{"id":748,"date":"2026-08-10T02:20:40","date_gmt":"2026-08-10T02:20:40","guid":{"rendered":"https:\/\/physicsfundamentalsinfo.com\/blog\/?p=748"},"modified":"2026-08-29T21:49:19","modified_gmt":"2026-08-29T21:49:19","slug":"lens-equation","status":"publish","type":"post","link":"https:\/\/physicsfundamentalsinfo.com\/blog\/waves\/lens-equation\/","title":{"rendered":"Lens Equation (1\/f = 1\/do + 1\/di)"},"content":{"rendered":"\n<div class=\"pf-citation\"><div class=\"eyebrow\">Definition<\/div><p>\n\nThe lens equation, written 1\/f = 1\/d<sub>o<\/sub> + 1\/d<sub>i<\/sub>, relates the focal length of a thin lens or spherical mirror to the object distance and the image distance. Rearranged, it locates an image from any two known quantities: a positive image distance means a real image, a negative one means a virtual image.\n\n<\/p><\/div>\n\n<p>Hold a magnifying glass over this page and the letters swell. Lift it away slowly and something odd happens \u2014 the text blurs, then flips upside down. Nothing about the glass changed. Only the distance did.<\/p>\n\n<p>That flip is not a quirk. It is one equation crossing zero, and the same equation tells a phone camera where to put your face, a projector where to throw a film, and an optician which lens will fix your vision.<\/p>\n\n<h2>What Is the Lens Equation?<\/h2>\n\n<p>The lens equation is the formula that links an optical element&#8217;s focal length to where an object sits and where its image forms. Give it any two of those three distances and it hands you the third.<\/p>\n\n<p>Its real power is that it does not care what the element is. A glass lens bends light by refraction; a curved mirror bends it by reflection. Two completely different physical mechanisms \u2014 and one identical piece of algebra.<\/p>\n\n<p>That is why physicists usually call it the <strong>lens and mirror equation<\/strong>, or the <strong>thin-lens equation<\/strong> when a lens is specifically meant. The name changes; the maths does not.<\/p>\n\n<figure class=\"pf-figure\" style=\"margin:1.6em 0;\"><img src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/08\/lens-equation-ray-object-distance-distance-focal.webp\" width=\"432\" height=\"254\" alt=\"Lens equation ray diagram showing object distance, image distance and focal length for a converging lens forming a real inverted image\" loading=\"lazy\" decoding=\"async\" style=\"width:100%;height:auto;max-width:216px;display:block;margin:0 auto;\" \/><\/figure>\n\n<p style=\"text-align:center;font-size:13px;font-style:italic;color:#1F2E47;margin-top:8px;\">The three quantities in the lens equation, drawn to scale for a converging lens with the object beyond twice the focal length.<\/p>\n\n<h3>Real images and virtual images<\/h3>\n\n<p>A <strong>real image<\/strong> is one that light actually reaches. Put a screen there and a picture appears \u2014 that is a cinema, a camera sensor, the retina at the back of your eye.<\/p>\n\n<p>A <strong>virtual image<\/strong> is where the light only <em>appears<\/em> to come from. Your brain traces the outgoing rays backwards and places an image behind the glass. Put a screen there and you get nothing at all.<\/p>\n\n<p>The lens equation tells you which one you have, and it tells you with a single minus sign.<\/p>\n\n<h2>The Lens Equation Formula<\/h2>\n\n<p>The equation in its standard form is short enough to memorise in one reading.<\/p>\n\n<div class=\"pf-formula\">1\/f = 1\/d_o + 1\/d_i<\/div>\n\n<p>Two companion formulas travel with it and you will need both. The first gives image size, the second converts a mirror&#8217;s curvature into a focal length.<\/p>\n\n<div class=\"pf-formula\">m = h_i \/ h_o = \u2212 d_i \/ d_o<\/div>\n\n<div class=\"pf-formula\">f = R \/ 2   (spherical mirror)<\/div>\n\n<div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\">\n<table style=\"width:100%;border-collapse:collapse;\">\n<thead>\n<tr style=\"background:#142139;color:#FAF6EE;\">\n<th style=\"padding:10px;text-align:left;border:1px solid #D9CFB8;\">Symbol<\/th>\n<th style=\"padding:10px;text-align:left;border:1px solid #D9CFB8;\">Quantity<\/th>\n<th style=\"padding:10px;text-align:left;border:1px solid #D9CFB8;\">SI unit<\/th>\n<th style=\"padding:10px;text-align:left;border:1px solid #D9CFB8;\">Sign rule<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>f<\/strong><\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Focal length<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">metre (m)<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Positive if converging, negative if diverging<\/td>\n<\/tr>\n<tr style=\"background:#F5F2EA;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>d<sub>o<\/sub><\/strong><\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Object distance<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">metre (m)<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Positive for any real object<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>d<sub>i<\/sub><\/strong><\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Image distance<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">metre (m)<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Positive if real, negative if virtual<\/td>\n<\/tr>\n<tr style=\"background:#F5F2EA;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>m<\/strong><\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Magnification<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">dimensionless<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Positive if upright, negative if inverted<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>h<sub>o<\/sub>, h<sub>i<\/sub><\/strong><\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Object and image height<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">metre (m)<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Positive above the optical axis<\/td>\n<\/tr>\n<tr style=\"background:#F5F2EA;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>R<\/strong><\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Radius of curvature (mirrors)<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">metre (m)<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Positive for a concave mirror<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\"><strong>P<\/strong><\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Optical power<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">dioptre (D) = m<sup>\u22121<\/sup><\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">P = 1\/f, with f in metres<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n<p>Note the units carefully. The equation itself is happy with centimetres, millimetres or metres \u2014 as long as <em>all three<\/em> distances use the same one. Optical power is the exception: dioptres demand metres.