You look down at a coin that lies at the bottom of a pool of liquid of depthand index of refraction(Fig. 34-57). Because you view with two eyes, which intercept different rays of light from the coin, you perceive the coin to bewhere extensions of the intercepted rays cross, at depth${\mathit{d}}_{\mathbf{a}}$instead of $\mathit{d}$. Assuming that the intercepted rays in Fig. 34-57 are close to a vertical axis through the coin, show that ${\mathit{d}}_{\mathbf{a}}\mathbf{=}\frac{\mathbf{d}}{\mathbf{n}}$.

1. Refraction index of water is$n$.
2. Refraction index air is$1$
3. Pool depth is$d$.
4. Intercepted rays at depth are.$d_a$

The laws of refraction state that the light coming from higher to lower refractive index medium moves away from the normal line while for the ray from lower to higher, the ray moves towards the normal line. Thus, considering the laws of refraction for the given media, the ray diagram of the given situation is traced.

Now, using the concept of trigonometric calculations for the drawn ray diagram, we can get the required relation for the case of small angle approximations.

Formula:

From the equation for small angle approximations,$\frac{\left(tan{\theta }_2\right)}{\left(tan{\theta }_1\right)}\approx \frac{\left(\sin {\theta }_2\right)}{\left(\sin {\theta }_1\right)}$(i)

The Snell’s law of refraction,$\frac{\sin {\theta }_2}{\sin {\theta }_1}=\frac{n_1}{n_2}$(ii)

The tangent equation of an angle of a right angled-triangle,

$\tan \theta =\frac{Perpendicular}{Base}$(iii)

Medium 1 is water and medium 2 is air. The light rays strike the water surface at point A and B.

Let, the midpoint between A and B be the point C. The penny$P$ is directly below point C. The location of apparent or virtual penny is V when the rays are traced back to the figure to get the position of the image in real.

Now, the angles $\angle CVB$are taken as ${\theta }_2$and angle$\angle CPB$as ${\theta }_1$. The triangles$CVB$and$CPB$share common horizontal side from $C$to$B$is$x.$

Now, using the given data in triangle CVB and equation (i), we get the tangent$\angle CVB$as follows:

localid="1663046799610" $tan{\theta }_2=\frac{x}{d_a}$

Now, using the given data in triangle CPB and equation (i), we get the tangent$\angle CPB$as follows:

localid="1663046327890" $tan{\theta }_1=\frac{x}{d}$

Using the above values in the condition of equation (i) for small angle approximation and the equation (ii) value, we get the required relation as follows:

$$\begin{gathered} \frac{x}{{\displaystyle \frac{d_a}{{\displaystyle \frac{x}{d}}}}\approx {\displaystyle \frac{n_1}{n_2}}} \\ \frac{d}{d_a}\approx n \\ d=\frac{d}{n} \end{gathered}$$

Hence, it is clearly shown that$d_a=\frac{d}{n}$.