Chapter 18, Problem 78A

(a)

To determine

The position of the image from the lens.

Answer to Problem 78A

The position of the image from the lens is $60\text{ mm}$ .

Explanation of Solution

Given:

The position of the onion cell is $d_{\text{o}}=12.0\text{ mm}$ .

The focal length is $f=10.0\text{ mm}$

Formula used:

The expression for the mirror equation is,

  $\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}$

Here, $d_{\text{o}}$ is the position of the object, $d_{\text{i}}$ is the position of the image, and $f$ is the focal length.

Calculation:

The position of the image is,

  $\begin{array}{l}\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\ \frac{1}{d_{\text{i}}}=\frac{1}{f}-\frac{1}{d_{\text{o}}}\\ d_{\text{i}}=\frac{f{d}_{\text{o}}}{d_{\text{o}}-f}\\ d_{\text{i}}=\frac{\left(10.0\text{ mm}\right)\left(12.0\text{ mm}\right)}{\left(12.0\text{ mm}\right)-\left(10.0\text{ mm}\right)}\end{array}$

  $d_{\text{i}}=60.0\text{ mm}$

Conclusion:

Thus, the position of the image from the lens is $60.0\text{ mm}$ .

(b)

To determine

The magnification of the image.

Answer to Problem 78A

The magnification of the image is $-5.0$ .

Explanation of Solution

Given:

The position of the onion cell is $d_{\text{o}}=12.0\text{ mm}$ .

The focal length is $f=10.0\text{ mm}$

Formula used:

The expression for the magnification equation is

  $m=\frac{h_{\text{i}}}{h_{\text{o}}}=-\frac{d_{\text{i}}}{d_{\text{o}}}$

Here, $h_{\text{i}}$ is the height of the image and $h_{\text{o}}$ is the height of the object.

Calculation:

From part (a), the position of the image from the lens is $d_{\text{i}}=60.0\text{ mm}$ .

The magnification is,

  $\begin{array}{l}m=-\frac{d_{\text{i}}}{d_{\text{o}}}\\ m=-\frac{30.0\text{ mm}}{6.0\text{ mm}}\\ m=-5.0\end{array}$

Conclusion:

Thus, the magnification of the image is $m=-5.0$ .

(c)

To determine

The position of the final image.

Answer to Problem 78A

The position of the final image is $-20.0\text{ mm}$ .

Explanation of Solution

Given:

The position of the eyepiece is $d_{\text{o}}=10.0\text{ mm}$ .

The focal length of the eyepiece is $f=20.0\text{ mm}$

Formula used:

The expression for the mirror equation is,

  $\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}$

Here, $d_{\text{o}}$ is the position of the object, $d_{\text{i}}$ is the position of the image, and $f$ is the focal length.

Calculation:

The position of the image is,

  $\begin{array}{l}\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\ \frac{1}{d_{\text{i}}}=\frac{1}{f}-\frac{1}{d_{\text{o}}}\\ d_{\text{i}}=\frac{f{d}_{\text{o}}}{d_{\text{o}}-f}\\ d_{\text{i}}=\frac{\left(20.0\text{ mm}\right)\left(10.0\text{ mm}\right)}{\left(10.0\text{ mm}\right)-\left(20.0\text{ mm}\right)}\end{array}$

  $d_{\text{i}}=-20.0\text{ mm}$

Conclusion:

Thus, the position of the final image is $d_{\text{i}}=-20.0\text{ mm}$ .

(d)

To determine

The magnification of the compound system.

Answer to Problem 78A

The magnification of the compound system is $-10.0$ .

Explanation of Solution

Given:

The position of the eyepiece is $d_{\text{o}}=10.0\text{ mm}$ .

The focal length of the eyepiece is $f=20.0\text{ mm}$

Formula used:

The expression for the magnification equation is

  $m=\frac{h_{\text{i}}}{h_{\text{o}}}=-\frac{d_{\text{i}}}{d_{\text{o}}}$

Here, $h_{\text{i}}$ is the height of the image and $h_{\text{o}}$ is the height of the object.

Calculation:

From part (b), the magnification of the image is $m_{\text{o}}=-5.0$ .

From part (c), the position of the final image is $d_{\text{i}}=-20.0\text{ mm}$ .

The magnification of the eyepiece is,

  $\begin{array}{l}{m}_e=-\frac{d_{\text{i}}}{d_{\text{o}}}\\ m_e=-\frac{\left(-20.0\text{ mm}\right)}{10.0\text{ mm}}\\ m_e=2.0\end{array}$

The magnification of the compound system is,

  $\begin{array}{l}{m}_{\text{total}}=m_o{m}_e\\ m_{\text{total}}=\left(-5.0\right)\times \left(2.0\right)\\ m_{\text{total}}=-10.0\end{array}$

Conclusion:

Thus, the magnification of the compound system is $m=-10.0$ .