Solved Problem on Dioptrics

At the top of the ceiling of a submarine, an isotropic light source was placed (i.e., one that emits light equally in all directions). When this submarine is submerged, what fraction of the light emerges from the surface of the seawater, whose refractive index is 1.34? Neglect the possibility of any light being absorbed by the water.

Problem data:

Index of refraction for seawater: n_w = 1.34;
Index of refraction for air: n_a = 1.

Problem diagram:

Figure 1

Solution:

The light the source emits spreads evenly over an imaginary sphere surrounding the source (Figure 1-A). There is a critical angle at which the light ray spreads parallel to the surface of the sea, if a ray emitted by the source has an angle smaller than this critical angle, it escapes into the air. However, if the angle is larger, the emitted ray will be reflected into the water. Let A₁ be the area of the sphere enclosing all the light emitted by the source, and A₂ the area of the spherical cap through which the light escapes into the air.
Given R as the radius of the sphere enclosing the source (Figure 1-B), the area of the sphere is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {A_1=4\pi R^2} \end{gather} \]

if h is the height of the cap that is above the water, its area will be

\[ \begin{gather} \bbox[#99CCFF,10px] {A_2=2\pi Rh} \end{gather} \]

the critical angle θ_L, and the radius is given as

\[ \begin{gather} R=L+h \tag{I} \end{gather} \]

Thus, the fraction of light f that will emerge is given by the area of the cap above water, A₂, divided by the total area of the sphere, A₁

\[ \begin{gather} f=\frac{A_2}{A_1}=\frac{\cancel 2\cancel{\pi}\cancel Rh}{\cancelto{2}{4}\cancel{\pi}R^{\cancel 2}}=\frac{h}{2R} \tag{II} \end{gather} \]

To determine the fraction of light that emerges, we only need to find h. In Figure 2, a ray is shown emerging from the light source at the critical angle θ_L. Drawing a vertical line N at the point where the ray emerges, the angle θ_L between R and L is the same as the angle between R and N because they are alternate angles.
Applying Snell’s Law

\[ \begin{gather} \bbox[#99CCFF,10px] {n_1\sin\theta_1=n_2\sin\theta_2} \end{gather} \]

Figure 2

\[ \begin{gather} n_m\sin\theta_{\small L}=n_a\sin90°\\[5pt] \sin\theta_{\small L}=\frac{n_a}{n_m}\sin90° \end{gather} \]

From Trigonometry \( \sin90°=1 \)

\[ \begin{gather} \sin\theta_{\small L}=\frac{n_a}{n_m}1\\[5pt] \sin\theta_{\small L}=\frac{n_a}{n_m}\tag{III} \end{gather} \]

From this figure, we can also determine the cosine of the critical angle

\[ \begin{gather} \cos \theta _L=\frac{\text{adjacent leg}}{\text{hypotenuse}}=\frac{L}{R}\\[5pt] R\cos \theta_{\small L}=L \end{gather} \]

from equation (I), we have L=R−h

\[ \begin{gather} R\cos \theta_{\small L}=R-h\\[5pt] h=R-R\cos\theta_{\small L}\\[5pt] h=R\left(1-\cos\theta_{\small L}\right) \tag{IV} \end{gather} \]

From Trigonometry

\[ \begin{gather} \cos^2\theta+\sin^2\theta =1\\[5pt] \cos \theta =\sqrt{1-\sin^2\theta \;} \end{gather} \]

substituting this cosine value into equation (IV)

\[ \begin{gather} h=R\left(1-\sqrt{1-\sin^2\theta_{\small L}\;}\right) \end{gather} \]

substituting equation (III)

\[ \begin{gather} h=R\left[1-\sqrt{1-\left(\frac{n_a}{n_m}\right)^2\;}\right] \tag{V} \end{gather} \]

substituting equation (V) into equation (II)

\[ \begin{gather} f=\frac{\cancel R\left[1-\sqrt{1-\left(\dfrac{n_a}{n_m}\right)^2\;}\right]}{2\cancel R}\\[5pt] f=\frac{1}{2}\left[1-\sqrt{1-\left(\frac{n_a}{n_m}\right)^2\;}\right]\\[5pt] f=\frac{1}{2}\left[1-\sqrt{1-\left(\frac{1}{1,34}\right)^2\;}\right]\\ f=0,167 \end{gather} \]

The fraction of light that emerges from the sea will be 16.7%.

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .