We define the following terms:
- Object distance ($u$) distance between the object and the center of the lens
- Image distance ($v$) distance between the focused image and the center of the lens
- Focal length ($f$) distance between the focal point and the center of the lens
When images are formed by a lens, they can be:
- Real or virtual
- Magnified or diminished
- Upright or inverted
Descriptions for images
Real image formed by the actual intersection of rays of light. It can be projected onto a screen
Virtual image formed by the apparent intersection of rays when traced backwards. It appears to be behind the lens
Magnified image has a magnitude of magnification (just ignore any negative sign) greater than 1 (image is larger than the object)
Diminished image has a magnitude of magnification (just ignore any negative sign) less than 1 (image is smaller than the object)
Upright image has the same orientation as the object (magnification is positive)
Inverted image has the orientation of the object flipped vertically (magnification is negative)
Sign convention
Focal length:
- For converging lens, focal length is always positive
- For diverging lens, focal length is always negative
Object distance:
- Positive if the object is on the same side the light is coming from
Image distance:
- Positive if the image is on the side opposite to where the light is coming (the image will be real)
Object height:
- Always positive because we consider the object to be upright
Image height:
- Positive if image is upright; negative for inverted images
Review video:
Loading video
Magnification
This is the ratio of the image distance ($v$) to the object distance ($u$) and due to the applications of similar triangles, the ratio of the image height ($h_i$) to the object height ($h_o$): $$ \begin{equation}\begin{aligned} m=-\frac{v}{u}=\frac{h_i}{h_o}\\ \end{aligned}\end{equation} $$ Magnification is positive for upright images and negative for inverted images.
The thin lens equation
For thin lens, there is a formula which relates the focal length ($f$) to the object distance ($u$) and image distance ($v$): $$ \begin{equation}\begin{aligned} \frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\ \end{aligned}\end{equation} $$ Images formed by converging lens will be real and inverted for object distances greater than $f$. These images will be magnified for object distances from $f$ to $2f$ and diminished for object distances greater than $2f$. For object distances less than $f$, the image will be upright and virtual.
All images formed by diverging lens will be virtual and upright, regardless of if the object distance is in the ranges $0$ to $f$, $f$ to $2f$ or greater than $2f$.
Run the following simulation and see if these rules are used when simulating the lens behaviour: https://ophysics.com/l12.html
Mission details
- An object 25 cm away from a lens produces a focused image on a film 15 cm away. What is the focal length of the converging lens. (Answer: 9.4 cm)
- If the focal length of the lens in your camera is 2 cm, at what distance must objects be placed so that a focused image is produced on a piece of film set 3 cm from the lens? (Answer: 6 cm)
- A tree 20 m high is located 40 m from a converging lens of focal length 8.0 cm.
- Calculate the distance from the lens to the image. (Answer: 8.0 cm)
- Calculate the magnification. (Answer: -0.0020)
- Calculate the height of the image of the tree. (Answer: -4 cm)
- A normal human eye has a focal length of about 2.3 cm. If you look at the tip of a pencil, 55.3 cm from your eye, how far is the image from the lens of your eye? (Answer: 2.4 cm)
- A converging lens produces an image twice the size of the original.
- If the object is placed 40 cm from the lens, where is the image produced? (Answer: 80 cm from the lens)
- What is the focal length of the lens? (Answer: 26.7 cm)
- If the image is 6 cm tall, how tall is the original object? (Answer: 3 cm)
Credits: FalconPhysics