Question

Optics help

Answer 1

Laser eye surgery is carried out by delivering highly intense bursts of energy using electromagnetic waves. A typical laser used in such surgery has a wavelength of 190 nm (ultraviolet light) and produces bursts of light that last for 1 ms. It delivers an energy of 0.5 mJ to a circular spot on the cornea with a diameter of 1 mm. (The light is well approximated by a plane wave for the short distance between the laser and the cornea.)

Assuming the energy of a single pulse is delivered to a volume of the cornea about 1 mm3, and assuming the pulses are delivered so quickly that the energy deposited has no time to flow out of that volume, how many pulses are required to raise the temperature of that volume from 20 C to 100 C? (Assume that the cornea has a heat capacity similar to that of water.)

Estimate the maximum strength of the electric field in one of these pulses.

Answer 2

@kropot72 @Mertsj @FibonacciChick666
anyone expert in optics

Answer 3

4.19 Joules raise the temperature of 1 gram of water by 1 degree C.
The density of water is 1000 kg / cubic meter. Therefore it is assumed that 1 mm^3 of cornea weighs 10^-3 grams.
4.19 mJ raise the temperature of 10^-3 g of cornea by 1 degree C.
80 * 4.19 mJ raise the temperature of 10^-3 g of cornea by 80 degrees C.
Each pulse delivers 0.5 mJ, therefore the required number of pulses is
\[\frac{80\times 4.19}{0.5}=670\ pulses\]

Answer 4

and for second part?

Answer 5

I know we are suppose to use plank's constant but how?

Answer 6

The energy of the pulse is 0.5 mJ = power (watts) * time (seconds) = W * 10^-3
Therefore the pulse power is
\[W=\frac{0.5\times 10^{-3}}{10^{-3}}=0.5\ watt\]
The power density is the pulse power divided by the area of the circular spot that is 1 mm in diameter.
\[P _{d}=\frac{0.5\times 4}{\pi \times10^{-6}}=\frac{2}{\pi}\times 10^{6}\ watts/m ^{2}\]
\[P _{d}=\frac{E ^{2}}{377}\]
where E is the electric field strength and 377 ohms is the characteristic impedance of free space.
\[E=10^{3}\sqrt{\frac{2\times 377}{\pi}}=15.5\ kV/m\]