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valentina_108 [34]
3 years ago
12

Biologists use optical tweezers to manipulate micron-sized objects using a beam of light. In this technique, a laser beam is foc

used 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?
Physics
1 answer:
LiRa [457]3 years ago
7 0

Answer:

The correct answer is:

(a) 3.31\times 10^{10} \ \omega/m^2

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

Explanation:

The given values are:

Power of Laser beam,

= 10 mW

on converting it, we get

= 10\times 10^{-3} \ \omega

= 10^{-2} \ \omega

Spot's diameter,

= 0.62 μm

= 0.62\times 10^{-6} \ m

Now,

(a)

The intensity of light will be:

⇒ I=\frac{P}{\pi  r^2}

or,

⇒    =\frac{4P}{\pi d^2}

On substituting the values, we get

⇒    =\frac{4\times 10^{-2}}{\pi(0.62)^2\times 10^{-12}}

⇒    =3.31\times 10^{10} \ \omega/m^2

(b)

The amplitude of electric field will be:

⇒ I=\frac{1}{2}\epsilon_0E_0.C

or,

⇒ E_0=\sqrt{\frac{2I}{CE_0} }

On substituting the values, we get

⇒       =\sqrt{\frac{3.31\times 10^{10}\times 2}{3\times 10^8\times 8.85\times 10^{-12}} }

⇒       =5\times 10^6 \ V/m

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