Chapter 36, Problem 50P

(a)

To determine

The value of $q_a$, $q_d$, ${h^{\prime }}_b$ and ${h^{\prime }}_c$.

Answer to Problem 50P

The value of $q_a$, $q_d$, ${h^{\prime }}_b$ and ${h^{\prime }}_c$ are $\underset{\_}{26.3\text{cm}}$, $\underset{\_}{46.7\text{cm}}$, $\underset{\_}{-8.75\text{cm}}$ and $\underset{\_}{-23.3\text{cm}}$ respectively.

Explanation of Solution

Write the expression for the relation between the object distance, image distance and focal length.

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

Here, $p$ is the object distance, $q$ is the image distance and $f$ is the focal length.

Write the expression for magnification.

    $M=\frac{-q}{p}$

Here, $M$ is the magnification.

Write the expression for the relation between height and magnification.

    $h^{\prime }=hM$

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

Substitute $\frac{-q}{p}$ for $M$ in the above equation.

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

Conclusion:

Substitute $30.0\text{cm}$ for $p$, $q_a$ for $q$ and $14.0\text{cm}$ for $f$ in equation (I).

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

Substitute $20.0\text{cm}$ for $p$, $q_d$ for $q$ and $14.0\text{cm}$ for $f$ in equation (I).

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

Substitute $30.0\text{cm}$ for $p$, $26.3\text{cm}$ for $q$ and $10.0\text{cm}$ for $h$ in equation (II).

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

Substitute $20.0\text{cm}$ for $p$, $46.7\text{cm}$ for $q$ and $10.0\text{cm}$ for $h$ in equation (II).

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

Therefore, the value of $q_a$, $q_d$, ${h^{\prime }}_b$ and ${h^{\prime }}_c$ are $\underset{\_}{26.3\text{cm}}$, $\underset{\_}{46.7\text{cm}}$, $\underset{\_}{-8.75\text{cm}}$ and $\underset{\_}{-23.3\text{cm}}$ respectively.

(b)

To determine

The sketch of the image.

Answer to Problem 50P

The sketch of the image is given in figure 1.

Explanation of Solution

The following figure gives the diagrammatic representation of the image formed by the thin lens.

Conclusion:

Therefore, the sketch of the image is given in figure 1.

(c)

To determine

To demonstrate that the area is $\left|h^{\prime }\right|=\left(10.0\text{cm}\right)q\left(\frac{1}{14\text{cm}}-\frac{1}{q}\right)$.

Answer to Problem 50P

It is demonstrated that the area is $\left|h^{\prime }\right|=\left(10.0\text{cm}\right)q\left(\frac{1}{14\text{cm}}-\frac{1}{q}\right)$.

Explanation of Solution

Substitute $14\text{cm}$ for $f$ in equation (I) and rearrange it to find $\frac{1}{p}$.

    $\frac{1}{p}=\frac{1}{14\text{cm}}-\frac{1}{q}$

Conclusion:

Substitute $\frac{1}{14\text{cm}}-\frac{1}{q}$ for $\frac{1}{p}$ and $10.0\text{cm}$ for $h$ in equation (II).

    $h^{\prime }=\left(10.0\text{cm}\right)\left(-q\right)\left(\frac{1}{14\text{cm}}-\frac{1}{q}\right)$

Write the expression for the modulus of the above equation.

    $\left|h^{\prime }\right|=\left(10.0\text{cm}\right)q\left(\frac{1}{14\text{cm}}-\frac{1}{q}\right)$

Therefore, it is demonstrated that the area is $\left|h^{\prime }\right|=\left(10.0\text{cm}\right)q\left(\frac{1}{14\text{cm}}-\frac{1}{q}\right)$.

(d)

To determine

The reason for the area to be ${\displaystyle {\int }_{q_a}^{q_d}\left|h^{\prime }\right|dq}$.

Answer to Problem 50P

The integral will add up the area of the whole image.

Explanation of Solution

The vertical dimension of the image is given by $\left|h^{\prime }\right|$ and the horizontal width of the image is $dq$. This dimensions whose product when integrated will give area of the whole image.

Therefore, the integral will add up the area of the whole image.

(e)

To determine

The area of the image.

Answer to Problem 50P

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

Explanation of Solution

Write the expression for the area of the image.

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

Conclusion:

Substitute $\left(10.0\text{cm}\right)q\left(\frac{1}{14\text{cm}}-\frac{1}{q}\right)$ for $\left|h^{\prime }\right|$ in the above equation and integrate within the limit $26.3\text{cm}\to 46.7\text{cm}$.

    $\begin{array}{c}A={\displaystyle {\int }_{26.3}^{46.7}\left(10.0\text{cm}\right)q\left(\frac{1}{14\text{cm}}-\frac{1}{q}\right)dq}\\ =328{\text{cm}}^2\end{array}$

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