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krok68 [10]
3 years ago
14

A diffraction grating with 270 lines per mm is used in an experiment to study the visible spectrum of a gas discharge tube. 1. A

t what angle from the beam axis will the fourth order peak occur if the tube emits light with wavelength of 665.0 nm?2. At what angle will the second order peak occur?
Physics
1 answer:
VikaD [51]3 years ago
6 0

Answer:

1) θ = 45.91°

2) θ = 21.04°

Explanation:

We are given;

Wavelength; λ = 665 nm = 6.65 × 10^(-7) m.

Distance between slits; d = 1mm/270 = 1/270 mm = (1/270) × 10^(-3) m

1) To find the angle, we will use the formula;

d sin θ = mλ

Where m is the order of peak which in this question is 4.

Thus, we have;

sin θ = mλ/d

sin θ = (4 × 6.65 × 10^(-7))/((1/270) × 10^(-3))

sin θ = 0.7183

θ = sin^(-1) 0.7183

θ = 45.91°

2) Similarly, d sin θ = mλ

Where m is the order of peak which in this question is 2. Thus;

sin θ = (2 × 6.65 × 10^(-7))/((1/270) × 10^(-3))

sin θ = 0.3591

θ = sin^(-1) 0.3591

θ = 21.04°

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3 years ago
Consider a situation of simple harmonic motion in which the distance between the endpoints is 2.39 m and exactly 8 cycles are co
aivan3 [116]

Answer:

1.195 m

2.8375 s

2.21433 rad/s

Explanation:

d = Distance = 2.39 m

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t = Time to complete 8 cycles = 22.7 s

Radius would be equal to the distance divided by 2

r=\frac{d}{2}\\\Rightarrow r=\frac{2.39}{2}\\\Rightarrow r=1.195\ m

The radius is 1.195 m

Time period would be given by

T=\frac{t}{N}\\\Rightarrow T=\frac{22.7}{8}\\\Rightarrow T=2.8375\ s

Time period of the motion is 2.8375 s

Angular speed is given by

\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{2.8375}\\\Rightarrow \omega=2.21433\ rad/s

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3 years ago
A particle executes simple harmonic motion with an amplitude of 1.69 cm. At what positive displacement from the midpoint of its
m_a_m_a [10]

Answer: 0.0146m

Explanation: The formula that defines the velocity of a simple harmonic motion is given as

v = ω√A² - x²

Where v = linear velocity, A = amplitude = 1.69cm = 0.0169m, x = displacement.

The maximum speed of a simple harmonic motion is derived when x = A, hence v = ωA

One half of maximum speed = speed of motion

3ωA/2 = ω√A² - x²

ω cancels out on both sides of the equation, hence we have that

A/2 = √A² - x²

(0.0169)/2 = √(0.0169² - x²)

0.00845 = √(0.0169² - x²)

By squaring both sides, we have that

0.00845² = 0.0169² - x²

x² = 0.0169² - 0.00845²

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x = √0.0002142

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