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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:

(a) dime

Explanation:

Convert all to metric unit:

0.5 cm = 0.005 m

1.8 cm = 0.018 m

71 cm = 0.71 m

In order to find out we would need to calculate the ratio R between the object diameter d and their distance s to our eyes:

R_m = \frac{d_m}{s_m} = \frac{3.48*10^6}{3.85*10^8} \approx 0.009

R_p = \frac{d_p}{s_p} = \frac{0.005}{0.71} \approx 0.007

R_d = \frac{d_d}{s_d} = \frac{0.018}{0.71} \approx 0.0253

Since the ratio of the dime is larger than the ratio of the moon, and the ratio of the pea is smaller than the ratio of the moon, only the (a) dime can cover your view of the moon.

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c) cubic centimetre is it's answer..

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The per-unit impedance of a single-phase electric load is 0.3. The base power is 500 kVA, and the base voltage is 13.8 kV. a. Fi
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Answer:

114.26

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Zpu2= Zpu1×(kV1/kV2)²×(kVA2/kVA1)

Zpu2: New per unit impedance

Zpu1: given per unit impedance

kV1: give base voltage

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Zpu2=??

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kV2=24kV

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b) to find ohmic impedance we will first calculate base value of impedance(Zbase). So,

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