Physics & Mechanics

Mirror/Lens Equation Calculator

Calculate image distance, object distance, or focal length using the standard mirror and thin lens equation.

m
m
m
Image Distance
0.15

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Predicting Image Location

The Mirror/Lens Equation is a simple yet powerful formula that allows you to predict exactly where an image will form when an object is placed in front of a mirror or a lens. It is used to design telescopes, microscopes, and cameras.

It relates three key distances:

  • Focal Length ($f$): The distance from the center of the lens/mirror to the point where parallel light rays converge.
  • Object Distance ($d_o$): The distance from the object (the thing you are looking at) to the lens.
  • Image Distance ($d_i$): The distance from the lens to where the image actually appears (like where you would place a sensor or film).

Sign Conventions

This is the trickiest part of the equation:

  • Converging (Convex Lens / Concave Mirror): $f$ is positive.
  • Diverging (Concave Lens / Convex Mirror): $f$ is negative.
  • Real Image: $d_i$ is positive (it forms on the opposite side of the lens or in front of a mirror).
  • Virtual Image: $d_i$ is negative (it appears 'inside' the lens or behind a mirror).

The Formula

1f=1do+1di\begin{aligned} \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \end{aligned}

Where:
f=
Focal Length
dod_o=
Distance to Object
did_i=
Distance to Image

Example Calculation

You place a candle $30 , \text{cm}$ ($0.3 , \text{m}$) in front of a converging lens with a focal length of $10 , \text{cm}$ ($0.1 , \text{m}$).

  1. Calculate 1/f and 1/do: $1 / 0.1 = 10$ and $1 / 0.3 \approx 3.33$.
  2. Subtract: $10 - 3.33 = 6.67$.
  3. Invert for di: $1 / 6.67 = 0.15 , \text{m}$.

A real, inverted image of the candle will form $15 , \text{cm} behind the lens.

Frequently Asked Questions

A real image can be projected onto a screen or a wall. Light rays actually meet at that point. A virtual image (like your reflection in a bathroom mirror) cannot be projected; light rays only appear to come from that point.

Whenever the image distance ($d_i$) is positive for a lens, the light rays have crossed over the center point, causing the image to be inverted. Most telescopes and microscopes produce inverted images, which is why they often need an extra lens to flip the image right-side up for viewing.

If $d_o = f$, then $1/d_i = 0$, which means $d_i$ is at infinity. The light rays leaving the lens are perfectly parallel and never meet. This is how searchlights and flashlights create a beam that travels a long distance.