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3 years ago

A laser with a power of 1.0 mW has a beam radius of 1.0 mm. What is the peak value of the electric field in that beam

Physics

1 answer:

Fynjy0 [20]3 years ago

4 0

Answer:

The peak value of the electric field is 489.64 V/m

Explanation:

Given;

power of the laser, P = 1.0 mW = 1 x 10⁻³ W

Radius of the beam, R = 1.0 mm = 1 x 10⁻³ m

Area of the beam = πr² = π(1 x 10⁻³ )² = 3.142 x 10⁻⁶ m²

The average intensity of the light = P / A

The average intensity of the light = ( 1 x 10⁻³) / (3.142 x 10⁻⁶)

The average intensity of the light = 318.27 W/m²

The peak value of the electric field is given by;

$E_o = \sqrt{\frac{2I_{avg}}{c\epsilon_o}}\\\\E_o = \sqrt{\frac{2(318.27)}{(3*10^8)(8.85*10^{-12})}}\\\\E_o = 489.64 \ V/m$

Therefore, the peak value of the electric field is 489.64 V/m.

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2 years ago

Read 2 more answers

You find an unmarked blue laser on your way to physics class. When you get to class you realize you can determine the wavelength

Ray Of Light [21]

Answer:

wavelength $\lambda$ = 437.27 nm

Explanation:

given data

first bright fringe = 2.96 mm

slit separation = 0.325 mm

distance D = 2.20 m

solution

we know that this is double slit experiment

so we apply here Fringe width formula that is

β = $\frac{D\lambda}{d}$ ....................1

$\lambda$ is Wavelength of light and D is Distance between screen and slit and d is slit width

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3 years ago

At t=0, a train approaching a station begins decelerating from a speed of 80 mi/hr according to the acceleration function a(t)=−

vredina [299]

Answer:

a) Δx = 49.23 mi , b) Δx = 5.77 mi

Explanation:

As we have an acceleration function we must use the definition of kinematics

a = dv / dt

∫dv = ∫ a dt

we integrate and evaluators

v - vo = ∫ (-1280 (1 + 8t)⁻³ dt = -1280 ∫ (1+ 8t)⁻³ dt

We change variables

1+ 8t = u

8 dt = du

v - v₀ = -1280 ∫ u⁻³ du / 8

v -v₀ = -1280 / 8 (-u⁻²/2)

v - v₀ = 80 (1+ 8t)⁻²

We evaluate between the initial t = 0 v₀ = 80 and the final instant t and v

v- 80 = 80 [(1 + 8t)⁻² - 1]

v = 80 (1+ 8t)⁻²

We repeat the process for defining speed is

v = dx / dt

dx = vdt

x-x₀ = 80 ∫ (1-8t)⁻² dt

x-x₀ = 80 ∫ u⁻² dt / 8

x-x₀ = 80 (-1 / u)

x-x₀ = -80 (1 / (1 + 8t))

We evaluate for t = 0 and x₀ and the upper point t and x

x -x₀ = -80 [1 / (1 + 8t) - 1]

We already have the function of time displacement

a) let's calculate the position at the two points and be

t = 0 h

x = x₀

t = 0.2 h

x-x₀ = -80 [1 / (1 +8 02) -1]

x-x₀ = 49.23

displacement is

Δx = x (0.2) - x (0)

Δx = 49.23 mi

b) in the interval t = 0.2 h at t = 0.4 h

t = 0.4h

x- x₀ = -80 [1 / (1+ 8 0.4) -1]

x-x₀ = 55 mi

Δx = x (0.4) - x (0.2)

Δx = 55 - 49.23

Δx = 5.77 mi

3 0

3 years ago