Chapter 18, Problem 65A

(a)

To determine

The location, size, and orientation of the image by the ray diagram.

Answer to Problem 65A

The location of the image is $10.5\text{ cm}$ .

The size of the image is $-1.8\text{ cm}$ .

The image is inverted.

Explanation of Solution

Given:

The height of the object is $h_{\text{o}}=2.4\text{ cm}$ .

The position of the object is $d_{\text{o}}=14.0\text{ cm}$ .

The focal length is $f=6.0\text{ cm}$

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

The vertical scale of 1 block is $0.4\text{ cm}$ .

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

  $\begin{array}{c}{d}_{\text{i}}=\left(\begin{array}{l}\text{Number}\text{of}\text{blocks }\\ \text{from the lens in }\\ \text{horizontal direction}\end{array}\right)\text{×}\left(\text{Horizontal}\text{Scale}\right)\\ =10.5\text{blocks}\times 1.\text{0}\text{cm}\\ =\text{10}\text{.5}\text{cm}\end{array}$

The vertical height $\left(h_{\text{i}}\right)$ of the image is,

  $\begin{array}{c}{h}_{\text{i}}=\left(\begin{array}{l}\text{Number}\text{of}\text{blocks }\\ \text{from the lens in }\\ \text{vertical direction}\end{array}\right)\text{×}\left(\text{Vertical}\text{Scale}\right)\\ =-4.5\text{blocks}\times \text{0}\text{.4}\text{cm}\\ =-\text{1}\text{.8}\text{cm}\end{array}$

The height of the image is $h_{\text{i}}=-1.8\text{ cm}$ . Hence, the image is an inverted image.

Conclusion:

Thus, the location of the image is $10.5\text{ cm}$ .

Thus, the size of the image is $-1.8\text{ cm}$ .

Thus, the image is inverted.

(b)

To determine

The location, size, and orientation of the image using equations.

Answer to Problem 65A

The location of the image is $10.5\text{ cm}$ .

The size of the image is $-1.8\text{ cm}$ .

The image is inverted.

Explanation of Solution

Given:

The height of the object is $h_{\text{o}}=2.4\text{ cm}$ .

The position of the object is $d_{\text{o}}=14.0\text{ cm}$ .

The focal length is $f=6.0\text{ cm}$

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 location of the image is,

  $\begin{array}{l}\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\ \frac{1}{d_{\text{i}}}=\frac{1}{f}-\frac{1}{d_{\text{o}}}\\ d_{\text{i}}=\frac{f{d}_{\text{o}}}{d_{\text{o}}-f}\\ d_{\text{i}}=\frac{\left(6.0\text{ cm}\right)\left(14.0\text{ cm}\right)}{\left(14.0\text{ cm}\right)-\left(6.0\text{ cm}\right)}\end{array}$

  $d_{\text{i}}=10.5\text{ cm}$

The height of the image is,

  $\begin{array}{l}m=\frac{h_{\text{i}}}{h_{\text{o}}}=-\frac{d_{\text{i}}}{d_{\text{o}}}\\ h_{\text{i}}=-\frac{d_{\text{i}}{h}_{\text{o}}}{d_{\text{o}}}\\ h_{\text{i}}=-\frac{\left(10.5\text{ cm}\right)\left(\text{2}\text{.4 cm}\right)}{\left(14.0\text{ cm}\right)}\\ h_{\text{i}}=-1.8\text{ cm}\end{array}$

Hence, the image is inverted.

Conclusion:

Thus, the location of the image is $10.5\text{ cm}$ .

Thus, the size of the image is $-1.8\text{ cm}$ .

Thus, the image is inverted.