This cheat sheet covers how to solve worked examples using the thin lens equation and the mirror equation. Students need it because image formation problems often combine formulas, signs, units, and ray diagram ideas. It helps organize the steps for finding image distance, focal length, magnification, and image type.
The focus is on clear setup and interpretation, not just calculation.
The central formula is , which works for thin lenses and spherical mirrors when the correct sign convention is used. Magnification is found with . A positive image distance usually means a real image, while a negative image distance usually means a virtual image.
The sign of tells whether the image is upright or inverted, and tells whether it is enlarged or reduced.
Key Facts
- The lens and mirror equation is , where is focal length, is object distance, and is image distance.
- Solving for image distance gives when .
- Magnification is , so image height is .
- For a converging lens or concave mirror, the focal length is usually positive, so .
- For a diverging lens or convex mirror, the focal length is usually negative, so .
- If , the image is real and forms where light rays actually meet.
- If , the image is virtual and appears to come from a location behind the lens or mirror.
- If , the image is inverted, and if , the image is upright.
Vocabulary
- Focal length
- The focal length is the distance from the lens or mirror to the focal point.
- Object distance
- The object distance is the distance from the object to the lens or mirror.
- Image distance
- The image distance is the distance from the lens or mirror to the image location.
- Magnification
- Magnification is the ratio of image height to object height, given by .
- Real image
- A real image forms where light rays actually converge and is usually indicated by .
- Virtual image
- A virtual image appears to form where light rays seem to originate and is usually indicated by .
Common Mistakes to Avoid
- Using the wrong sign for focal length is wrong because converging lenses and concave mirrors usually have , while diverging lenses and convex mirrors usually have .
- Forgetting the negative sign in is wrong because it changes whether the image is identified as upright or inverted.
- Substituting distances in centimeters and meters together is wrong because all distances in must use the same unit.
- Calling every negative answer impossible is wrong because often means a virtual image, not a calculation failure.
- Assuming a larger image distance always means a larger image is wrong because image size depends on , not on alone.
Practice Questions
- 1 A converging lens has and an object is placed at . Find , , and state whether the image is real or virtual.
- 2 A concave mirror has and forms an image when an object is at . Calculate and .
- 3 A diverging lens has and an object is placed at . Find the image distance and describe the image orientation.
- 4 Explain why an image with and is described as virtual and upright, even though the calculated image distance is negative.
Understanding Lens & Mirror Equation Worked Examples
A reliable worked example starts before any numbers go into a calculator. Draw a simple principal axis, then mark the lens or mirror, the focal point, and the object. This sketch makes the physical situation easier to check later.
For a converging device, an object placed farther than one focal length can produce an image on the far side of a lens or in front of a concave mirror. If the object moves closer to the focal point, the image moves farther away.
At exactly the focal point, outgoing rays are parallel, so no image forms at a finite screen position. This explains why a calculation can produce an extremely large image distance when the object distance is close to the focal length.
Rearranging the equation deserves care because fractions create common errors. A useful method is to multiply through by the focal length, object distance, and image distance before isolating the unknown. Keep the original signs attached to each distance during every algebra step.
Do not treat a negative focal length as an error that needs fixing. It describes how a diverging lens or convex mirror spreads light. Convert all lengths to one unit before calculating.
Centimetres work well when every given value uses centimetres. Metres work well for larger setups. A final answer with mixed units can look reasonable while being physically wrong.
The calculated image distance tells where the image belongs, but magnification completes the description. Its size tells how many times taller or shorter the image is than the object. For example, a magnification with size one half means the image height is half the object height.
A negative result means the top and bottom are reversed. This matches a real image formed by a projector or a camera lens. A positive result means the image keeps the same orientation.
Virtual images are often upright, such as the image seen in a bathroom mirror. An upright image can still be enlarged or reduced, so orientation alone does not identify image size.
Ray diagrams provide an independent check on every numerical answer. For lenses, a ray parallel to the axis refracts through the focal region for a converging lens. For mirrors, that ray reflects through the focal region for a concave mirror.
A ray through the centre of a thin lens travels nearly straight, while a ray aimed at the centre of curvature of a spherical mirror returns along its path. Two rays are enough to locate an image. When rays do not physically meet, extend them backward with dotted lines to locate a virtual image.
Pay close attention to cases near the focal point, to negative signs, and to whether the predicted image could be caught on a screen. These checks connect equation work to real optical devices such as glasses, magnifying glasses, security mirrors, cameras, and telescopes.