°) (3)

\lambda=0.00074559mm (4)

Knowing 1mm=10^{6}nm:

\lambda=745.59nm  >>>This is the wavelength of the light, wich corresponds to red.

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Explanation:

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IF YOU ANSWER THESE 3 QUESTIONS! I WILL GIVE YOU BRAINEST!!! 17 POINTS!!!
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Answer:

1. a

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Explanation:

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2 years ago
A jet is circling an airport control tower at a distance of 20.6 km. An observer in the tower watches the jet cross in front of
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Answer:

197.76 m

Explanation:

r = Radius of the path = 20.6 km = 20.6\times 10^3\ m

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3 years ago
A solid cylinder has a mass of 5 kg and radius of 2 m and is fixed so that it is able to rotate freely around its center without
kari74 [83]

Answer:

0.893 rad/s in the clockwise direction

Explanation:

From the law of conservation of angular momentum,

angular momentum before impact = angular momentum after impact

L₁ = L₂

L₁ = angular momentum of bullet = + 9 kgm²/s (it is positive since the bullet tends to rotate in a clockwise direction from left to right)

L₂ = angular momentum of cylinder and angular momentum of bullet after collision.

L₂ = (I₁ + I₂)ω where I₁ = rotational inertia of cylinder = 1/2MR² where M = mass of cylinder = 5 kg and R = radius of cylinder = 2 m, I₂ = rotational inertia of bullet about axis of cylinder after collision = mR² where m = mass of bullet = 0.02 kg and R = radius of cylinder = 2m and ω = angular velocity of system after collision

So,

L₁ = L₂

L₁ = (I₁ + I₂)ω

ω = L₁/(I₁ + I₂)

ω = L₁/(1/2MR² + mR²)

ω = L₁/(1/2M + m)R²

substituting the values of the variables into the equation, we have

ω = L₁/(1/2M + m)R²

ω = + 9 kgm²/s/(1/2 × 5 kg + 0.02 kg)(2 m)²

ω = + 9 kgm²/s/(2.5 kg + 0.02 kg)(4 m²)

ω = + 9 kgm²/s/(2.52 kg)(4 m²)

ω = +9 kgm²/s/10.08 kgm²

ω = + 0.893 rad/s

The angular velocity of the cylinder bullet system is 0.893 rad/s in the clockwise direction-since it is positive.

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your answer is phenomena which are too large


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