Question

power densities used i Typical values of a) For a laser with a ( n Photo Dynamic Therapy PDT are in the range of 120 mW/em a beam radius of 0.5 mm (assume negligible divergence), issue if the laser head is held 1.0 cm from the patient? what is the power density on the patient's t b) What radius of the spot size would you use to achieve a power density of 120 mWlem 21 c) What would you do to get this desired spot size? d) The fluence for this procedure is 140 joules/em2. How long must the exposure time be to deposit this

Answer 1

Here, the best part i know.

Given, power density (PD) = 120 mW/cm² = 0.12 W cm^-2

(a) Power density (PD) = Power / Area .

Power is given P = 3 W, therefore, we can calculate area of the spot using the above formula.

Area = Power / Power density = 3 W / 0.12 W cm^-2 = 25 cm²

Now, area of the spot is A = πR². Therefore, R² = 25 cm² / 3.14 = 7.961 cm².

Which implies that, Radius of the spot size is R = 2.821 cm.

(d) Given that the total fluence of tissue was 130 joules/cm², that means, energy per area is given.

We know that from power density PD = Power / area and Power = energy / time.

Therefore, PD = Energy/ (area x time), which implies that time t = 140 joules cm^-2 / 0.12 W cm^-2 = 1166.6 sec

since joules / W = sec.

Therefore, exposure time t = 1166.6 sec (or) 19.44 min

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