A horizontal meter stick is centered at the bottom of a $3.0\mathrm{m}$-deep, $3.0\mathrm{m}$-wide pool of water. How long does the meter stick appear to be as you look at it from the edge of the pool?

We have given,

height of the pool = $3\mathrm{m}$

Wide = $3\mathrm{m}$

We have to find the appear length of the meter stick in the pole.

Using Snell's law at the surface of water, let us the angle of incident of the ray is ${\mathrm{\theta }}_1$and angle of refraction is ${\mathrm{\theta }}_2$and also considerer their refractive index as of air and water is ${\mathrm{n}}_1,{\mathrm{n}}_2.$Then,

${\mathrm{n}}_1{\mathrm{sin\theta }}_1={\mathrm{n}}_2{\mathrm{sin\theta }}_2$

As we know in general that

$$\begin{gathered} {\mathrm{n}}_1=1 \\ {\mathrm{n}}_2=1.3 \end{gathered}$$

Then, as given that the meter stick is at the center of the pole as shown in the figure and we will analyze by considering the light coming from the water to the eye.

Now using the Snell law we can write, for the equations

${\mathrm{n}}_1{\mathrm{sin\theta }}_1={\mathrm{n}}_2\mathrm{sin\theta }{\text{'}}_1$

From the geometry we can write,

${\mathrm{\theta }}_1={\tan }^{-1}\frac{1}{3}$

then, similarly using the geometry, we can write,

$$\begin{gathered} {\mathrm{\theta }}_1^{\text{'}}={\tan }^{-1}\frac{BC}{3} \\ \mathrm{BC}=3\tan \left({\sin }^{-1}\left(\frac{{\mathrm{n}}_1}{{\mathrm{n}}_2}{\mathrm{sin\theta }}_1\right)\right) \\ \mathrm{BC}=1.39\mathrm{m} \end{gathered}$$

Now let us consider that light ray is coming from the air to water,

Now by geometry we can say that ,

${\mathrm{\theta }}_2={\tan }^{-1}\frac{2}{3}$

then ,

$$\begin{gathered} {\mathrm{\theta }}_2\text{'}={\tan }^{-1}\frac{\mathrm{BD}}{3} \\ \mathrm{BD}=3\tan \left({\sin }^{-1}(\frac{1}{1.3}{\mathrm{sin\theta }}_2)\right) \\ \mathrm{BD}=3.28\mathrm{m} \end{gathered}$$

Then the length will be,

$\mathrm{L}=3.28-1.39=1.9\mathrm{m}$