Respuesta :
Answer: (a) The depth at which a diver will experience half the surface intensity of light is 0.81 m.
(b) The depth at which a diver will experience one-tenth the surface intensity is 2.69 m.
Explanation:
(a) We know that Lambert-Beer's law is as follows.
[tex]log_{10} = \frac{I_{o}}{I_{t}} = \epsilon \times c \times l[/tex]
As it is given that,
[tex]\frac{I_{o}}{I_{t}} = 2[/tex]
and, [tex]\epsilon = 6.2 \times 10^{-3}[/tex]
We know that molarity of sea water is 599 mM.
[tex]log_{10}(2) = 6.2 \times 10^{-5} \times 599 \times 10^{-3} \times l[/tex]
l = [tex]\frac{0.301}{6.2 \times 10^{-5} \times 599 \times 10^{-3}}[/tex]
= 81 cm
= 0.81 m
Therefore, the depth at which a diver will experience half the surface intensity of light is 0.81 m.
(b) We are given that,
[tex]\frac{I}{I_{o}} = 10[/tex]
[tex]log_{10}(10) = 6.2 \times 10^{-5} \times 599 \times 10^{-3} \times l[/tex]
l = [tex]\frac{1}{6.2 \times 10^{-5} \times 599 \times 10^{-3} \times l}[/tex]
= 269 cm
or, = 2.69 m (as 1 m = 100 cm)
Therefore, the depth at which a diver will experience one-tenth the surface intensity is 2.69 m.
