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Biologists use optical tweezers to manipulate micron-sized objects using a beam of light. In this technique, a laser beam is focused to a very small-diameter spot. Because small particles are attracted to regions of high light intensity, the focused beam can be used to "grab" onto particles and manipulate them for various experiments. In one experiment, a 10 mW laser beam is focused to a spot that has a diameter of 0.67 μm

Part A: What is the intensity of the light in this spot?

Part B: What is the amplitude of the electric field?

Respuesta :

Answer:

Explanation:

Part A) Using

light intensity I= P/A

A= Area= π (Radius)^2= π((0.67*10^-6m)/(2))^2= 1.12*10^-13 m^2

Radius= Diameter/2

P= power= 10*10^-3=0.01 W

light intensity I= 0.01/(1.12*10^-13)= 9*10^10 W/m^2

Part B)  Using

I=c*ε*E^2/2

rearrange to solve for E= [tex]\sqrt{[/tex]((I*2)/(c*ε))

c is the speed of light which is 3*10^8 m/s^2

ε=permittivity of free space or dielectric constant= 8.85* 10^-12 F⋅m−1

I= the already solved light intensity= 8.85*10^10 W/m^2

amplitude of the electric field E= [tex]\sqrt{[/tex](9*10^10 W/m^2)*(2) / (3*10^8 m/s^2)*(8.85* 10^-12 F⋅m−1)

---> E= [tex]\sqrt{[/tex](1.8*10^11) / (2.66*10^-3) = [tex]\sqrt{[/tex](6.8*10^13) = 8.25*10^6 V/m    

 

The definition of intensity and its relationship with the electromagnetic fields allows to find the results for the questions are:

      A) The intensity at the focus point is: I = 2.84 10¹⁰ W / m²

      B) The electric field at this point is: E = 1.35 10⁵ V / m

 

Given parameters

  • Laser power P = 10 mW = 10 10-3 W
  • Focusing diameter d = 0.67 um = 0.67 10-6 m

To find

    A. Intensity.

    B. Amplitude of the electric field.

Part A.

The intensity of a wave is defined by the relationship between the power and the area of ​​the ray.

         I = P / A

Where I is the intensity, P is the power and A is the area.

Let's find the area of ​​a focus circle,

       A = π r² =  [tex]\pi \frac{d^2}{4}[/tex]  

       A = π (0.67 10⁻⁶)² / 4

       A = 3,526 10⁻¹³ m²

We calculate the intensity.

        I = [tex]\frac{10 \ 10^{-3}}{3.526 \ 10^{-13}}[/tex]

        I = 2.84 10¹⁰ W / m²

Part B

Intensity is the vector product of the electric and magnetic field.

        I = [tex]\frac{1}{\mu_o }[/tex]  E x B

The two fields are related.

        E =c  B

if we want the average field.

       I = [tex]\frac{1}{2 \mu_o \ c} \ E^2[/tex]  

       

       E = [tex]\sqrt{2 \mu_o \ c} \ \sqrt{I }[/tex]  

     

let's calculate.

       E = [tex]\sqrt{2 \ 2\pi \ 10^{-7} \ 3 \ 10^8 \ 2.84 \ 10^{10}}[/tex]  

       E = [tex]\sqrt{1.84357 \ 10^{10}}[/tex]  

       E = 1.35 10⁵ V / m

In conclusion using the definition of intensity and its relationship with electromagnetic fields we can find the results for the questions are:

      A) The intensity at the focus point is: I = 2.84 1010 W / m²

      B) The electric field at this point is: E = 1.35 10⁵ V / m

Learn more here:  brainly.com/question/9195922

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