Chapter 35, Problem 30P

In Figure P35.30, a thin converging lens of focal length 14.0 cm forms an image of the square abed, which is h_e = h_b = 10.0 cm high and lies between distances of p_d = 20.0 cm and p_a = 30.0 cm from the lens. Let a′, b′, c′. and d′ represent the respective corners of the image. Let q_a represent the image distance for points a′ and b′, q_d represent the image distance for points c′ and d′, ${h^{\prime }}_b$, represent the distance from point b′ to the axis, and ${h^{\prime }}_c$ represent the height of c′. (a) Find q_a, q_d, ${h^{\prime }}_b$, and ${h^{\prime }}_c$. (b) Make a sketch of the image. (c) The area of the object is 100 cm². By carrying out the following steps, you will evaluate the area of the image. Let q represent the image distance of any point between a′ and d′, for which the object distance is p. Let h′ represent the distance from the axis to the point at the edge of the image between b′ and c′ at image distance q. Demonstrate that

$\left|h^{\prime }\right|=10.0q\left(\frac{1}{14.0}-\frac{1}{q}\right)$

where h′ and q are in centimeters. (d) Explain why the geometric area of the image is given by

${\displaystyle {\int }_{q_a}^{q_d}\left|h^{\prime }\right|}dq$

(e) Carry out the integration to find the area of the image.

Figure P35.30

To determine

The position of the given constants. $q_a$, $q_d$, ${h^{\prime }}_b$ and ${h^{\prime }}_c$.

Answer to Problem 30P

The values of $q_a$, $q_d$ ${h^{\prime }}_b$ and ${h^{\prime }}_c$ is $26.3\text{cm}$, $46.7\text{cm}$, $8.75\text{cm}$ and $23.3\text{cm}$ respectively

Explanation of Solution

Given info: The focal length of the lens is $14.0\text{cm}$. The object is square $abcd$ of height $10.0\text{cm}$ and the positioned between $20\text{cm}$ and $30.0\text{cm}$.

Formula to calculate the image of any object in a thin lens is,

$\frac{1}{f}=\frac{1}{q}+\frac{1}{p}$ (1)

Here,

$f$ is the focal length of the lens.

$p$ is the object position.

$q$ is the image position.

Substitute $q_a$ for $q$, $p_a$ for $p$ in the equation (1),

$\frac{1}{f}=\frac{1}{p_a}+\frac{1}{q_a}$ (2)

Here,

$p_a$ is the object distance corresponding to $ab$.

$q_a$ is the image distance corresponding to $ab$.

Substitute $30.0\text{cm}$ for $p_a$ and $14.0\text{cm}$ for $f$ in equation (2)

$\begin{array}{c}\frac{1}{f}=\frac{1}{p_a}+\frac{1}{q_a}\\ \frac{1}{14.0\text{cm}}=\frac{1}{30.0\text{cm}}+\frac{1}{q_a}\\ q_a=26.3\text{cm}\end{array}$

Substitute $p_d$ for $p$ and $q_d$ for $q$ in the equation (1),

$\frac{1}{f}=\frac{1}{q_d}+\frac{1}{p_d}$ (3)

Here,

$q_d$ is the image distance corresponding to the $cd$.

$p_d$ is the object position corresponding to $cd$.

Substitute, $20.0\text{cm}$ for $p_d$ and $14.0\text{cm}$ for $f$ in equation (3),

$\begin{array}{c}\frac{1}{f}=\frac{1}{q_d}+\frac{1}{p_d}\\ \frac{1}{14.0\text{cm}}=\frac{1}{q_d}+\frac{1}{20.0\text{cm}}\\ p_d=46.7\text{cm}\end{array}$

Formula to calculate the height of the corresponding image point $ab$

${h^{\prime }}_b=h_b\left(\frac{-q_a}{p_a}\right)$ (4)

Here,

$h_b{}^{\prime }$ is to the height of image corresponding to $ab$

$h_b$ is the height of the object corresponding to $ab$.

Substitute $10.0\text{cm}$ for $h_b$, $26.3\text{cm}$ for $q_a$ and $30.0\text{cm}$ in equation (4).

$\begin{array}{c}{h^{\prime }}_b=h_b\left(\frac{-q_a}{p_a}\right)\\ {h^{\prime }}_b=10.0\text{cm}\left(\frac{-26.3\text{cm}}{30.0\text{cm}}\right)\\ {h^{\prime }}_b=-8.75\text{cm}\end{array}$

 For calculating the image height corresponding to $cd$,

${h^{\prime }}_c=h_c\left(\frac{-q_d}{p_d}\right)$ (5)

Substitute $10.0\text{cm}$ for $h_d$, $46.7\text{cm}$ for $q_a$ and $20.0\text{cm}$ in equation (5).

$\begin{array}{c}{h^{\prime }}_c=h_c\left(\frac{-q_d}{p_d}\right)\\ {h^{\prime }}_c=10.0\text{cm}\left(\frac{-46.7\text{cm}}{20.0\text{cm}}\right)\\ {h^{\prime }}_c=-23.3\text{cm}\end{array}$

Conclusion:

Therefore, the values of $q_a$, $q_d$ ${h^{\prime }}_b$ and ${h^{\prime }}_c$ is $26.3\text{cm}$, $46.7\text{cm}$, $8.75\text{cm}$ and $23.3\text{cm}$ respectively.

