Chapter 17, Problem 85A

To determine

To sketch:The rays on the given diagram to obtain the height and location of the image.

Answer to Problem 85A

The location of the image is $2.7\text{ cm}$ from the mirror.

The height of the image is $1.0\text{ cm}\left(\text{erect}\right)$ .

The sketch of the given diagram along with rays is shown in Figure 2.

Explanation of Solution

Given:

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

The position of the object is $d_{\text{o}}=8.0\text{ cm}$ .

The focal length is $f=-4.0\text{ cm}$

The given diagram is shown below.

  

Figure 1

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 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.

  

The horizontal scale is 1 block is equal to $1.0\text{ m}$ .The vertical scale is 1 block is equal to $0.50\text{ m}$ .Measure the distance of the image as per the horizontal scale as $2.7\text{block}\times \text{1}\text{cm}=-2.7\text{ cm}$ from the mirror. Negative sign is included since the image is other side of mirror.Measure the vertical height of the image as per the vertical scale as $2\text{block}\times 0.50\text{cm}=\text{1}\text{.0 cm}$ .

Conclusion:

Thus, the location of the image is $2.7\text{ cm}$ from the mirror.

Thus, the height of the image is $\text{1}\text{.0 cm}\left(\text{erect}\right)$ .

The complete ray diagram is shown in Figure 1.