Chapter 18, Problem 62A

To determine

To draw: A ray diagram for the location of the image.

Answer to Problem 62A

A ray diagram for the location of image is shown in Figure 1.

The location of the image is $34\text{ cm}$ .

Explanation of Solution

Given:

The position of the candle is $d_{\text{o}}=34.0\text{ cm}$ .

The focal length is $f=17.0\text{ cm}$

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 using mirror equation as follows:

  $\begin{array}{l}\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\ \frac{1}{d_{\text{i}}}=\frac{1}{f}-\frac{1}{d_{\text{o}}}\\ d_{\text{i}}=\frac{f{d}_{\text{o}}}{d_{\text{o}}-f}\\ d_{\text{i}}=\frac{\left(17.0\text{ cm}\right)\left(34.0\text{ cm}\right)}{\left(34.0\text{ cm}\right)-\left(17.0\text{ cm}\right)}\end{array}$

  $d_{\text{i}}=34\text{ cm}$

The height of object using magnification equation is,

  $\begin{array}{l}\frac{h_{\text{i}}}{h_{\text{o}}}=-\frac{d_{\text{i}}}{d_{\text{o}}}\\ \frac{h_{\text{i}}}{h_{\text{o}}}=-\frac{34\text{ cm}}{34\text{ cm}}\\ h_{\text{i}}=-h_{\text{o}}\end{array}$

Consider the height of the object as $h_{\text{o}}=6.0\text{cm}$ .

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 ray diagram as shown below.

  

Figure 1

The horizontal scale of 1 block is $2.0\text{ cm}$ .The vertical scale of 1 block is $1.0\text{ cm}$ .

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

  $\begin{array}{c}{d}_{\text{i}}=\left(\begin{array}{l}\text{Number}\text{of}\text{blocks }\\ \text{from the lens in }\\ \text{horizontal direction}\end{array}\right)\text{×}\left(\text{Horizontal}\text{Scale}\right)\\ =17\text{blocks}\times \text{2}\text{.0}\text{cm}\\ =\text{34}\text{.0}\text{cm}\end{array}$

Conclusion:

Thus, the location of the image is $34.0\text{ cm}$ .and ray diagram is shown in Figure 1.