Mirror Equation Calculator
Solve the mirror and thin-lens equation for image distance and magnification, from the focal length and the object distance.
Mirror Equation Calculator
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Formula: 1/f = 1/dₒ + 1/dᵢ | m = -dᵢ/dₒ
Worked example — dᵢ=15cm, m=-0.5×
Formula
1/f = 1/dₒ + 1/dᵢ | m = -dᵢ/dₒ
Relates focal length, object distance, and image distance for mirrors and thin lenses.
Worked Examples
Example 1: Converging lens
Problem:f=10cm, dₒ=30cm
Solution:1/dᵢ=1/10-1/30=2/30, dᵢ=15cm, m=-0.5
Result:dᵢ=15cm, m=-0.5×
Frequently Asked Questions
What is the mirror/lens equation?
The mirror and thin-lens equation is 1/f = 1/dₒ + 1/dᵢ, where f is the focal length, dₒ is the object distance, and dᵢ is the image distance. Magnification is m = −dᵢ/dₒ. All distances are typically in the same unit (cm or m).
How do you find the image distance and magnification?
Rearrange to 1/dᵢ = 1/f − 1/dₒ, then dᵢ = 1/(1/f − 1/dₒ). Example: f = 10 cm, dₒ = 30 cm → 1/dᵢ = 0.1 − 0.0333 = 0.0667, so dᵢ = 15 cm, m = −15/30 = −0.5× (inverted, reduced).
What units and sign conventions are used?
Distances are in centimetres (cm) or metres (m). For mirrors: positive dᵢ means real image (in front of mirror); negative means virtual (behind). For lenses: positive dᵢ means real image (opposite side from object). Positive magnification means upright image.
What is a real vs. virtual image in optics?
A real image forms where reflected or refracted rays actually converge and can be projected on a screen (dᵢ > 0 for mirrors). A virtual image forms where rays appear to diverge from and cannot be projected (dᵢ < 0). Plane mirrors always produce virtual images.
What is the difference between a converging and a diverging element?
Converging elements - concave mirrors and convex lenses - have positive focal length and can form real, projectable images. Diverging elements, meaning convex mirrors and concave lenses, have negative focal length and always give a virtual, upright, reduced image of a real object. Take f = -10 cm with an object at 30 cm: 1/d_i = -0.1 - 0.0333 = -0.1333, so d_i = -7.5 cm and m = +0.25, exactly that pattern.
What happens when the object sits at or inside the focal point?
Placed exactly at the focal point, 1/d_i = 1/f - 1/f = 0, so the image distance runs off to infinity: the rays emerge parallel and no image forms. Move closer than f and the image distance turns negative, giving a magnified upright virtual image - the ordinary magnifying-glass case. With f = 10 cm and the object at 5 cm, d_i = -10 cm and m = +2, a twofold enlargement.
How does focal length relate to a mirror's radius of curvature?
For a spherical mirror in the paraxial approximation, f = R/2, so a concave mirror ground to a 40 cm radius has a 20 cm focal length. Thin lenses instead follow the lensmaker's equation, 1/f = (n - 1) x (1/R_1 - 1/R_2), where n is the refractive index of the glass. That is why the same lens shape has a different focal length when made from a higher-index material.
What does the magnification value tell you about the image?
Magnification m = -d_i/d_o also equals the image height divided by the object height. Its magnitude sets the size - greater than 1 means enlarged, less than 1 means reduced - while its sign sets the orientation, negative for inverted and positive for upright. A 4 cm object imaged at m = -0.5 becomes a 2 cm inverted image, which for a mirror also means the image is real.
Why do these results assume paraxial rays?
The mirror and thin-lens equations hold for rays close to the optical axis and making small angles with it, and for elements whose thickness is negligible. Wide apertures introduce spherical aberration, where marginal rays focus nearer than central rays. Lenses additionally suffer chromatic aberration because refractive index varies with wavelength, and thick lenses require principal planes rather than a single measurement point.
How do you find the image formed by two lenses in sequence?
Apply the equation twice. The image produced by the first lens becomes the object for the second, with its object distance measured from the second lens. If that intermediate image would fall beyond the second lens, it acts as a virtual object and takes a negative object distance. Overall magnification is the product m_total = m_1 x m_2, which is how microscopes and telescopes are analysed.
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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