Question

A swimmer enters a gloomier world (in one sense) on diving to greater depths. Given that the mean molar absorption coefficient of sea water in the visible region is 6.2 x 10-5- L moI-1 cm-1, calculate the depth at which a diver will experience (a) half the surface intensity of light, (b) one-tenth the surface intensity

Answer 1

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.

    $log_{10} = \frac{I_{o}}{I_{t}} = \epsilon \times c \times l$

As it is given that,

           $\frac{I_{o}}{I_{t}} = 2$                

and,   $\epsilon = 6.2 \times 10^{-3}$

We know that molarity of sea water is 599 mM.

   $log_{10}(2) = 6.2 \times 10^{-5} \times 599 \times 10^{-3} \times l$

       l = $\frac{0.301}{6.2 \times 10^{-5} \times 599 \times 10^{-3}}$

        = 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,

              $\frac{I}{I_{o}} = 10$

    $log_{10}(10) = 6.2 \times 10^{-5} \times 599 \times 10^{-3} \times l$

                l = $\frac{1}{6.2 \times 10^{-5} \times 599 \times 10^{-3} \times l}$

                  = 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.