Respuesta :
Answer:
(a). The power per unit area at the retina is [tex]13.8\times10^{14}\ W/m^2[/tex].
(b). The energy delivered per pulse to an area of molecular size is [tex]0.174\times10^{-12}\ J[/tex]
Explanation:
Given that,
Emitted energy = 2.20 mJ
Time t = 1.00 ns
Diameter D= 45.0 μm
Diameter of circular area d= 0.400 nm
(a). We need to calculate the power per unit area at the retina
Using formula of power
[tex]Power\ per\ area = \dfrac{E}{t\times A}[/tex]
[tex]Power\ per\ area = \dfrac{E}{t\times \pi r^2}[/tex]
Put the value into the formula
[tex]Power\ per\ area = \dfrac{2.20\times10^{-3}}{1.00\times10^{-9}\times \pi (22.5\times10^{-6})^2}[/tex]
[tex]Power\ per\ area = 13.8\times10^{14}\ W/m^2[/tex]
(b). We need to calculate the energy is delivered per pulse to an area of molecular size
Using formula of energy
[tex]E'=E\times(\dfrac{d}{D})^2[/tex]
Put the value in to the formula
[tex]E'=2.20\times10^{-3}\times(\dfrac{0.400\times10^{-9}}{45.0\times10^{-6}})^2[/tex]
[tex]E'=1.738\times10^{-13}\ J[/tex]
[tex]E'=0.174\times10^{-12}\ J[/tex]
Hence, (a). The power per unit area at the retina is [tex]13.8\times10^{14}\ W/m^2[/tex].
(b). The energy delivered per pulse to an area of molecular size is [tex]0.174\times10^{-12}\ J[/tex]
