Chapter 25, Problem 48P

(a)

To determine

Show that the ratio of height of the container to the width is $\frac{h}{d}=\sqrt{\frac{n^2-1}{4-n^2}}$.

Answer to Problem 48P

Showed that the ratio of height of the container to the width is $\underset{\_}{\frac{h}{d}=\sqrt{\frac{n^2-1}{4-n^2}}}$.

Explanation of Solution

Figure 1 represent the path of the ray before the container is filled.

From the Figure 1, write the expression for angle of incidence.

    $\sin {\theta }_1=\frac{d}{s_1}$        (I)

Here, ${\theta }_1$ is the angel of incidence of the ray before the container is filled, $d$ is the diameter of the container, and $s_1$ is the distance travelled by the ray inside the container before the container is filled.

In the Figure, $d$, $h$, and $s_1$ forms a right angle triangle. So $s_1$ can be calculated using the expression $\sqrt{h^2+d^2}$.

Use $\sqrt{h^2+d^2}$ for $s_1$ in the equation (I).

    $\begin{array}{c}\sin {\theta }_1=\frac{d}{\sqrt{h^2+d^2}}\\ =\frac{1}{\sqrt{{\left(\frac{h}{d}\right)}^2+1}}\end{array}$        (II)

Here, $h$ is the height of the container.

Figure 2 represent the path of the ray after the container is filled.

From the Figure 2, write the expression for angle of incidence.

    $\sin {\theta }_2=\frac{d/2}{s_2}$        (III)

Here, ${\theta }_2$ is the angel of refraction of the ray after the container is filled, and $s_2$ is the distance travelled by the ray inside the container after the container is filled.

In the Figure, $d/2$, $h$, and $s_1$ forms a right angle triangle. So $s_2$ can be calculated using the expression $\sqrt{h^2+{\left(d/2\right)}^2}$.

Use $\sqrt{h^2+{\left(d/2\right)}^2}$ for $s_2$ in the equation (I).

    $\begin{array}{c}\sin {\theta }_2=\frac{d/2}{\sqrt{h^2+{\left(d/2\right)}^2}}\\ =\frac{1}{\sqrt{4{\left(\frac{h}{d}\right)}^2+1}}\end{array}$        (IV)

Apply Snell’s law at the boundary.

    $1.00\sin {\theta }_1=n\sin {\theta }_2$        (V)

Here, $n$ is the refractive index of the fluid.

Use equation (II) and (IV) in equation (V).

    $\begin{array}{c}\frac{1.00}{\sqrt{{\left(\frac{h}{d}\right)}^2+1}}=\frac{n}{\sqrt{4{\left(\frac{h}{d}\right)}^2+1}}\\ n^2{\left(\frac{h}{d}\right)}^2+n^2=4{\left(\frac{h}{d}\right)}^2+1\\ {\left(\frac{h}{d}\right)}^2\left(4-n^2\right)=n^2-1\\ \frac{h}{d}=\sqrt{\frac{n^2-1}{4-n^2}}\end{array}$        (VI)

Conclusion:

Therefore, the ratio of height of the container to the width is $\underset{\_}{\frac{h}{d}=\sqrt{\frac{n^2-1}{4-n^2}}}$

(b)

To determine

The height of the container.

Answer to Problem 48P

The height of the container is $\underset{\_}{4.73\text{cm}}$.

Explanation of Solution

Conclusion:

Substitute, $1.333$ for $n$, and $8.00\text{cm}$ for $d$ in the equation (VI), to find $h$.

    $\begin{array}{c}\frac{h}{8.00\text{cm}}=\sqrt{\frac{{\left(1.333\right)}^2-1}{4-{\left(1.333\right)}^2}}\\ =0.5907\\ =0.5907\times 8.00\text{cm}\\ h=4.73\text{cm}\end{array}$

Therefore, the height of the container is $\underset{\_}{4.73\text{cm}}$.

(c)

To determine

The range of values of $n$ will be the center of the coin not be visible for any values of $h$ and $d$.

Answer to Problem 48P

The range of values of $n$ will be the center of the coin not be visible for any values of $h$ and $d$ is $\underset{\_}{1,2,\text{and n}>2}$.

Explanation of Solution

From the equation (VI), $\frac{h}{d}=\sqrt{\frac{n^2-1}{4-n^2}}$ it is clear that $h$ will be zero and infinity for the value $n=1$ and $n=2$, and also it has no real solution for $n>2$.

Conclusion:

Substitute, $1$ for $n$ in the equation (VI).

    $\begin{array}{c}\frac{h}{d}=\sqrt{\frac{1-1}{4-1}}\\ h=0\end{array}$

Substitute, $2$ for $n$ in the equation (VI).

    $\begin{array}{c}\frac{h}{d}=\sqrt{\frac{4-1}{4-4}}\\ h=\infty \end{array}$

Therefore, the range of values of $n$ will be the center of the coin not be visible for any values of $h$ and $d$ is $\underset{\_}{1,2,\text{and n}>2}$.