Chapter 35, Problem 50AP

(a)

To determine

The position of the objects image $x^{\prime }$ as a function of the objects position $x$.

Answer to Problem 50AP

The position of the objects image $x^{\prime }$ as a function of the objects position $x$ is $x^{\prime }=\frac{1024-58.0x}{6.0-x}$.

Explanation of Solution

Given info: The initial position of the object is $0\text{cm}$, the focal length of the converging lens is $26.0\text{cm}$, the position of the lens is $32.0\text{cm}$ and the final position of the object is $12.0\text{cm}$.

The formula to calculate the focal length is,

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

Here,

$p$ is the position of the object.

$q$ is the position of the image.

Consider the position of the image is $x$ as the object moves and the position of the lens is $32.0\text{cm}$. So, the position of the object as any instant is,

$p=32\text{cm}-x$

Substitute $32\text{cm}-x$ for $p$ and $26.0\text{cm}$ for $f$ in above equation to find the value of $q$.

$\begin{array}{c}\frac{1}{26.0\text{cm}}=\frac{1}{32\text{cm}-x}+\frac{1}{q}\\ q=\frac{832-26x}{6.0-x}\end{array}$

Thus, the position of the image for the minimum magnification is $\frac{832-26x}{6.0-x}$.

The formula to calculate the image position with respect to the object position is,

$x^{\prime }=32.0+q$

Substitute $\frac{832-26x}{6.0-x}$ for $q$ in above equation to find the value of $x^{\prime }$.

$\begin{array}{c}{x}^{\prime }=32.0+\frac{832-26x}{6.0-x}\\ x^{\prime }=\frac{1024-58.0x}{6.0-x}\end{array}$

Conclusion

Therefore, the position of the objects image $x^{\prime }$ as a function of the objects position $x$ is $x^{\prime }=\frac{1024-58.0x}{6.0-x}$.

(b)

To determine

The pattern of the images motion with reference to a table of values.

Answer to Problem 50AP

The pattern of the images motion with reference to a table of values is shown below.

$x$ Object position $\left(\text{cm}\right)$	$x^{\prime }$ Image position $\left(\text{cm}\right)$
$0$	$170.7$
$1$	$193.2$
$2$	$227.0$
$3$	$283.3$
$4$	$396.0$
$5$	$734.0$
$6$	$\infty$
$7$	$-618.0$
$8$	$-280.0$
$9$	$-167.3$
$10$	$-111.0$
$11$	$-77.2$
$12$	$-54.7$

Explanation of Solution

Given info: The initial position of the object is $0\text{cm}$, the focal length of the converging lens is $26.0\text{cm}$, the position of the lens is $32.0\text{cm}$ and the final position of the object is $12.0\text{cm}$.

The formula to calculate the image distance is,

$x^{\prime }=\frac{1024-58.0x}{6.0-x}$

Taking the integer value of the position of the object as the integer values between $0$ to $12$ the corresponding values of the position of the image is provided in the table shown below.

$x$ Object position $\left(\text{cm}\right)$	$x^{\prime }$ Image position $\left(\text{cm}\right)$
$0$	$170.7$
$1$	$193.2$
$2$	$227.0$
$3$	$283.3$
$4$	$396.0$
$5$	$734.0$
$6$	$\infty$
$7$	$-618.0$
$8$	$-280.0$
$9$	$-167.3$
$10$	$-111.0$
$11$	$-77.2$
$12$	$-54.7$

Conclusion

Therefore, the pattern of the images motion with reference to a table of values is shown below.

$x$ Object position $\left(\text{cm}\right)$	$x^{\prime }$ Image position $\left(\text{cm}\right)$
$0$	$170.7$
$1$	$193.2$
$2$	$227.0$
$3$	$283.3$
$4$	$396.0$
$5$	$734.0$
$6$	$\infty$
$7$	$-618.0$
$8$	$-280.0$
$9$	$-167.3$
$10$	$-111.0$
$11$	$-77.2$
$12$	$-54.7$

(c)

To determine

The distance that the image moves when the object moves $12.0\text{cm}$ to the right.

Answer to Problem 50AP

The image moves from infinity to beyond when the object moves $12.0\text{cm}$ to the right.

Explanation of Solution

Given info: The initial position of the object is $0\text{cm}$, the focal length of the converging lens is $26.0\text{cm}$, the position of the lens is $32.0\text{cm}$ and the final position of the object is $12.0\text{cm}$.

The formula to calculate the image distance is,

$x^{\prime }=\frac{1024-58.0x}{6.0-x}$

Substitute $12$ for $x$ in above equation to find the value of $x^{\prime }$.

$\begin{array}{c}{x}^{\prime }=\frac{1024-58.0\left(12\right)}{6.0-\left(12\right)}\\ =-54.7\end{array}$

The image position at $x$ at zero is $170.7$ and at $x$ equal $12$ the position of the image is $-54.7$. So the image moves first in the in the forward direction to infinity in the right and then jumps back to minus infinity on the left and then proceeds again in the forward direction.

Conclusion

Therefore, the image moves from infinity to beyond when the object moves $12.0\text{cm}$ to the right.

(d)

To determine

The direction of the image when the object moves $12.0\text{cm}$ to the right.

Answer to Problem 50AP

The direction of the image movement is right but is opposite during the jump when the object moves $12.0\text{cm}$ to the right.

Explanation of Solution

Given info: The initial position of the object is $0\text{cm}$, the focal length of the converging lens is $26.0\text{cm}$, the position of the lens is $32.0\text{cm}$ and the final position of the object is $12.0\text{cm}$.

The formula to calculate the image distance is,

$x^{\prime }=\frac{1024-58.0x}{6.0-x}$

The direction of the movement of the image is always right but the direction is left during the time when the image jumps to a negative infinite value form the positive infinite value. The image first moves in the positive $x$ direction but between the value of $6\text{cm}$ and $7\text{cm}$ position of the object the image jumps in the negative infinite direction.

Conclusion

Therefore, the direction of the image movement is right but is opposite during the jump when the object moves $12.0\text{cm}$ to the right.