The meter stick in FIGURE P34.48 lies on the bottom of a 100-cm long tank with its zero mark against the left edge. You look into the tank at a 30° angle, with your line of sight just grazing the upper left edge of the tank. What mark do you see on the meter stick if the tank is (a) empty, (b) half full of water, and (c) completely full of water?

We need to find out :

a)The point seen by the person on the meter stick if the tank is empty.

b)The point seen by the person on the meter stick if the tank is half full of water.

c)The point seen by the person on the meter stick if the tank is completely full of water.

The ray moves in a straight line and strikes the meter stick at a distance of x cm as shown below :

$$\begin{gathered} \tan 60°=\frac{x}{50cm} \\ x=(50cm)\times \tan 60° \\ x=87cm \end{gathered}$$

The rays strikes the water at ${\theta }_1=60°$with respect to normal and refracts inwards at an angle of ${\theta }_2$ and strikes the meter scale at a distance of x₁ + x₂ from zero point as shown below.

$$\begin{gathered} \tan 60°=\frac{x_1}{25cm} \\ {x}_1=(25cm)\times \tan 60° \\ {x}_1=25\sqrt{3}cm \end{gathered}$$

Applying snell's law to find ${\theta }_2$

$$\begin{gathered} n_1\sin {\theta }_1=n_2\sin {\theta }_2 \\ \sin {\theta }_2=\frac{n_1\sin {\theta }_1}{n_2} \\ {\theta }_2={\sin }^{-1}\left(\frac{n_1\sin {\theta }_1}{n_2}\right) \\ {\theta }_2={\sin }^{-1}\left(\frac{(1.0)\sin 60°}{1.33}\right) \\ {\theta }_2=40.6° \end{gathered}$$

We know that,

$$\begin{gathered} \tan {\theta }_2=\frac{x_2}{25cm} \\ {x}_2=(25cm)\times \tan (40.6)° \\ {x}_2=21.45cm \\ Thereforelightstrikesat{x}_1+x_{2}cm=(21.45+25\sqrt{3})cm=64.8cm \end{gathered}$$

The ray strikes at the surface at an angle of ${\theta }_1$with respect to normal and refracts at an angle of ${\theta }_2$and strikes at the distance of x cm as shown below :

$$\begin{gathered} \tan {\theta }_2=\frac{x}{50cm} \\ x=(50cm)\times \tan (40.6)° \\ x=42.9cm \end{gathered}$$