Physics, 05.08.2020 18:01 delphinelilly2846

Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visible light with no greater than 1.0 mW total power. They're relatively safe because the eye's blink reflex limits exposure time to 250 ms. Requried:
a. Find the intensity of a 1-mW class 2 laser with beam diameter 2.0 mm .
b. Find the total energy delivered before the blink reflex shuts the eye.
c. Find the peak electric field in the laser beam.

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Answer from: awgggg

(a) The intensity of laser beam is $318.2 \;\rm W/m^{2}$ .

(b) The total energy delivered before the blink reflex shuts the eye is $2.5 \times 10^{-4} \;\rm J$ .

Given data:

The power of laser is, $P=1 \;\rm mW = 1 \times 10^{-3} \;\rm W$ .

The exposure time is, $t = 250\;\rm ms = 250 \times 10^{-3} \;\rm s$ .

The beam diameter is, $d = 2 \;\rm mm = 2 \times 10^{-3} \;\rm m$ .

The standard expression for the intensity of beam is given as,

I = P/A

Here, P is the power of the laser and A is the cross-sectional area of the beam. And its value is,

$A =\pi /4 \times d^{2}\\\\A =\pi /4 \times (2 \times 10^{-3})^{2}\\\\A =3.142 \times 10^{-6} \;\rm m^{2}$

Then intensity is,

$I = (1 \times 10^{-3})/(3.142 \times 10^{-6})\\\\I =318.2 \;\rm W/m^{2}$

Thus, the intensity of laser beam is $318.2 \;\rm W/m^{2}$ .

(b)

The expression for the total energy delivered is given as,

E = Pt

Solving as,

$E = 1 \times 10^{-3} \times (250 \times 10^{-3})\\\\E = 2.5 \times 10^{-4} \;\rm J$

Thus, the total energy delivered before the blink reflex shuts the eye is $2.5 \times 10^{-4} \;\rm J$ .

(c)

The expression for the peak electric field is given as,

$E = \sqrt{\dfrac{2I}{c \times \epsilon_{0}}}$

Solving as,

$E = \sqrt{\dfrac{2 \times 318.2}{(3 \times 10^{8}) \times (8.85 \times 10^{9})}}\\\\E =1.55 \times 10^{-8} \;\rm V/m$

Thus, the required value of peak electric field in the laser beam is $1.55 \times 10^{-8} \;\rm V/m$ .

Learn more about the laser intensity here:

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Answer from: Benitez0212

a) 318.2 W/m^2

b) 2.5 x 10^-4 J

c) 1.55 x 10^-8 v/m

Explanation:

Power of laser P = 1 mW = 1 x 10^-3 W

exposure time t = 250 ms = 250 x 10^-3 s

If beam diameter = 2 mm = 2 x 10^-3 m

then

cross-sectional area of beam A = $\pi d^{2} /4$ = (3.142 x $(2*10^{-3} )^{2}$ )/4

A = 3.142 x 10^-6 m^2

a) Intensity I = P/A

where P is the power of the laser

A is the cros-sectional area of the beam

I = ( 1 x 10^-3)/(3.142 x 10^-6) = 318.2 W/m^2

b) Total energy delivered E = Pt

where P is the power of the beam

t is the exposure time

E = 1 x 10^-3 x 250 x 10^-3 = 2.5 x 10^-4 J

c) The peak electric field is given as

E = $\sqrt{2I/ce_{0} }$

where I is the intensity of the beam

E is the electric field

c is the speed of light = 3 x 10^8 m/s

$e_{0}$ = 8.85 x 10^9 m kg s^-2 A^-2

E = $\sqrt{2*318.2/3*10^8*8.85*10^9}$ = 1.55 x 10^-8 v/m

Answer from: Quest

Because the dept in the ocean

Answer from: Quest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

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Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many l...

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