Chapter 36, Problem 36P

(a)

To determine

The image of the fish located $5.00\text{cm}$ and $25.0\text{cm}$ from the front wall of the aquarium.

Answer to Problem 36P

The image of the fish located $5.00\text{cm}$ and $25.0\text{cm}$ from the front wall of the aquarium are $\underset{\_}{-3.77\text{cm}}$ and $\underset{\_}{-19.3\text{cm}}$ respectively.

Explanation of Solution

Write the relation between the refractive indices, object distance, image distance and the radius of curvature.

    $\frac{n_1}{p}+\frac{n_2}{q}=\frac{n_2-n_1}{R}$                                                                                                       (I)

Here, $n_1$ is the refractive index of ice, $n_2$ is the refractive index of air, $R$ is the radius of curvature, $p$ is the object distance and $q$ is the image distance.

Conclusion:

Substitute $-225\text{cm}$ for $R$, $5.00\text{cm }$ for $p$, $1.000$ for $n_2$ and $1.333$ for $n_1$ in the above equation.

    $\begin{array}{c}\frac{1.333}{5.00\text{cm }}+\frac{1.000}{q}=\frac{1.00-1.333}{-225\text{cm}}\\ q=-3.77\text{cm}\end{array}$

Substitute $-225\text{cm}$ for $R$, $25.0\text{cm }$ for $p$, $1.000$ for $n_2$ and $1.333$ for $n_1$ in the equation (I).

    $\begin{array}{c}\frac{1.333}{25.0\text{cm }}+\frac{1.000}{q}=\frac{1.00-1.333}{-225\text{cm}}\\ q=-19.3\text{cm}\end{array}$

Therefore, the image of the fish located $5.00\text{cm}$ and $25.0\text{cm}$ from the front wall of the aquarium are $\underset{\_}{-3.77\text{cm}}$ and $\underset{\_}{-19.3\text{cm}}$ respectively.

(b)

To determine

The magnification of the image of the fish located $5.00\text{cm}$ and $25.0\text{cm}$ from the front wall of the aquarium.

Answer to Problem 36P

The magnification of the image of the fish located $5.00\text{cm}$ and $25.0\text{cm}$ from the front wall of the aquarium are $\underset{\_}{+1.01}$ and $\underset{\_}{+1.03}$ respectively.

Explanation of Solution

Write the expression for the magnification of the image of the fish.

    $M=-\frac{n_{1}q}{n_{2}p}$                                                                                                               (II)

Conclusion:

Substitute $-3.77\text{cm}$ for $q$, $5.00\text{cm }$ for $p$, $1.000$ for $n_2$ and $1.333$ for $n_1$ in the equation (II).

    $\begin{array}{c}M=-\frac{1.333\left(-3.77\text{cm}\right)}{1.000\left(5.00\text{cm }\right)}\\ =+1.01\end{array}$

Substitute $-19.2\text{cm}$ for $q$, $25.0\text{cm }$ for $p$, $1.000$ for $n_2$ and $1.333$ for $n_1$ in the equation (II).

    $\begin{array}{c}M=-\frac{1.333\left(-19.2\text{cm}\right)}{1.000\left(25.00\text{cm }\right)}\\ =+1.03\end{array}$

Therefore, the magnification of the image of the fish located $5.00\text{cm}$ and $25.0\text{cm}$ from the front wall of the aquarium are $\underset{\_}{+1.01}$ and $\underset{\_}{+1.03}$ respectively.

(c)

To determine

The reason that we don’t need to know the refractive index of plastic in the problem.

Answer to Problem 36P

The reason that we don’t need to know the refractive index of plastic in the problem is that the plastic will not change the direction of the ray passing through it.

Explanation of Solution

The uniform thickness of the plastic will make sure that the entry and exit rays to be nearly parallel. There will be a slight displacement only but no change in the direction of the ray passing through the plastic.

Therefore, the reason that we don’t need to know the refractive index of plastic in the problem is that the plastic will not change the direction of the ray passing through it.

(d)

To determine

Whether the image of the fish is farther from the front surface than the fish itself if this aquarium were very long from front to back.

Answer to Problem 36P

Yes, the image of the fish will be farther from the front surface than the fish itself if this aquarium were very long from front to back.

Explanation of Solution

The image of the fish will be farther from the front surface than the fish itself if this aquarium were very long from front to back.

Therefore, the image of the fish will be farther from the front surface than the fish itself if this aquarium were very long from front to back.

(e)

To determine

The magnification.

Answer to Problem 36P

The magnification is $\underset{\_}{+2.00}$.

Explanation of Solution

Assume that the image distance is $-3.00\left|R\right|$ and the object distance is $2.00\left|R\right|$.

Conclusion:

Substitute $-19.2\text{cm}$ for $q$, $25.0\text{cm }$ for $p$, $1.000$ for $n_2$ and $1.333$ for $n_1$ in the equation (II).

    $\begin{array}{c}M=-\frac{1.333\left(-3.00\left|R\right|\right)}{1.000\left(2.00\left|R\right|\right)}\\ =+2.00\end{array}$

Therefore, the magnification is $\underset{\_}{+2.00}$.