To determine

To draw: The sketch of the ray diagram.

Answer to Problem 30P

 The sketch of the ray diagram is,

Explanation of Solution

The image of the square $abcd$ will be a trapezoid of $a^{\prime }{b}^{\prime }{c}^{\prime }{d}^{\prime }$.

Figure (1)

To determine

To show: The relation, $\left|h^{\prime }\right|=10q\left(\frac{1}{14.0\text{cm}}-\frac{1}{q}\right)$.

Answer to Problem 30P

The relation between height of the image and the image distance is $\left|h^{\prime }\right|=10q\left(\frac{1}{14.0\text{cm}}-\frac{1}{q}\right)$.

Explanation of Solution

the focal length of the given lens is $14.0\text{cm}$. The formula to calculate the magnification is

$-\frac{h^{\prime }}{h}=\frac{q}{p}$ (6)

Here,

$h^{\prime }$ is the image height.

$h$ is the object height.

$q$ is the image distance.

$p$ is the object distance.

From the lens,

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$ (7)

Here,

$f$ is the focal length of the lens.

Substitute $14.0\text{cm}$ for $f$ in equation (7).

$\begin{array}{l}\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\ \frac{1}{14.0\text{cm}}=\frac{1}{p}+\frac{1}{q}\\ \frac{1}{p}=\frac{1}{14.0\text{cm}}-\frac{1}{q}\end{array}$ (8)

Substitute $\frac{1}{14.0\text{cm}}-\frac{1}{q}$ for $p$ and $10.0\text{cm}$ for $h$ in equation (6),

$\begin{array}{l}-\frac{h^{\prime }}{h}=\frac{q}{p}\\ -\frac{h^{\prime }}{h}=q\left(\frac{1}{p}\right)\\ h^{\prime }=-10\left(\frac{1}{14.0\text{cm}}-\frac{1}{q}\right)q\\ \left|h^{\prime }\right|=10q\left(\frac{1}{14.0\text{cm}}-\frac{1}{q}\right)\end{array}$ (9)

Conclusion:

Therefore the relation between height of the image and the image distance is $\left|h^{\prime }\right|=10q\left(\frac{1}{14.0\text{cm}}-\frac{1}{q}\right)$.

To determine

To write: The explanation that the geometric area of image is ${\displaystyle {\int }_{q_a}^{q_d}\left|h^{\prime }\right|dq}$

Answer to Problem 30P

The geometric area of a image is explained by the integral ${\displaystyle {\int }_{q_a}^{q_d}\left|h^{\prime }\right|dq}$.

Explanation of Solution

Given info: The geometric area of a image is,

$\text{Area}={\displaystyle {\int }_{q_a}^{q_d}\left|h^{\prime }\right|dq}$. (10)

Here,

$h^{\prime }$ is the height of the image.

$dq$ is the smaller region of image distance.

From equation (10) the integral sums up the small areas of region covered by the image itself. The height of the small regions is $h^{\prime }$ and the small portion of image distance is $dq$.

Therefore area of that small region is

$dA=h^{\prime }dq$

Therefore the integration from $q_a$  to $q_d$ of $dA$ will give the geometric area of the image.

Conclusion:

Therefore, the geometric area of the image is given by integral ${\displaystyle {\int }_{q_a}^{q_d}\left|h^{\prime }\right|dq}$.

To determine

The geometric area of the image.

Answer to Problem 30P

The geometric area is $328{\text{cm}}^2$.

Explanation of Solution

Given info: The geometric area of the image is given by the integral,

${\displaystyle {\int }_{q_a}^{q_d}\left|h^{\prime }\right|dq}$ . (11)

From equation (9) substitute $10q\left(\frac{1}{14.0\text{cm}}-\frac{1}{q}\right)$ for $h^{\prime }$ and $26.3\text{cm}$ for $q_a$, $46.7\text{cm}$ for $q_d$ in equation (11)

$\begin{array}{l}{\displaystyle {\int }_{q_a}^{q_d}\left|h^{\prime }\right|dq}\\ {\displaystyle \underset{26.3cm}{\overset{46.7\text{cm}}{\int }}10q\left(\frac{1}{14.0\text{cm}}-\frac{1}{q}\right)}dq\end{array}$

Integrate the above equation with respect to $dq$ to get the area ,

$\begin{array}{c}Area={\displaystyle \underset{26.3cm}{\overset{46.7\text{cm}}{\int }}10q\left(\frac{1}{14.0\text{cm}}-\frac{1}{q}\right)}dq\\ =10.0\text{cm}{\left[\frac{q^2}{28.0\text{cm}}-q\right]}_{26.3\text{cm}}^{46.7\text{cm}}\\ =10.0\text{cm}\left[\frac{\left[{\left(46.7\text{cm}\right)}^2-26.3{\text{cm}}^2\right]}{28.0\text{cm}}-46.7\text{cm+26}\text{.3}\text{cm}\right]\\ =328{\text{cm}}^2\end{array}$

Conclusion:

Therefore, the geometric area of the image is $328{\text{cm}}^2$ .