I=0.5E0EO²C where E0 is the permitivity of the vacuum and C is the velocity of light. a)Show that the units of the equetion are consistant b)A certain helium-neon laser emits 0.20mW in a beam of cross sectional area 3.0mm.Determine the peak values of the electric field associated with this beam.

Question

I=0.5E0EO²C where E0 is the permitivity of the vacuum and C is the velocity of light. a)Show that the units of the equetion are consistant b)A certain helium-neon laser emits 0.20mW in a beam of cross sectional area 3.0mm.Determine the peak values of the electric field associated with this beam.

Answer 1

a) To show that the units of the equation are consistent, we need to check that the units on both sides of the equation match.

Starting with the given equation:
I = 0.5 * E₀ * E₀ * C

Let's break down the units of each term:
I: Intensity = watts (W)
E₀: Permittivity of vacuum = coulombs per volt per meter (C/V/m)
C: Velocity of light = meters per second (m/s)

Substituting the units into the equation:
W = 0.5 * (C/V/m) * (C/V/m) * (m/s)

Simplifying:
W = 0.5 * (C/V/m)² * (m/s)

Now let's examine the units:
W = 0.5 * (m/s)² * (m/s) => W = 0.5 * m²/s²

The unit of power (watt) is consistent with the right-hand side of the equation (0.5 * m²/s²). Therefore, the units of the equation are consistent.

b) In this part, we need to determine the peak value of the electric field associated with the given beam of a helium-neon laser.

Given information:
Power (P) of the laser beam = 0.20 mW
Beam cross-sectional area (A) = 3.0 mm²

We will use the equation for intensity (I) of the laser beam:
I = P / A

First, we need to convert the area from mm² to m²:
1 mm² = (1 mm)² = (0.001 m)² = 0.000001 m²

Substituting the values into the equation:
I = (0.20 mW) / (0.000001 m²)

Next, we need to convert milliwatts to watts:
1 mW = 0.001 W

Substituting the new value:
I = (0.001 W) / (0.000001 m²)

Simplifying:
I = 1000 W/m²

The intensity (I) is equal to the peak value of the electric field (E₀) squared times the velocity of light (C):
I = 0.5 * E₀ * E₀ * C

Rearranging the equation:
E₀ = √(2I / C)

Substituting the calculated intensity and the velocity of light:
E₀ = √(2 * 1000 W/m² / (3 × 10⁸ m/s))

Calculating the square root and simplifying:
E₀ ≈ √(0.00667) ≈ 0.082 V/m

Therefore, the peak value of the electric field associated with this beam is approximately 0.082 V/m.