Chapter 18, Problem 79A

(a)

To determine

The position, height, and orientation of the image as formed by the lens.

To identify: Whether the image is real or virtual.

Answer to Problem 79A

The position of the image is $21\text{ cm}$ .

The height of the image is $-0.494\text{ cm}$ .

The image is a real image and is inverted compared to the object.

Explanation of Solution

Given:

The position of the telescope is $d_{\text{o}}=425\text{ cm}$ .

The height of the telescope is $h_{\text{o}}=10.0\text{ cm}$ .

The focal length is $f=20.0\text{ cm}$

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.

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:

The position of the image is,

  $\begin{array}{l}\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\ \frac{1}{d_{\text{i}}}=\frac{1}{f}-\frac{1}{d_{\text{o}}}\\ d_{\text{i}}=\frac{f{d}_{\text{o}}}{d_{\text{o}}-f}\\ d_{\text{i}}=\frac{\left(20.0\text{ cm}\right)\left(\text{425 cm}\right)}{\left(\text{425 cm}\right)-\left(20.0\text{ cm}\right)}\end{array}$

  $d_{\text{i}}=21.0\text{ cm}$

The height of the image is,

  $\begin{array}{l}m=\frac{h_{\text{i}}}{h_{\text{o}}}=-\frac{d_{\text{i}}}{d_{\text{o}}}\\ h_i=-\frac{d_{\text{i}}}{d_{\text{o}}}{h}_o\\ h_i=-\frac{\left(21.0\text{ cm}\right)}{\left(\text{425 cm}\right)}\times \left(10.0\text{ cm}\right)\\ h_i=-0.494\text{ cm}\end{array}$

Since height is negative, image is inverted compared to the object but it is real.

Conclusion:

Thus, the position of the image is $21.0\text{ cm}$ , the height of the image is $h_{\text{i}}=-0.494\text{ cm}$ .

The image is a real image and is inverted compared to the object.

(b)

To determine

The position, height, and orientation of the image that people sees when looking into the telescope.

To identify: Whether the image is real or virtual.

Answer to Problem 79A

The position of the image is $-3.2\times {10}^2\text{ cm}$ .

The height of the image is $-40\text{ cm}$ .

The image is a virtual image and is inverted compared to the object.

Explanation of Solution

Given:

The position of the telescope is $d_{\text{o}}=425\text{ cm}$ .

The height of the telescope is $h_{\text{o}}=10.0\text{ cm}$ .

The focal length is $f=20.0\text{ cm}$

The focal length of eye piece is $f_{\text{new}}=4.05\text{ cm}$ .

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.

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 is $d_{\text{i}}=21.0\text{ mm}$ .

The new position of the object is,

  $\begin{array}{l}{d}_{\text{o, new}}=25.0\text{ cm}-d_i\\ d_{\text{o, new}}=25.0\text{ cm}-21.0\text{ cm}\\ d_{\text{o, new}}=4.0\text{ cm}\end{array}$

The new height of the object is,

  $\begin{array}{l}{h}_{\text{o, new}}=h_i\\ h_{\text{o, new}}=-0.494\text{ cm}\end{array}$

The position of the image is,

  $\begin{array}{l}\frac{1}{d_{\text{o, new}}}+\frac{1}{d_{\text{i, new}}}=\frac{1}{f_{\text{new}}}\\ \frac{1}{d_{\text{i, new}}}=\frac{1}{f_{\text{new}}}-\frac{1}{d_{\text{o, new}}}\\ d_{\text{i, new}}=\frac{f_{\text{new}}{d}_{\text{o, new}}}{d_{\text{o, new}}-f_{\text{new}}}\\ d_{\text{i, new}}=\frac{\left(\text{4}\text{.05 cm}\right)\left(\text{4}\text{.0 cm}\right)}{\left(\text{4}\text{.0 cm}\right)-\left(4.05\text{ cm}\right)}\end{array}$

  $d_{\text{i, new}}=-3.2\times {10}^2\text{ cm}$

The height of the image is,

  $\begin{array}{l}m=\frac{h_{\text{i}}}{h_{\text{o}}}=-\frac{d_{\text{i}}}{d_{\text{o}}}\\ h_{\text{i, new}}=-\frac{d_{\text{i, new}}}{d_{\text{o, new}}}{h}_{\text{o, new}}\\ h_{\text{i, new}}=-\frac{\left(-3.2\times {10}^2\text{ cm}\right)}{\left(\text{4}\text{.0 cm}\right)}\times \left(-0.494\text{ cm}\right)\\ h_{\text{i, new}}=-40\text{ cm}\end{array}$

Hence, the image is a virtual image and is inverted compared to the object.

Conclusion:

Thus, the position of the image is $-3.2\times {10}^2\text{ cm}$ , the height of the image is $-40\text{ cm}$ .

The image is a virtual image and is inverted compared to the object.

(c)

To determine

The magnification of the telescope.

Answer to Problem 79A

The magnification of the telescope is $-4.0$ .

Explanation of Solution

Given:

The position of the telescope is $d_{\text{o}}=425\text{ cm}$ .

The height of the telescope is $h_{\text{o}}=10.0\text{ cm}$ .

The focal length is $f=20.0\text{ cm}$

The focal length of eye piece is $f_{\text{new}}=4.05\text{ cm}$ .

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 new height of the image is $h_{\text{i, new}}=-40\text{ cm}$ .

The magnification of the telescope is,

  $\begin{array}{l}m=\frac{h_{\text{i}}}{h_{\text{o}}}=-\frac{d_{\text{i}}}{d_{\text{o}}}\\ m=\frac{h_{\text{i, new}}}{h_{\text{o}}}\\ m=\frac{\left(-40\text{ cm}\right)}{\left(\text{10}\text{.0 cm}\right)}\\ m=-4.0\end{array}$

Conclusion:

Thus, the magnification of the telescope is $m=-4.0$ .