Chapter 18.2, Problem 29SSC

To determine

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

To verify: The result using thin lens and magnification equations.

Answer to Problem 29SSC

The location of the image is $20.0\text{ cm}$ .

The size of the image is $-24.0\text{ cm}$ .

Explanation of Solution

Given:

The height of the object is $h_{\text{o}}=6.0\text{ cm}$ .

The position of the object is $d_{\text{o}}=5.0\text{ cm}$ .

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

Formula used:

The expression for the lens 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 lens as a horizontal line.
• Place the lens at the center, place C at the left side of the lens and place F at the other side.
• Sketch a vertical line at the lens point to represent the lens. 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 be $I_1$

Sketch the ray diagram as shown below.

  

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

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

  $\begin{array}{c}{d}_{\text{i}}=\left(\begin{array}{l}\text{Number}\text{of}\text{blocks }\\ \text{from the lens in }\\ \text{horizontal direction}\end{array}\right)\text{×}\left(\text{Horizontal}\text{Scale}\right)\\ =10\text{blocks}\times 2.\text{0}\text{cm}\\ =2\text{0}\text{.0}\text{cm}\end{array}$

Hence, the position of the image is $d_{\text{i}}=20.0\text{ cm}$ . The vertical height $\left(h_{\text{i}}\right)$ of the image is,

  $\begin{array}{c}{h}_{\text{i}}=\left(\begin{array}{l}\text{Number}\text{of}\text{blocks of image }\\ \text{in vertical direction from the axis}\end{array}\right)\text{×}\left(\text{Vertical}\text{Scale}\right)\\ =-12\text{blocks}\times 2.0\text{cm}\\ =-24\text{cm}\end{array}$

Using lens and magnification equation.

The location of the image is,

  $\begin{array}{l}\frac{1}{d_{\text{o}}}+\frac{1}{d_{\text{i}}}=\frac{1}{f}\\ \frac{1}{d_{\text{i}}}=\frac{1}{f}-\frac{1}{d_{\text{o}}}\\ d_{\text{i}}=\frac{f{d}_{\text{o}}}{d_{\text{o}}-f}\\ d_{\text{i}}=\frac{\left(4.0\text{ cm}\right)\left(5.0\text{ cm}\right)}{\left(5.0\text{ cm}\right)-\left(4.0\text{ cm}\right)}\end{array}$

  $d_{\text{i}}=20.0\text{ cm}$

The height of the image is,

  $\begin{array}{l}m=\frac{h_{\text{i}}}{h_{\text{o}}}=-\frac{d_{\text{i}}}{d_{\text{o}}}\\ h_{\text{i}}=-\frac{d_{\text{i}}{h}_{\text{o}}}{d_{\text{o}}}\\ h_{\text{i}}=-\frac{\left(20.0\text{ cm}\right)\left(6.0\text{ cm}\right)}{\left(5.0\text{ cm}\right)}\\ h_{\text{i}}=-24.0\text{ cm}\end{array}$

Conclusion:

Thus, the location of the image is $20.0\text{ cm}$ .Thus, the size of the image is $-24.0\text{ cm}$ .