Question

Light from an aquarium (Fig. 9.27) goes from water $(n=\frac{4}{3})$through a plane of glass $\mathbf{(}\mathbf{n}\mathbf{=}\frac{\mathbf{3}}{\mathbf{2}}\mathbf{)}$into the air $\mathbf{n}\mathbf{=}\mathbf{1}$. 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?

Short answer

The minimum and maximum transmission coefficients are$T_{min}=0.93455$andlocalid="1657453230176" $T_{max}=0.987959$respectively, and it is clearly seen that fish sees you just as well as you see it.

Step-by-step solution

Given Information:

Given data:

The refractive index of water is $n_1=\frac{4}{3}$

The refractive index of glass is $n_2=\frac{3}{2}$

The refractive index of air is$n_3=1$

Determine the minimum and maximum transmission coefficients:

Write the expression for the inverse of the transmission coefficient.

${\mathit{T}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{n}}_{\mathbf{1}}{\mathbf{n}}_{\mathbf{3}}}\left[{\left(n_1+n_3\right)}^2+\frac{\left(n_1^2-n_2^2\right)\left(n_3^2-n_2^2\right)}{n_2^2}si{n}^2\frac{n_2\omega d}{C}\right]$

Substitute localid="1657452458494" $n_1=\frac{4}{3},n_2=\frac{3}{2}$, and $n_3=1$in the above expression.

$T^{-1}=\frac{1}{4\left(\frac{4}{3}\right)\left(1\right)}\left[1+{\left(\frac{4}{3}\right)}^2+\frac{\left[{\left(\frac{4}{3}\right)}^2-{\left(\frac{3}{2}\right)}^2\left(r^2-{\left(\frac{3}{2}\right)}^2\right)\right]}{{\left(\frac{3}{2}\right)}^2}{\sin }^2\left(\frac{3\omega d}{2C}\right)\right]$

$T^{-1}=\frac{3}{16}\left[\frac{49}{9}+\frac{\left(\frac{-17}{36}\right)\left(-\frac{5}{4}\right)}{\left(\frac{9}{4}\right)}{\sin }^2\frac{3\omega d}{2C}\right]$

$T^{-1}=\frac{3}{16}\left[\frac{49}{9}+\left(\frac{17}{36}\right)\times \left(\frac{5}{4}\right)\times \left(\frac{4}{9}\right){\sin }^2\left(\frac{3\omega d}{2C}\right)\right]$

$T^{-1}=\frac{49}{48}+\frac{85}{\left(48\right)\left(36\right)}{\sin }^2\left(\frac{3\omega d}{2C}\right)$ …… (1)

For transmission coefficient to be minimum,localid="1657452777022" ${\sin }^2\frac{3\omega d}{2C}=1$

Substitute ${\sin }^2\frac{3\omega d}{2C}=1$in equation (1).

localid="1657453325025" role="math" $$\begin{gathered} T_{\min }=\frac{48}{49+\frac{85}{36}} \\ {T}_{min}=0.93455 \end{gathered}$$

For transmission coefficient to be minimum, ${\sin }^2\frac{3\omega d}{2C}=0$

Substitute ${\sin }^2\frac{3\omega d}{2C}=0$in equation (1).

$T_{\max }=\frac{48}{49}$

localid="1657453029338" $T_{\max }=0.987959$

On interchanging the values of $n_1$and $n_3$, there will be no change in the transmission coefficient equation. Hence, it is clearly seen that fish sees you just as well as you see it.

Therefore, the minimum and maximum transmission coefficients are localid="1657453345104" $T_{min}=0.93455$and$T_{\max }=0.987959$respectively, and it is clearly seen that fish sees you just as well as you see it.