Sunligth on the lake bottom

At noon on the summer solstice on the equator, the sun is at its zenith. The sunlight shines on a crystal clear lake with the depth d = 4 m d = 4\,\text{m} What is the relative light intensity I / I 0 I / I_0 of the sun's rays reaching the bottom of the water? Here, I 0 I_0 denotes the intensity of the incident sunlight. Returns the result in percent and rounds it to the nearest integer.

Assumptions: Water has a refractive index of n = ε = 1.33 n = \sqrt {\varepsilon} = 1.33 , whereas for the air we can assume n = 1 n = 1 . At a depth of d 0 = 9.9 m d_0 = 9.9 \, \text {m} the intensity of the transmitted light decreases by half by absorption.

Hint: You can solve this problem with the help of the Fresnel equations, but for real challenge you can also derive the solution from the scratch: For light of wavelength λ \lambda , the incident light beam has an electric field of the form E ( i ) ( z , t ) = E 0 e i ( ω t + k z ) E_{(i)}(z,t) = E_0 e^{i (\omega t + k z)} with the freqeuncy ω = c k \omega = c k , the wave number k = 2 π / λ k = 2 \pi/\lambda and light velocity c c . The total electric field has the form E ( z , t ) = { E ( i ) ( z , t ) + E ( r ) ( z , t ) z > 0 E ( t ) ( z , t ) z < 0 E(z,t) = \begin{cases} E_{(i)}(z,t) + E_{(r)}(z,t) & z > 0 \\ E_{(t)}(z,t) & z < 0 \end{cases} with the contributions E ( r ) ( z , t ) E_{(r)}(z,t) and E ( t ) ( z , t ) E_{(t)}(z,t) of the reflected and transmitted beams. Find a solution for which functions E ( z ) E(z) and E ( z ) E'(z) can be continued steadily at z = 0 z = 0 (water surface) . The intensity of the light is calculated according to I = 1 2 c ε ε 0 E 2 I = \frac{1}{2} c \varepsilon \varepsilon_0 | E |^2 .


The answer is 74.

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1 solution

Markus Michelmann
Nov 17, 2017

In the water wave number and wave length change according to k = n k k' = n k and λ = λ / n \lambda' = \lambda/n . Furthermore, absorption causes an exponential decay E ( t ) e α z E_{(t)} \propto e^{\alpha z} of the transmitted beam. The decay coeffizient α \alpha is determined by I ( d 0 ) I ( 0 ) = e 2 α d 0 = 1 2 α = log 2 2 d 0 = 0.035 1 m \frac{I(-d_0)}{I(0)} = e^{- 2\alpha d_0} = \frac{1}{2} \quad \Rightarrow \quad \alpha = \frac{\log 2}{2 d_0} = 0.035 \,\frac{1}{\text{m}} Therefore, the electric field can be written as E ( z , t ) = { E 0 e i ( ω t + k z ) + A e i ( ω t k z ) z > 0 B e i ( ω t + n k z ) e α z z < 0 \begin{aligned} E(z,t) = \begin{cases} E_0 e^{i (\omega t + k z)} + A e^{i (\omega t - k z)} & z > 0 \\ B e^{i (\omega t + n k z)} e^{\alpha z} & z < 0 \end{cases} \end{aligned} with unknown coefficients A A and B B . The connection condition for the water surface results to E ( z = 0 , t ) = ( E 0 + A ) e i ω t = B e i ω t E ( z = 0 , t ) = i k ( E 0 A ) e i ω t = ( i n k + α ) B e i ω t i n k B e i ω t \begin{aligned} E(z = 0,t) &= (E_0 + A ) e^{i \omega t} = B e^{i \omega t} \\ E'(z= 0,t) &= i k(E_0 - A ) e^{i \omega t} = (i n k + \alpha) B e^{i \omega t} \approx i n k B e^{i \omega t} \end{aligned} In the last line α \alpha was neglected, because the eletric field changes only slightly by the absorption on the long scale of a wavelength ( λ d 0 \lambda \ll d_0 or k α k \gg \alpha ). Therefore, a system of linear equations results ( 1 1 1 n ) ( A B ) = ( E 0 E 0 ) ( A B ) = 1 1 + n ( n 1 1 1 ) ( E 0 E 0 ) = E 0 1 + n ( 1 n 2 ) \begin{aligned} \left( \begin{array}{cc} -1 & 1 \\ 1 & n \end{array} \right) \left( \begin{array}{c} A \\ B \end{array} \right) &= \left( \begin{array}{c} E_0 \\ E_0 \end{array} \right) \\ \Rightarrow \qquad \left( \begin{array}{c} A \\ B \end{array} \right) &= \frac{-1}{1+n}\left( \begin{array}{cc} n & -1 \\ -1 & -1 \end{array} \right) \left( \begin{array}{c} E_0 \\ E_0 \end{array} \right) = \frac{E_0}{1+n}\left( \begin{array}{c} 1-n \\ 2 \end{array} \right) \end{aligned} The light intensity on the lake bottom results I = 1 2 c ε ε 0 E ( d ) 2 = 1 2 n c 0 ε 0 B 2 e 2 α d = 4 n ( 1 + n ) 2 e 2 α d I 0 I I 0 = 4 n ( 1 + n ) 2 e 2 α d 0.741 \begin{aligned} I &= \frac{1}{2} c \varepsilon \varepsilon_0 |E(-d)|^2 = \frac{1}{2} n c_0 \varepsilon_0 |B|^2 e^{-2 \alpha d} = \frac{4 n}{(1 + n)^2} e^{-2 \alpha d} I_0 \\ \Rightarrow \quad \frac{I}{I_0} &= \frac{4 n}{(1 + n)^2} e^{-2 \alpha d} \approx 0.741 \end{aligned} with the Intensity I 0 = 1 2 c 0 ε 0 E 0 2 I_0 = \frac{1}{2} c_0 \varepsilon_0 |E_0|^2 .of the incident light.

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