Chapter 35, Problem 19P

(a)

To determine

The image position of the fishes that are located at $5.00\text{cm}$ and $25.0\text{cm}$ in front of the aquarium wall.

Answer to Problem 19P

The image is $3.77\text{cm}$ from the front of the wall when the fish is at $5.00\text{cm}$ and when the fish is at $25.0\text{cm}$ the image is $19.3\text{cm}$ form the front of the wall

Explanation of Solution

Given info: The radius of curvature of the curved plastic is $2.25\text{m}$. The refractive index of water is $1.33$ and that of air is $1.00$.

The formula to calculate image position of the fish inside the aquarium is,

$\frac{n_1}{p}+\frac{n_2}{q}=\frac{n_2-n_1}{R}$ (1)

Here,

$n_1$ is the refractive index of the water.

$n_2$ is the refractive index of the air.

$p$ is the object distance.

$q$ is the image distance.

$R$ is the radius of curvature

The radius of curvature will be negative the centre of curvature lies on the object side.

For part (i): when the fish is at $5.00\text{cm}$.

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

Substitute  $1.33$ for $n_1$, $1.00$ for $n_2$ and $-2.25\text{m}$ for $R$ and $25.0\text{cm}$ for $p$. in the equation (1),

$\begin{array}{c}\frac{n_1}{p}+\frac{n_2}{q}=\frac{n_2-n_1}{R}\\ \frac{1.33}{5.00\text{cm}}+\frac{1.00}{q}=\frac{1.000-1.333}{-2.25\text{m}}\\ \frac{1.33}{5.00\text{cm}}+\frac{1.000}{q}=\frac{-0.3333}{-2.25\text{m}\times \frac{\text{1}\text{cm}}{{10}^{-2}\text{m}}}\\ q=-3.77\text{cm}\end{array}$ (2)

The image position for the fish at $5.00\text{cm}$ is $3.77\text{cm}$ from the front of the wall in the water.

For part (ii) when the fish is at $25.0\text{cm}$.

From equation (2) the image position is,

$\begin{array}{c}\frac{n_1}{p}+\frac{n_2}{q}=\frac{n_2-n_1}{R}\\ \frac{1.33}{25.0\text{cm}}+\frac{1.00}{q}=\frac{1.000-1.333}{-2.25\text{m}}\\ \frac{1.33}{25.0\text{cm}}+\frac{1.000}{q}=\frac{-0.3333}{-2.25\text{m}\times \frac{\text{1}\text{cm}}{{10}^{-2}\text{m}}}\\ q=-19.3\text{cm}\end{array}$ (3)

The image is at $19.3\text{cm}$ for the fish at $25.0\text{cm}$ from the front of the wall.

Conclusion:

Therefore, the image is $3.77\text{cm}$ from the front of the wall when the fish is at $5.00\text{cm}$ and when the fish is at $25.0\text{cm}$ the image is $19.3\text{cm}$ form the front of the wall.

(b)

To determine

The magnification of the images for part (a)

Answer to Problem 19P

The magnification of the image is $+1.01$ when the fish is at $5.00\text{cm}$ and when the fish is at $25.0\text{cm}$ the magnification of the image is $+1.03$.

Explanation of Solution

Given info: The radius of curvature of the curved plastic is $2.25\text{m}$. The refractive index of water is $1.33$ and that of air is $1.00$.

The formula to calculate the magnification of the image is,

$M=-\frac{n_{1}q}{n_{2}p}$ (4)

For part (i): when the fish is at $5.00\text{cm}$.

Substitute $1.33$ for $n_2$, $1.00$ for $n_1$, $5.00\text{cm}$ for $p$ and $-3.77\text{cm}$ for $q$ in equation (4).

$\begin{array}{c}M=-\frac{n_{1}q}{n_{2}p}\\ =-\left(\frac{1.333\left(-3.77\text{cm}\right)}{1.00\left(5.00\text{cm}\right)}\right)\\ =+1.01\end{array}$

Thus when the fish is at $5.00\text{cm}$ the magnification of the image is $+1.01$.

For part (ii): when the fish is at $25.0\text{cm}$.

Substitute $1.33$ for $n_2$, $1.00$ for $n_1$, $25.0\text{cm}$ for $p$ and $-19.3\text{cm}$ for $q$ in equation(4).

$\begin{array}{c}M=-\frac{n_{1}q}{n_{2}p}\\ =-\left(\frac{1.333\left(-19.3\text{cm}\right)}{1.00\left(25.0\text{cm}\right)}\right)\\ =+1.03\end{array}$

Thus when the fish is at $25.0\text{cm}$ the magnification of the image is $+1.03$.

Conclusion:

Therefore, when the fish is at $5.00\text{cm}$ the magnification of the image is $+1.01$ and when the fish is at $25.0\text{cm}$ the magnification of the image is $+1.03$.

(c)

To determine

The reason refractive index of the plastic is not required to solve the problem.

Answer to Problem 19P

The refractive index of plastic is not playing any major role in light propagation

Explanation of Solution

The plastic has uniform thickness and the surface from which the ray is entering and the surface from which is leaving are parallel to each other. The ray might get slightly displaced, but it will not change the direction of its propagation by going through plastic air interface. The only difference will be due to water-air interface.

Conclusion:

Therefore, the ray might get slightly displaced, but it will not change the direction of its propagation by going through plastic air interface. So the refractive index of plastic is not playing any major role in light propagation.

