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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 0.62 μm.

a. What is the intensity of the light in this spot?
b. What is the amplitude of the electric field?

User Iamkrillin
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1 Answer

4 votes

Answer:

The correct answer is:

(a)
3.31* 10^(10) \ \omega/m^2

(b)
5* 10^6 \ N/C

Step-by-step explanation:

The given values are:

Power of Laser beam,

= 10 mW

on converting it, we get

=
10* 10^(-3) \ \omega

=
10^(-2) \ \omega

Spot's diameter,

= 0.62 μm

=
0.62* 10^(-6) \ m

Now,

(a)

The intensity of light will be:


I=(P)/(\pi r^2)

or,


=(4P)/(\pi d^2)

On substituting the values, we get


=(4* 10^(-2))/(\pi(0.62)^2* 10^(-12))


=3.31* 10^(10) \ \omega/m^2

(b)

The amplitude of electric field will be:


I=(1)/(2)\epsilon_0E_0.C

or,


E_0=\sqrt{(2I)/(CE_0) }

On substituting the values, we get


=\sqrt{(3.31* 10^(10)* 2)/(3* 10^8* 8.85* 10^(-12)) }


=5* 10^6 \ V/m

User Jenay
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4.3k points