Lens & Mirror Equation Worked Examples Cheat Sheet

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 $\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$ , which works for thin lenses and spherical mirrors when the correct sign convention is used. Magnification is found with $m=\frac{h_i}{h_o}=-\frac{d_i}{d_o}$ . A positive image distance usually means a real image, while a negative image distance usually means a virtual image.

The sign of $m$ tells whether the image is upright or inverted, and $|m|$ tells whether it is enlarged or reduced.

Key Facts

The lens and mirror equation is $\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$ , where $f$ is focal length, $d_o$ is object distance, and $d_i$ is image distance.
Solving for image distance gives $d_i=\frac{f d_o}{d_o-f}$ when $d_o \ne f$ .
Magnification is $m=\frac{h_i}{h_o}=-\frac{d_i}{d_o}$ , so image height is $h_i=m h_o$ .
For a converging lens or concave mirror, the focal length is usually positive, so $f>0$ .
For a diverging lens or convex mirror, the focal length is usually negative, so $f<0$ .
If $d_i>0$ , the image is real and forms where light rays actually meet.
If $d_i<0$ , the image is virtual and appears to come from a location behind the lens or mirror.
If $m<0$ , the image is inverted, and if $m>0$ , the image is upright.

Vocabulary

Focal length: The focal length $f$ is the distance from the lens or mirror to the focal point.
Object distance: The object distance $d_o$ is the distance from the object to the lens or mirror.
Image distance: The image distance $d_i$ is the distance from the lens or mirror to the image location.
Magnification: Magnification $m$ is the ratio of image height to object height, given by $m=\frac{h_i}{h_o}=-\frac{d_i}{d_o}$ .
Real image: A real image forms where light rays actually converge and is usually indicated by $d_i>0$ .
Virtual image: A virtual image appears to form where light rays seem to originate and is usually indicated by $d_i<0$ .

Common Mistakes to Avoid

Using the wrong sign for focal length is wrong because converging lenses and concave mirrors usually have $f>0$ , while diverging lenses and convex mirrors usually have $f<0$ .
Forgetting the negative sign in $m=-\frac{d_i}{d_o}$ 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 $\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$ must use the same unit.
Calling every negative answer impossible is wrong because $d_i<0$ 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 $|m|=\left|\frac{d_i}{d_o}\right|$ , not on $d_i$ alone.

Practice Questions

1 A converging lens has $f=10\text{ cm}$ and an object is placed at $d_o=30\text{ cm}$ . Find $d_i$ , $m$ , and state whether the image is real or virtual.
2 A concave mirror has $f=15\text{ cm}$ and forms an image when an object is at $d_o=45\text{ cm}$ . Calculate $d_i$ and $m$ .
3 A diverging lens has $f=-12\text{ cm}$ and an object is placed at $d_o=24\text{ cm}$ . Find the image distance and describe the image orientation.
4 Explain why an image with $d_i<0$ and $m>0$ is described as virtual and upright, even though the calculated image distance is negative.

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Lens & Mirror Equation Worked Examples Cheat Sheet

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