Chapter 26, Problem 38P

In Figure P26.38, a thin converging lens of focal length 14.0 cm forms an image of the square abcd, which is h_c = 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′_b represent the distance from point b′ to the axis, and h′_c represent the height of c′. (a) Find q_a, q_d, h′_b, and h′_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 P26.38

To determine

The values of given parameters.

Answer to Problem 38P

The values of the given parameters are

    $\begin{array}{c}{q}_a=26.3\text{cm}\\ q_d=46.7\text{cm}\\ h_b{}^{\prime }=-8.75\text{cm}\\ h_c{}^{\prime }=-23.3\text{cm}\end{array}$

Explanation of Solution

Write the mirror equation for the side a.

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

Here $f$ is the focal length, $p_a$ is the object distance and $q_a$ is the image distance.

Rewrite (I) in terms of $q_a$.

    $q_a=\frac{p_{a}f}{p_a-f}$        (II)

Similarly, write the image distance for side d.

    $q_d=\frac{p_{d}f}{p_d-f}$        (III)

Write the equation for magnification for side b

    $\begin{array}{c}M=\frac{{h^{\prime }}_b}{h}\\ =\frac{-q_a}{p_a}\end{array}$        (IV)

Here ${h^{\prime }}_b$ is the image height, $h$ is the object height.

Rewrite (IV) in terms of $h_b{}^{\prime }$.

    ${h^{\prime }}_b=\frac{-q_a}{p_a}h$        (V)

Similarly,

    ${h^{\prime }}_b=\frac{-q_d}{p_d}h$        (VI)

Conclusion:

Substitute $30\text{cm}$ for $p_a$, $14\text{cm}$ for $f$ in (II)

    $\begin{array}{c}{q}_a=\frac{\left(14\text{cm}\right)\left(30\text{cm}\right)}{30\text{cm}-14\text{cm}}\\ =26.3\text{cm}\end{array}$

Substitute $14\text{cm}$ for $f$ and $20\text{cm}$ for $p_d$ in (III)

    $\begin{array}{c}{q}_d=\frac{\left(14\text{cm}\right)\left(20\text{cm}\right)}{20\text{cm}-14\text{cm}}\\ =46.7\text{cm}\end{array}$

Substitute $10\text{cm}$ for $h$, $-26.3\text{cm}$ for $q_a$ and $30\text{cm}$ for $p_a$ in (V)

    $\begin{array}{c}{h^{\prime }}_b=\frac{-\left(-26.3\text{cm}\right)}{\left(30\text{cm}\right)}\left(10\text{cm}\right)\\ =-8.75\text{cm}\end{array}$

Substitute $-46.7\text{cm}$ for $q_d$, $10\text{cm}$ for $h$ and $20\text{cm}$ for $p_d$ in(VI)

    $\begin{array}{c}{h^{\prime }}_b=\frac{-\left(46.7\text{cm}\right)}{\left(20\text{cm}\right)}\left(10\text{cm}\right)\\ =-23.3\text{cm}\end{array}$

The values of the given parameters are

    $\begin{array}{c}{q}_a=26.3\text{cm}\\ q_d=46.7\text{cm}\\ h_b{}^{\prime }=-8.75\text{cm}\\ h_c{}^{\prime }=-23.3\text{cm}\end{array}$

To determine

Sketch of the image.

Answer to Problem 38P

The image has been drawn

Explanation of Solution

Conclusion:

The image has been drawn

To determine

Prove the given equation.

Answer to Problem 38P

The equation has been proved.

Explanation of Solution

Write the equation for magnification

    $\begin{array}{c}M=\frac{h^{\prime }}{h}\\ =\frac{-q}{p}\end{array}$        (VII)

Rewrite (VII) in terms of $h^{\prime }$.

    $h^{\prime }=h\left(\frac{-q}{p}\right)$        (VIII)

Write the mirror equation in terms of $p$

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

Substitute (IX) in (VIII)

    $h^{\prime }=h\left(-q\right)\left(\frac{1}{f}-\frac{1}{q}\right)$        (X)

Conclusion:

Substitute $14\text{cm}$ for $f$ and $10\text{cm}$ for $h$ in (X)

    $h^{\prime }=\left(10\text{cm}\right)q\left(\frac{1}{14\text{cm}}-\frac{1}{q}\right)$        (XI)

Hence proved.

To determine

Geometric area of the image.

Answer to Problem 38P

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

Explanation of Solution

The integral sign suggests that the areas of the small regions are added up to get the whole area. The $h^{\prime }$ indicates the vertical dimension of the image. And $dq$ is the horizontal width of the image.

Conclusion:

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

To determine

Area of the image.

Answer to Problem 38P

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

Explanation of Solution

Write the equation for area

    $A={\displaystyle {\int }_{q_a}^{q_d}\left|h^{\prime }\right|}dq$        (XII)

Substitute (XI) in (XII)

    $\begin{array}{c}A={\displaystyle {\int }_{q_a}^{q_d}10\text{cm}\left(q\right)\left(\frac{1}{14}-\frac{1}{q}\right)}dq\\ =10{\left[\frac{q^2}{28\text{cm}}-q\right]}_{26.3\text{cm}}^{46.7\text{cm}}\\ =328{\text{cm}}^{\text{2}}\end{array}$

Conclusion:

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