View a distant object through a converging lens. The image will be a perverted and inverted image that is smaller than the object. Now look through the lens at an object that is very close to the lens. The image will be a normal and erect image that is larger than the object.

# Subsection1.2Measuring Object and Image Positions¶ permalink

Arrange a lens and a light box (the object) on the optical bench. Place the lens at about the middle of the bench, and the object at one end of the bench. Look for the image of the object, and locate its position through parallax. Also locate the position of the image by using a screen.

With our light box at 0cm and our lens at 50cm we found the image to be at 84.2cm using parallax and at 81.8cm using a screen. Using the screen was easier and more reliable.

Move the lens to different points on the bench, recording the positions of object and image (measured from the lens) as you go. Denote the distance of the object from the lens $d_o\text{,}$ and the distance of the image from the lens $d_i\text{.}$ Continue this procedure until the image moves off the end of the bench. Our measurements can be seen in Figure 1.1.

It should be noted that the image the image is erect and as the distance to the lens increases, the size of the image decreases.

Find the image of a very distant object. The location of the image when the object is very far away is called the focal point of the lens. We can convince ourselves that the lens has two focal points, one on each side at the same distance from the, by finding the image and then flipping the lens and noting that the image is at the same position. However, as we have seen images appear at places other than our focal point, we know that not all images appear at the focal point.

The distance between the lens and the focal point is called the focal length of the lens, which we will denote $f\text{.}$ Experimentally we determined the focal length of our mirror to be 26cm, however it should have been 20cm.

Now place the bulb at various locations with $d_o \lt f\text{,}$ especially when the lens is as close to the lens as possible, and locate the image. We can find these images by parallax, but not by using a screen as they are virtual image. These images are erect and the same size as the object. When the object is placed at the focal point there is no image created as the light travels out parallel, thus never converging.

In Subsection 1.2, we measured $d_o$ and $d_i\text{,}$ the distances of the object and image from the lens. It turns out that analysis will be easier if we measure the distance of the object from the focal point; let's call this distance $x_o = d_o − f\text{.}$ In the same way, let us measure the distance of the real image from the focal point on its side of the lens; let's call this distance $x_i = d_i - f\text{.}$ Our linearized data can be found in Figure 1.2.

Figure 1.3 shows an erect virtual image that is slightly larger than the object.

Figure 1.4 shows an erect virtual image that is twice the size of the object.

Figure 1.5 shows an erect virtual image that is four times the size of the object.

Figure 1.6 does not create an image due to the lack of intersecting lines as caused by the object being at the focal point.

Figure 1.7 shows an inverted real image that is twice the size of the object.

Figure 1.8 shows an inverted real image that is the same size as the object.

Figure 1.9 shows an inverted real image that is half the size of the object.

# Subsection1.5Theoretical Investigation of the Lens¶ permalink

We find that the relationship between $x_o\text{,}$ $x_i\text{,}$ and $f$ to be \begin{equation*} \frac{1}{x_o} + \frac{1}{x_i} = \frac{1}{f}. \end{equation*}

Further, we can express this relationship as being between $d_o$ and $d_i\text{.}$ This relationship is \begin{equation*} \frac{1}{d_o - f} + \frac{1}{d_i - f} = \frac{1}{f} \end{equation*}

# Subsection1.6Investigating a Diverging Lens¶ permalink

With diverging lenses the image is always a smaller, upright, and virtual image.

Diverging lenses also follow the equation \begin{equation*} \frac{1}{x_o} + \frac{1}{x_i} = \frac{1}{f} \end{equation*} and additionally we can calculate magnification, $M\text{,}$ using the equation \begin{equation*} M = \frac{-x_i}{x_o} \end{equation*}