(e)

To determine

The image distance of the fish is greater than the fish itself and the magnification

Answer to Problem 19P

Yes, the image distance of the fish can be larger than the distance where the fish is itself.

Explanation of Solution

For the object distance greater than the radius of curvature the image distance will greater than the distance at which fish is itself. If the aquarium were very long the radius of curvature will not increase therefore if the object distance is more than the radius of curvature the image of the fish will be at even farther distance away from the fish itself.

Conclusion:

Therefore, If the fish is present at distance larger than the radius of curvature the image of the fish would be even farther than that of fish itself.

(d)

To determine

The magnification of the image when the image of the fish is even farther than the position of fish itself.

Answer to Problem 19P

.

Explanation of Solution

Let the object distance is greater than the radius of curvature.

For the condition

$p=\left|R\right|$

Formula to calculate the image distance from Lens formula

$\frac{n_1}{p}+\frac{n_2}{q}=\frac{n_1-n_2}{\left|R\right|}$

Substitute $\left|R\right|$ for $p$ in the above equation

$\begin{array}{c}\frac{n_1}{p}+\frac{n_2}{q}=\frac{n_1-n_2}{\left|R\right|}\\ \frac{n_1}{\left|R\right|}+\frac{n_2}{q}=\frac{n_1-n_2}{\left|R\right|}\\ \frac{n_2}{q}=\frac{-n_2}{\left|R\right|}\\ q=-\left|R\right|\end{array}$

For the condition

$p>\left|R\right|$

Take reciprocal of the above question

$\begin{array}{c}p>\left|R\right|\\ \frac{1}{R}>\frac{1}{p}\\ \frac{1}{R}-\frac{1}{p}>0\end{array}$ (5)

Formula to calculate the image distance from Lens formula,

$\frac{n_1}{p}+\frac{n_2}{q}=\frac{n_1-n_2}{\left|R\right|}$

$\begin{array}{c}\frac{n_1}{p}+\frac{n_2}{q}=\frac{n_1-n_2}{\left|R\right|}\\ \frac{n_2}{q}=\frac{n_1-n_2}{R}-\frac{n_1}{p}\\ \frac{n_2}{q}=\frac{n_1}{R}-\frac{n_2}{R}-\frac{n_1}{p}\end{array}$

Divide by $n_2$ both sides the equation.

$\frac{1}{q}=-\frac{1}{\left|R\right|}+\frac{n_1}{n_2}\left(\frac{1}{\left|R\right|}-\frac{1}{p}\right)$

The condition is when the image distance is greater than the radius of curvature take the magnitude of the equation.

$\frac{1}{\left|q\right|}=\frac{1}{\left|R\right|}-\frac{n_1}{n_2}\left(\frac{1}{\left|R\right|}-\frac{1}{p}\right)$ (6)

Substitute $1.33$ for $n_1$ and $1.00$ for $n_2$ in equation (6).

$\begin{array}{c}\frac{1}{\left|q\right|}=\frac{1}{\left|R\right|}-\frac{n_1}{n_2}\left(\frac{1}{\left|R\right|}-\frac{1}{p}\right)\\ \frac{1}{\left|q\right|}=\frac{1}{\left|R\right|}-\frac{1.33}{1.00}\left(\frac{1}{\left|R\right|}-\frac{1}{p}\right)\\ \frac{1}{\left|q\right|}<\frac{1}{\left|R\right|}\end{array}$ (7)

The reciprocal of the equation is

$\left|q\right|>\left|R\right|$

Thus the image of the fish will also be at greater distance than that of radius of curvature.

An example for the above case is let the fish is at twice the distance of the magnitude of radius of curvature.

The image of the fish is calculated from the formula from equation (7).

$\frac{1}{\left|q\right|}=\frac{1}{\left|R\right|}-\frac{1.33}{1.00}\left(\frac{1}{\left|R\right|}-\frac{1}{p}\right)$

Substitute $2.00\left|R\right|$ for $p$

$\begin{array}{l}\frac{1}{\left|q\right|}=\frac{1}{\left|R\right|}-\frac{1.33}{1.00}\left(\frac{1}{\left|R\right|}-\frac{1}{p}\right)\\ \frac{1}{\left|q\right|}=\frac{1}{\left|R\right|}-\frac{1.33}{1.00}\left(\frac{1}{\left|R\right|}-\frac{1}{\left|2R\right|}\right)\\ q=-3.00\left|R\right|\end{array}$

Thus the image of the fish is $3\left|R\right|$ when the object is at $2\left|R\right|$, Hence for the obj4ect at greater than the radius of curvature the image is farther than the object itself.

The formula to calculate the magnification of the image is,

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

Substitute  $1.33$ for $n_1$ and $1.00$ for $n_2$ $2\left|R\right|$ for $p$ and $-3\left|R\right|$ for $q$

$\begin{array}{c}M=-\frac{n_{1}q}{n_{2}p}\\ M=-\frac{1.33\left(-3.00\left|R\right|\right)}{1.00\left(\left|R\right|\right)}\\ =+2.00\end{array}$

Thus the magnification of the fish image is $+2.00$ if the fish is present at the distance of $2\left|R\right|$ and the image formed is at $3\left|R\right|$.

Conclusion:

Therefore, if the fish is present at distance larger than the radius of curvature the image of the fish would be even farther than that of fish itself.