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love history [14]

2 years ago

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Very short pulses of high-intensity laser beams are used to repair detached portions of the retina of the eye. The brief pulses

of energy absorbed by the retina welds the detached portion back into place. In one such procedure, a laser beam has a wavelength of 810 nm and delivers 250 mW of power spread over a circular spot 510 ?m in diameter. The vitreous humor (the transparent fluid that fills most of the eye) has an index of refraction of 1.34.

1 answer:

Tom [10]2 years ago

5 0

Answer:

a) U = 0.375 mJ

b) p_rad = 4.08 mPa

c) λ_med = 604 nm ; f_med = 3.7 * 10^14 Hz

d) E_o = 3.04 * 10^4 V / m , B _o = 1.013 *10^-4 Nm/Amp

Explanation:

Given:

λ_air : wavelength = 810 * 10^(-9) m

P: Power delivered = 0.25 W

d : Diameter of circular spot = 0.00051 m

c : speed of light vacuum = 3 * 10^8 m/s

n_air : refraction Index of light in air = 1

n_med : refraction Index of light in medium = 1.34

ε_o : permittivity of free space = 8.85 * 10^-12 C / Vm

part a

The Energy delivered to retina per pulse given that laser pulses are 1.50 ms long:

U = P*t

U = (0.25 ) * (0.0015 )

U = 0.375 mJ

Answer : U = 0.375 mJ

part b

What average pressure would the pulse of the laser beam exert at normal incidence on a surface in air if the beam is fully absorbed?

p_rad = I / c

Where I : Intensity = P / A

p_rad = P / A*c

Where A : Area of circular spot = pi*d^2 / 4

p_rad = 4P / pi*d^2*c

p_rad = 4(0.25) / pi*0.00051^2*(3.0 * 10^8)

p_rad = 0.00408 Pa

Answer : p_rad = 4.08 mPa

part c

What are the wavelength and frequency of the laser light inside the vitreous humor of the eye?

λ_med = n_air*λ_air / n_med

λ_med = (1) * (810 nm) / 1.34

λ_med = 604 nm

f_med = f_air

f_med = c / λ_air

f_med = (3*10^8) / (810 * 10^-9)

f_med = 3.7 * 10^14 Hz

Answer : λ_med = 604 nm ; f_med = 3.7 * 10^14 Hz

d)

What is the electric and magnetic field amplitude in the laser beam?

I = P / A

I = 0.5*ε_o*c*E_o ^2

I = 4P / pi*d^2

Hence, E_o = ( 8 P / ε_o*c*pi*d^2 ) ^ 0.5

E_o = ( 8 * 0.25 / (8.85*10^-12) * (3*10^8) * π * (0.00051)^2) ^ 0.5

E_o = 3.04 * 10^4 V / m

For maximum magnetic field strength:

B_o = E_o / c

B_o = 3.04 * 10^4 / (3*10^8)

B _o = 1.013 *10^-4 Nm/Amp

Answer: E_o = 3.04 * 10^4 V / m , B _o = 1.013 *10^-4 Nm/Amp

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The diagram for this question is shown on the first uploaded image

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