Very short pulses of high-intensity laser beams are used to repairdetached portions of the retina of the eye. the brief pulses ofenergy absorbed by the retina welds the detached portion back intoplace. in one such procedure, a laser beam has a wavelength of 900nm, and delivers 258 mw, of power spread over a circular spot 595{\mu}m, in diameter. the vitreous humor (the transparent fluid thatfills most of the eye) has an index of refraction of 1.34.part a if the laser pulses are each 1.5 ms, long, how much energy isdelivered to the retina with each pulse? part b what average pressure does the pulse of the laser beam exert on theretina as it is fully absorbed by the circular spot? part c what is the wavelength of the laser light inside the vitreous humorof the eye? part d what is the frequency of the laser light inside the vitreous humorof the eye? part e what is the maximum value of the electric field in the laserbeam? part f what is the maximum value of the magnetic field in the laserbeam?

a)  W = 3.87 10⁻⁴ J , b)  P = 3.10 10⁻³ Pa , c)  λ = 671. 6 nm , d) frequency does not change , e)  Emax = 1.39 C / m  and f) Bmax = 4.7 10⁻⁹ T

Explanation:

a) Let's use the concepts of power that is work for the unit of time and work is the change of kinetic energy

    P = W / t

An electromagnetic wave has an intensity

    I = P / A =

   W = P t

   W = 258 10⁻³ 1.5 10⁻³

   W = 3.87 10⁻⁴ W s

   W = 3.87 10⁻⁴ J

b) the radiation pressure is given by the ratio

    P = I / c

Where I is the intensity

     I = Powers / A

    A = π r² = π (d/2)²

     I = 258 10⁻³ / π (297.5 10⁻⁶)²

     I = 9.29 10 5 W / m²

     P = 9.29 10⁵/3 10⁸

     P = 3.10 10⁻³ Pa

c) The wavelength when passing a measured of different refractive index changes in the way

     λ = λ₀ / n

     λ = 900 10⁻⁹ / 1.34

     λ = 671.6 10⁻⁹ m

     λ = 671. 6 nm

d) when the light strikes a medium creates a forced oscillation in the electrons of the medium, this is a resonance phenomenon, so the frequency does not change

e) The maximum electric field is

      I = Emax2 / 2 μ₀ c

     Emax = Ra (2 μ₀ c I)

     Emax = Ra (2 4 π 10⁻⁷ 3 10⁸ 258 10⁻³)

     Emax = 1.39 C / m

f) the elective and magnetic fields are related

     c = Emax Bmax

     Bmax = Emax / c

     Bmax = 1.39 / 3 10⁸

     Bmax = 4.7 10⁻⁹ T

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west