Question

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 weld the detached portion back into place. In one such procedure, a laser beam has a wavelength of 810 nm and delivers 255 mW of power spread over a circular spot 163 μm in diameter. The vitreous humor (the transparent fluid that fills most of the eye) has an index of refraction of 1.34.

Part A: If the laser pulses are each 1.30 ms long, how much energy is delivered to the retina with each pulse Express your answer in millijoules to three significant figures.

Part B: What average pressure does the pulse of the laser beam exert on the retina as it is fully absorbed by the circular spot? Express your answer in Pascals to three significant figures.

Part C: What is the wavelength of the laser light inside the vitreous humor of the eye? Express your answer in nanometers to three significant figures.

Part D: What is the frequency of the laser light inside the vitreous humor of the eye? Express your answer in Hertz to three significant figures.

Part E: What is the maximum value of the electric field in the laser beam? Express your answer in volts per meter to three significant figures.

Part F: What is the maximum value of the magnetic field in the laser beam? Express your answer in teslas to three significant figures.

Answer 1

A)

P = Power of the laser beam = 255 mW = 0.255 W

t = duration = 1.30 ms = 1.30 x 10^-3 s

Energy is given as

E = P t

E = (0.255) (1.30 x 10^-3)

E = 3.32 x 10^-4 J

B)

r = radius of the beam = diameter/2 = 163 x 10^-6/2 = 81.5 x 10^-6 m

Intensity is given as

I = P/(r²)

I = (0.255)/((3.14) ((0.5) (163 x 10^-6))²)

I = 1.22 x 10⁷ W/m²

Pressure is given as

Pressure = I/c

Pressure = (1.22 x 10⁷)/(3 x 10⁸)

Pressure = 0.0407 Pa

C)

= wavelength in air = 810 nm

_n = wavelength in vitreous humor = ?

n = index of refraction of vitreous humor =1.34

wavelength in vitreous humor is given as

_n = /n

_n = 810/1.34

_n = 604 nm

D)

frequency is given as

f = c/

f = (3 x 10⁸)/(810 x 10^-9)

f = 3.70 x 10¹⁴ Hz

E)

E_o = maximum electric field

Intensity is given as

I = (0.5) E_o² c

1.22 x 10⁷ = (0.5) (8.85 x 10^-12) E_o² (3 x 10⁸)

E_o = 9.59 x 10⁴ N/C

F)

maximum magnetic field is given as

B_o = E_o/c = (9.59 x 10⁴)/(3 x 10⁸) = 3.19 x 10^-4 T