Chapter 18, Problem 105A

To determine

The position and size of the image.

To identify: Whether the orientation of the image is changed or not.

Answer to Problem 105A

The position of the new image is $30.0\text{ cm}$ .

The size of the new image is $12.0\text{ cm}$ .

The orientation of the image remainsunchanged.

Explanation of Solution

Given:

The position of the image is $x_{\text{i}}=25.0\text{ cm}$ .

The position of the image is $h_{\text{i}}=4.0\text{ cm}$ .

The focal length of the lens is $f=-15.0\text{ cm}$ .

The position of the new image is $x_{\text{o}}=-10.0\text{ cm}$ .

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 as follows:

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

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

Calculation:

Consider the position of original image as $h_{\text{o}}=4.0\text{ cm}$ . The position of the new image is,

  $\begin{array}{l}\frac{1}{x_{\text{o}}}+\frac{1}{x_{\text{i}}}=\frac{1}{f}\\ \frac{1}{x_{\text{i}}}=\frac{1}{f}-\frac{1}{x_{\text{o}}}\\ x_{\text{i}}=\frac{f{x}_{\text{o}}}{x_{\text{o}}-f}\\ x_{\text{i}}=\frac{\left(-15.0\text{ cm}\right)\left(-10.0\text{ cm}\right)}{\left(-10.0\text{ cm}\right)-\left(-15.0\text{ cm}\right)}\end{array}$

  $x_{\text{i}}=30.0\text{ cm}$

The height of the new image is calculated using the magnification formula.

  $\begin{array}{l}m=\frac{h_{\text{i}}}{h_{\text{o}}}=-\frac{d_{\text{i}}}{d_{\text{o}}}\\ h_{\text{i}}=-\frac{x_{\text{i}}{h}_{\text{o}}}{x_{\text{o}}}\\ h_{\text{i}}=-\frac{\left(30.0\text{ cm}\right)\left(\text{4}\text{.0 cm}\right)}{\left(-10.0\text{ cm}\right)}\\ h_{\text{i}}=12.0\text{ cm}\end{array}$

From the above calculations, it is observed that the orientation of the image is not changed.

Conclusion:

Thus, the position of the new image is $30.0\text{ cm}$ and the size of the new image is $12.0\text{ cm}$ .

Also, the orientation of the image is not changed.