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finlep [7]
3 years ago
8

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

Part B: What is the amplitude of the electric field?
Physics
1 answer:
vekshin13 years ago
8 0

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= \sqrt{((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= \sqrt{(9*10^10 W/m^2)*(2) / (3*10^8 m/s^2)*(8.85* 10^-12 F⋅m−1)

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

 

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Answer:

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Explanation:

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According to the first law of thermodynamics

dQ = dU + dW

dQ = 500 - 1600

dQ = - 1100 J

As heat is negative so it flows out.

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suppose you want to determine the surface area of this sugar cube. it has edges that are each 2 cm long. if you cut the cube in
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Answer:

Half: 6 cm^2    Whole: 12 cm^2

Explanation:

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Five identical quintuplets leave earth when they reach the age of 21, in the year 2121. Each quintuplet goes on a spaceship jour
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Answer:

Explanation:

This is a problem based on time dilation , a theory given by Albert Einstein .

The formula of time dilation is as follows .

t₁ = \frac{t}{\sqrt{1-\frac{v^2}{c^2} } }

t is time measured on the earth and t₁ is time measured by man on ship .

A ) Given t = 20 years , t₁ = ? v = .4c

\frac{20}{\sqrt{1-\frac{.16c^2}{c^2} } }

=1.09 x 20

t₁= 21.82 years

B ) Given t = 5 years , t₁ = ? v = .2c

\frac{5}{\sqrt{1-\frac{.04c^2}{c^2} } }

=1.02 x 5

t₁= 5.1 years

C ) Given t = 10 years , t₁ = ? v = .8c

\frac{10}{\sqrt{1-\frac{.64c^2}{c^2} } }

=1.67 x 10

t₁= 16.7  years

D ) Given t = 10 years , t₁ = ? v = .4c

\frac{10}{\sqrt{1-\frac{.16c^2}{c^2} } }

=1.09 x 10

t₁= 10.9  years

E ) Given t = 20 years , t₁ = ? v = .8c

\frac{20}{\sqrt{1-\frac{.64c^2}{c^2} } }

=1.67 x 20

t₁= 33.4   years

7 0
2 years ago
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