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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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Sound intensity :

The power carried by sound waves per unit area in the direction  perpendicular to that region is known as sound intensity or acoustic intensity. The watt per square meter (W/m2) is the SI unit of intensity, which also covers sound intensity. Sound intensity is a measure of how quickly energy moves across a given space. The unit area in the SI measurement system is 1 m2. So Watts per square meter are used to measure sound intensity. As there will be energy flow in certain directions but not in others, sound intensity also provides a measure of direction.

To learn more about Sound intensity visit: brainly.com/question/12899113

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A person wants to fire a water balloon cannon such that it hits a target 100m100m away. if the cannon can only be launched at 45
vladimir2022 [97]
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3 years ago
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Light rays in a material with index of refrection 1.35 1.35 can undergo total internal reflection when they strike the interface
nekit [7.7K]

Answer:

1.30

Explanation:

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where theta_c is the critical angle, n1 is the index of refraction of the material where the light is totally reflected, and n2 is the refractive index of the other material.

By taking n_2 and replacing we obtain:

n_2=n_1sin\theta_c=(1.35)sin75.1\°=1.30

hope this helps!!

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