Chapter 17.2, Problem 18PP

To determine

The image position using a scale diagram and the mirror equation.

Answer to Problem 18PP

The position of the image is $-8.57\text{ cm}$ .

Explanation of Solution

Given:

The position of the object is $d_{\text{o}}=20.0\text{cm}$ .

The focal length is $f=-15.0\text{ cm}$

Formula used:

The expression for the mirror equation is as follow

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

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 procedure to draw the ray diagram is,

• Sketch the principal axis of the mirror as a horizontal line.
• Place the mirror at the center, place C at the left side of the mirror and place F at the other side.
• Sketch a vertical line at the mirror point to represent the mirror. This is the principal plane.
• Sketch the object as an arrow and label its top $O_1$ .
• Draw ray 1, the parallel ray. It is parallel to the principal axis and reflects the principal plane and passes through F .
• Draw ray 2, the focus ray. It passes through F , reflects the principal plane, and is reflected parallel to the principal axis.
• The image is located where rays 1 and 2 cross after reflection. Label the point as $I_1$ .

Sketch the image position using a scale diagram as shown below.

  

The vertical scale of 1 block is $1.0\text{cm}$ .

The horizontal scale of 1 block is $1.558\text{ cm}$ .

The horizontal distance $\left(d_{\text{i}}\right)$ of the image from the mirror is,

  $\begin{array}{c}{d}_{\text{i}}=\left(\begin{array}{l}\text{Number}\text{of}\text{blocks }\\ \text{from the mirror in }\\ \text{horizontal direction}\end{array}\right)\text{×}\left(\text{Horizontal}\text{Scale}\right)\\ =-5.5\text{blocks}\times 1.\text{558}\text{cm}\\ =-\text{8}\text{.57}\text{cm}\end{array}$

The position of the image using mirror equation is,

  $\begin{array}{l}\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\ \frac{1}{d_{\text{i}}}=\frac{1}{f}-\frac{1}{d_{\text{o}}}\\ d_{\text{i}}=\frac{f{d}_{\text{o}}}{d_{\text{o}}-f}\\ d_{\text{i}}=\frac{\left(-15.0\text{ cm}\right)\left(20.0\text{ cm}\right)}{\left(20.0\text{ cm}\right)-\left(-15.0\text{ cm}\right)}\end{array}$

  $d_{\text{i}}=-8.57\text{ cm}$

Conclusion:

Thus, the position of the image is $-8.57\text{ cm}$ .