Question 9.38: Light from an aquarium (Fig. 9.27) goes from water (n = 4 /3...

Light from an aquarium (Fig. 9.27) goes from water \left(n=\frac{4}{3}\right) through a plane of glass \left(n=\frac{3}{2}\right) into air (n = 1). Assuming it’s a monochromatic plane wave and that it strikes the glass at normal incidence, find the minimum and maximum transmission coefficients (Eq. 9.199). You can see the fish clearly; how well can it see you?

T^{-1}=\frac{1}{4 n_{1} n_{3}}\left[\left(n_{1}+n_{3}\right)^{2}+\frac{\left(n_{1}^{2}-n_{2}^{2}\right)\left(n_{3}^{2}-n_{2}^{2}\right)}{n_{2}^{2}} \sin ^{2}\left(\frac{n_{2} \omega d}{c}\right)\right]                            (9.199)

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From Eq. 9.199,

T^{-1}=\frac{1}{4(4 / 3)(1)}\left\{[(4 / 3)+1]^{2}+\frac{[(16 / 9)-(9 / 4)][1-(9 / 4)]}{(9 / 4)} \sin ^{2}(3 \omega d / 2 c)\right\}

 

=\frac{3}{16}\left[\frac{49}{9}+\frac{(-17 / 36)(-5 / 4)}{(9 / 4)} \sin ^{2}(3 \omega d / 2 c)\right]=\frac{49}{48}+\frac{85}{(48)(36)} \sin ^{2}(3 \omega d / 2 c).

T=\frac{48}{49+(85 / 36) \sin ^{2}(3 \omega d / 2 c)} .

Since \sin ^{2}(3 \omega d / 2 c) ranges from 0 to 1, T_{\min }=\frac{48}{49+(85 / 36)}=0.935 ; T_{\max }=\frac{48}{49}=0.980 Not much variation, and the transmission is good (over 90%) for all frequencies. Since Eq. 9.199 is unchanged when you switch 1 and 3, the transmission is the same either direction, and the fish sees you just as well as you see it.

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