Chapter 24, Problem 10E

(a)

To determine

The location of the image graphically.

Answer to Problem 10E

The location of the image is shown in the figure below.

Explanation of Solution

Write the expression for the focal length of the convex lens.

  $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

Here, $f$ is the focal length of the lens, $v$ is the image distance and $u$ is the object distance.

Rearrange the above expression in terms of $1/v$.

  $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$        (I)

Conclusion:

Substitute $0.2\text{m}$ for $f$ and $-0.08\text{m}$ for $u$ in equation (I).

  $\begin{array}{c}\frac{1}{v}=\frac{1}{0.2\text{m}}-\frac{1}{0.08\text{m}}\\ =\left(\frac{20-50}{4}\right)\text{m}\\ =-\left(\frac{30}{4}\right)\text{m}\end{array}$

Simplify the above expression in terms of $v$.

  $\begin{array}{c}v=-\left(\frac{2}{15}\right)\text{m}\\ =-0.133\text{m}\end{array}$

The ray diagram for the required condition is drawn below.

Here, the image distance from the lens is $-0.133\text{m}$, virtual and on the same side as the object.

Thus, the location of the image formed by the lens is shown above.

(b)

To determine

The location of the image algebraically.

Answer to Problem 10E

The location of the image is $-0.133\text{m}$, virtual and on the same sideas the object.

Explanation of Solution

Write the expression for the focal length of the convex lens.

  $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

Here, $f$ is the focal length of the lens, $v$ is the image distance and $u$ is the object distance.

Rearrange the above expression for the image distance.

  $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$        (I)

Conclusion:

Substitute $0.2\text{m}$ for $f$ and $-0.08\text{m}$ for $u$ in equation (I).

  $\begin{array}{c}\frac{1}{v}=\frac{1}{0.2\text{m}}-\frac{1}{0.08\text{m}}\\ =\left(\frac{20-50}{4}\right)\text{m}\\ =-\left(\frac{30}{4}\right)\text{m}\end{array}$

Simplify the above expression in terms of $v$.

  $\begin{array}{c}v=-\left(\frac{2}{15}\right)\text{m}\\ =-0.133\text{m}\end{array}$

Thus, the location of the image is $-0.133\text{m}$, virtual and on the same sideas the object.

(c)

To determine

The magnification of the lens.

Answer to Problem 10E

The magnification of the lens is $+1.66$.

Explanation of Solution

Write the expression for the focal length of the convex lens.

  $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

Here, $f$ is the focal length of the lens, $v$ is the image distance and $u$ is the object distance.

Rearrange the above expression for the image distance.

  $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$        (I)

Write the expression for the magnification of the lens.

  $m=\frac{v}{u}$        (II)

Here, $m$ is the magnification of the lens.

Conclusion:

Substitute $0.2\text{m}$ for $f$ and $-0.08\text{m}$ for $u$ in equation (I).

  $\begin{array}{c}\frac{1}{v}=\frac{1}{0.2\text{m}}-\frac{1}{0.08\text{m}}\\ =\left(\frac{20-50}{4}\right)\text{m}\\ =-\left(\frac{30}{4}\right)\text{m}\end{array}$

Simplify the above expression in terms of $v$.

  $\begin{array}{c}v=-\left(\frac{2}{15}\right)\text{m}\\ =-0.133\text{m}\end{array}$

Substitute $-0.133\text{m}$ for $v$ and $-0.08\text{m}$ for $u$ in equation (II).

  $\begin{array}{c}m=\frac{\left(-0.133\text{m}\right)}{\left(-0.08\text{m}\right)}\\ =+1.66\end{array}$

Thus, the magnification of the lens is $+1.66$.