Chapter 17, Problem 84A

To determine

To draw:A ray diagram that will show the position of the image and height of the image.

Answer to Problem 84A

A ray diagram is shown in Figure 1.

The position of the image is $14.4\text{ m}$ .

The height of the image is $\text{2}\text{.4 m}\left(\text{inverted}\right)$ .

Explanation of Solution

Given:

The height of the object is $h_{\text{o}}=3.0\text{ cm}$ .

The position of the object is $d_{\text{o}}=18.0\text{ cm}$ .

The focal length is $f=8.0\text{ cm}$

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 procedure to draw the ray diagram is

• Sketch the principal axis of the mirror as a horizontal line.
• Place the mirror at the right side, place C at the center and F halfway between the mirror and C .
• 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 $I_1$ .

The ray diagram as shown below.

  

Figure 1

The horizontal scale is 1 block which is equal to $1.0\text{ m}$ .

The vertical scale is 1 block whichis equal to $0.50\text{ m}$ .

Measure the horizontal distance of the object from the mirror as per the horizontal scale as $14.4\text{block}\times \text{1}\text{m}=14.4\text{ m}$ .Measure the height of the image of the object as per the vertical scale as $4.8\text{block}\times 0.50\text{m}=-2.4\text{ m}$

Negative sign is included since the image is inverted.

Conclusion:

Therefore, a ray diagram is shown in Figure 1. The position of the image is $14.4\text{ m}$ and the height of the image is $\text{2}\text{.4 m}\left(\text{inverted}\right)$ .