Question

Laser beams are sometimes used to burn away cancerous tissue. show answer Incorrect Answer What is the intensity, in watts per square meter, of a laser beam that is 90.0% absorbed by a 1.95-mm diameter spot of cancerous tissue and must deposit 495 J of energy to it in a time period of 3.8 s

Answer 1

Answer:

The intensity of the beam is $ 1.21\times10^{7} \frac{W}{m^2}$

Explanation:

We can find the intensity absorbed by the spot cancerous tissue with the equation:

$I_{abs}= \frac{P}{A}$(1)

with P average power and A the effective area. Average power is the energy (E) absorbed in a certain time (t) so:

$P=\frac{E}{t}$ (2)

Using (2) on (1)

$I_{abs}=\frac{E}{tA}$(3)

We can assume the area of the spot as a circumference of radius (r) 1.95 mm or $1.95\times10^{-3}m$ so the area of a circumference with this radius is:

$A=\pi r^2 = \pi (1.95\times10^{-3})^(2)=1.19\times10^{-5} m^2$

Using this value on (3)

$I_{abs}=\frac{(495)}{3.8(1.19\times10^{-5})} = 1.09\times10^{7} \frac{W}{m^2}$ (3)

Because the tissue only absorbs 90% of the intensity the total intensity, we can use the relation:

$\frac{I}{100}=\frac{1.09\times10^{7}}{90}$

to find the intensity of the beam (I):

$I=\frac{(1.09\times10^{7})(100)}{90}=1.21\times10^{7} \frac{W}{m^2}$