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Q71PE
Expert-verified(a) What is the intensity in W/m2 of a laser beam used to burn away cancerous tissue that, when 90.0% absorbed, puts 500 J of energy into a circular spot 2.00 mm in diameter in 4.00 s?
(b) Discuss how this intensity compares to the average intensity of sunlight (about 700 W/m2) and the implications that would have if the laser beam entered your eye. Note how your answer depends on the time duration of the exposure.
(a) The intensity of a laser beam is \(4.4 \times {10^7}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\).
(b) The intensity of the laser beam is \(6.32 \times {10^4}\)times that the intensity of the sunlight.
The laser beam can damage the eye in some seconds.
The given data can be listed below as:
The intensity is referred to as the power that is transmitted per unit area of an object. The SI unit of intensity is watt per square meter.
The equation of intensity is expressed as:
\({\rm{I}} = \frac{{\rm{P}}}{{{\rm{\pi }}{{\left( {\frac{{\rm{d}}}{{\rm{2}}}} \right)}^{\rm{2}}}}}\) (1)
Here, \({\rm{I}}\) is the intensity, \({\rm{P}}\) is the rate of the consumption of energy, and \({\rm{d}}\) is the diameter of the circular spot.
The equation of the rate of the consumption of energy is expressed as:
\({\rm{P}} = \frac{{\rm{E}}}{{\rm{t}}}\)
Here, \({\rm{E}}\) is the energy has been put and \({\rm{t}}\) is the time taken to put the energy.
Let the intensity of the laser beam is \({\rm{I}}\) and as the intensity absorbed is \(90.0\% \), then the intensity of the laser beam will be \({\rm{I}} \times \frac{{90}}{{100}} = 0.{\rm{9I}}\).
Substitute 0.9I for I and \(\frac{{\rm{E}}}{{\rm{t}}}\) for P in the above equation.
\(\begin{aligned}{}{\rm{0}}{\rm{.9I}} &= \frac{{\rm{E}}}{{{\rm{\pi }}{{\left( {\frac{{\rm{d}}}{{\rm{2}}}} \right)}^{\rm{2}}}{\rm{t}}}}\\{\rm{I}} &= \frac{{\rm{E}}}{{{\rm{0}}{\rm{.9\pi }}{{\left( {\frac{{\rm{d}}}{{\rm{2}}}} \right)}^{\rm{2}}}{\rm{t}}}}\end{aligned}\)
Substitute the values in the above equation.
\(\begin{aligned}{\rm{}} &= \frac{{500\;{\rm{J}}}}{{0.9\left( {3.14} \right){{\left( {\frac{{{\rm{2 \times 1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{m}}}}{2}} \right)}^2}\left( {4.00\;{\rm{s}}} \right)}}\\ &= \frac{{500\;{\rm{J}}}}{{\left( {{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{{\rm{m}}^2}} \right)\left( {11.304\;{\rm{s}}} \right)}}\\ &= \frac{{500\;{\rm{J}}}}{{\left( {11.304{\rm{ \times 1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{{\rm{m}}^{\rm{2}}} \cdot {\rm{s}}} \right)}}\\ &= 4.4 \times {10^7}\;{\rm{J/}}{{\rm{m}}^{\rm{2}}} \cdot {\rm{s}}\end{aligned}\)
Hence, further as:
\(\begin{aligned}{\rm{I}} &= 4.4 \times {10^7}\;{\rm{J/}}{{\rm{m}}^{\rm{2}}} \cdot {\rm{s}}\\ &= 4.4 \times {10^7}\;{\rm{J/}}{{\rm{m}}^{\rm{2}}} \cdot {\rm{s}} \times \frac{{1\;{\rm{W}}}}{{1\;{\rm{J/s}}}}\\ &= 4.4 \times {10^7}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\end{aligned}\)
Thus, the intensity of a laser beam is \(4.4 \times {10^7}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\).
The equation of the ratio of the intensity of the laser beam to the sunlight is expressed as:
\({\rm{r}} = \frac{{\rm{I}}}{{{{\rm{I}}_{\rm{1}}}}}\)
Here, \({\rm{r}}\) is the ratio of the intensity of the laser beam to the sunlight and \({{\rm{I}}_{\rm{1}}}\) is the average intensity of the sunlight.
Substitute the values in the above equation.
\(\begin{aligned}{\rm{r}} &= \frac{{4.4 \times {{10}^7}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}}}{{700\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}}}\\ &= 6.32 \times {10^4}\end{aligned}\)
Thus, the intensity of the laser beam is \(6.32 \times {10^4}\)times that of the intensity of the sunlight.
As the intensity of the laser beam is very high, in only some seconds, it can severely damage the eye.
Thus, the laser beam can damage the eye in some seconds.
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