Physics, 19.03.2020 01:22 joemoe15jr

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 μm in diameter. The vitreous humor (the transparent fluid that fills most of the eye) has an index of refraction of 1.34.A) If the laser pulses are each 1.5 ms, long, how much energy is delivered to the retina with each pulse?B) What average pressure does the pulse of the laser beam exert on the retina as it is fully absorbed by the circular spot?C) What is the wavelength of the laser light inside the vitreous humor of the eye?D) What is the frequency of the laser light inside the vitreous humor of the eye?E) What is the maximum value of the electric field in the laser beam?F) What is the maximum value of the magnetic field in the laser beam?

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Answer from: kainagoicoowntpi

See attachment for complete solving.

Explanation:

Given that: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 μm in diameter. The vitreous humor (the transparent fluid that fills most of the eye) has an index of refraction of 1.34.A) If the laser pulses are each 1.5 ms, long, how much energy is delivered to the retina with each pulse?B) What average pressure does the pulse of the laser beam exert on the retina as it is fully absorbed by the circular spot?C) What is the wavelength of the laser light inside the vitreous humor of the eye?D) What is the frequency of the laser light inside the vitreous humor of the eye?E) What is the maximum value of the electric field in the laser beam?F) What is the maximum value of the magnetic field in the laser beam.

See attachment for further solving.

Very short pulses of high-intensity laser beams are used to repair detached portions of the retina o

Answer from: AniyaT05

Part A

energy is delivered to the retina with each pulse is $U= 3.75*10^{-4}J$

Part B

The average pressure is $p=4.0*10^{-3}Pa$

Part C

The wavelength $\lambda_Y= 604.48nm$

Part D

The frequency $f=3.704*10^{14}Hz$

Part E

maximum value of the electric field is $E_{maximum} ==30.361\ KV/m$

Part F

maximum value of the magnetic field is $B_{maximum}=1.012*10^{-7}T$

Explanation:

From the question we are given that

The wavelength of the laser beam is $\lambda = 810nm$

The power of the laser beam is $P = 250mW = 250*10^{-3}W$

The spread diameter $d = 510\mu m = 510 *10^{-6}m$

The refractive index of the eye is i = 1.34

Generally the mathematical representation of the energy delivered to the retina is

U = Pt

Where P is the power $= 250*10^{-3}W$

t is the time $= 1.5*10^{-3} s$

$U = (250 *10^{-3})(1.5*10^{-3} ) = 3.75*10^{-4}J$

Generally the mathematical representation of pressure of pulse is

$p = \frac{I}{c}$

Where I is the intensity of the pulse given as

$I = \frac{P}{A}$ where A is the area $A= \pi r^2$ and P is the power of the laser beam

and c is the speed of light

$p = \frac{P}{\pi r^2c} = \frac{250*10^{-3}W}{\pi[\frac{510*10^{-6}m}{2} ]^2 (3.0*10^8 m/s)}$

$=4.0*10^{-3}Pa$

Generally the mathematical representation of the wave length of the light inside the vitreous humor of the eye is

$\lambda_Y = \frac{\lambda_0}{n_Y}$

Where $\lambda_Y$ is the wave length of the light inside the vitreous humor of the eye

$\lambda_0$ is the wavelength of the beam

n_Y is the index of refraction of vitreous humor of the

So,

$\lambda_Y = \frac{810*10^{-9}}{1.34}$

$\lambda_Y= 604.48nm$

Generally the mathematically representation for the frequency of the laser light is

$f= \frac{c}{\lambda_0}$

$f = \frac{3*10^8 m/s}{810*10^{-9}}$

$f=3.704*10^{14}Hz$

Generally the mathematically representation for the maximum value of the electric field is given as

$E_{maximum} = \sqrt{\frac{2P}{\pi r^2 \epsilon_0c} }$

$=\sqrt{\frac{2(250*10^{-3})W}{\pi (2.55*10^{-4})^2(8.85*10^{-12}C^2/N \cdot m^2)(3*10^{8})} }$

$=30.361\ KV/m$

Generally the mathematically representation for the maximum value of the magnetic field is given as

$B_{maximum} = \frac{E_{maximum}}{c}$

$B_{maximum} = \frac{30.361*10^3}{3.0*10^8} \\$

$B_{maximum}=1.012*10^{-7}T$

Answer from: Quest

see earlier answer

Answer from: Quest

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg

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