<\/p>\n\n<h2>The Three Sign Rules That Decide Every Answer<\/h2>\n\n<p>Nearly every wrong answer in geometric optics is a sign error, not an arithmetic one. Three rules prevent almost all of them.<\/p>\n\n<ol>\n<li><strong>Object distance is always positive.<\/strong> For any real object sitting in front of the lens or mirror, d<sub>o<\/sub> is greater than zero. You will rarely meet the exception (a &#8220;virtual object&#8221; inside a multi-lens system).<\/li>\n<li><strong>Focal length carries the element&#8217;s character.<\/strong> Converging elements \u2014 a convex lens, a concave mirror \u2014 take a positive f. Diverging elements \u2014 a concave lens, a convex mirror \u2014 take a negative f.<\/li>\n<li><strong>Image distance is an output, never an input.<\/strong> You do not decide its sign. The equation returns it, and its sign is the answer: positive means real, negative means virtual.<\/li>\n<\/ol>\n\n<p>Rule three is the one students fight. A negative d<sub>i<\/sub> feels like a mistake, so they quietly drop the minus sign and lose the whole result. It is not a mistake \u2014 it is the equation telling you the image is virtual.<\/p>\n\n<figure class=\"pf-figure\" style=\"margin:1.6em 0;\"><img src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/08\/lens-equation-sign-convention-map-where-object.webp\" width=\"396\" height=\"216\" alt=\"Sign convention map for the lens equation showing where object distance and image distance are positive or negative for a thin lens and for a spherical mirror\" loading=\"lazy\" decoding=\"async\" style=\"width:100%;height:auto;max-width:198px;display:block;margin:0 auto;\" \/><\/figure>\n\n<p style=\"text-align:center;font-size:13px;font-style:italic;color:#1F2E47;margin-top:8px;\">Where each quantity is positive and negative, for a thin lens and for a spherical mirror.<\/p>\n\n<h3>Why your textbook may print a different formula<\/h3>\n\n<p>Search the lens equation and you will meet a rival version: <strong>1\/v \u2212 1\/u = 1\/f<\/strong>. Many UK and Indian syllabuses use it, and so do our companion guides to the <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/waves\/concave-convex-lens\/\">concave and convex lens<\/a> and the <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/waves\/concave-convex-mirror\/\">concave and convex mirror<\/a>.<\/p>\n\n<p>Neither version is wrong. They are the same physics with different bookkeeping \u2014 the Cartesian convention measures every distance as a signed coordinate along the direction light travels, so a real object to the left of the lens gets a <em>negative<\/em> distance. Georgia State University&#8217;s <a href=\"http:\/\/hyperphysics.phy-astr.gsu.edu\/hbase\/geoopt\/lenseq.html\" target=\"_blank\" rel=\"noopener\">HyperPhysics reference on the thin lens equation<\/a> sets both forms side by side and notes the Cartesian one is favoured in advanced texts because it scales better to multi-lens instruments.<\/p>\n\n<p>Here is the exact translation.<\/p>\n\n<div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\">\n<table style=\"width:100%;border-collapse:collapse;\">\n<thead>\n<tr style=\"background:#142139;color:#FAF6EE;\">\n<th style=\"padding:10px;text-align:left;border:1px solid #D9CFB8;\">Quantity<\/th>\n<th style=\"padding:10px;text-align:left;border:1px solid #D9CFB8;\">This article (real-is-positive)<\/th>\n<th style=\"padding:10px;text-align:left;border:1px solid #D9CFB8;\">Cartesian convention<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Lens equation<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">1\/f = 1\/d<sub>o<\/sub> + 1\/d<sub>i<\/sub><\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">1\/v \u2212 1\/u = 1\/f<\/td>\n<\/tr>\n<tr style=\"background:#F5F2EA;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Mirror equation<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">1\/f = 1\/d<sub>o<\/sub> + 1\/d<sub>i<\/sub> (identical)<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">1\/v + 1\/u = 1\/f<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Real object<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">d<sub>o<\/sub> positive<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">u = \u2212d<sub>o<\/sub> (negative)<\/td>\n<\/tr>\n<tr style=\"background:#F5F2EA;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Real image, lens<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">d<sub>i<\/sub> positive<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">v = +d<sub>i<\/sub> (positive)<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Real image, mirror<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">d<sub>i<\/sub> positive<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">v = \u2212d<sub>i<\/sub> (negative)<\/td>\n<\/tr>\n<tr style=\"background:#F5F2EA;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Magnification<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">m = \u2212d<sub>i<\/sub>\/d<sub>o<\/sub><\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">lens: m = v\/u &nbsp;&nbsp; mirror: m = \u2212v\/u<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n<p>Substitute the middle column into the right column and the two collapse into one another exactly. In practice, pick whichever convention your course uses and never mix them inside a single problem \u2014 that is where the marks go missing.<\/p>\n\n<h2>How to Use the Lens Equation in Four Steps<\/h2>\n\n<p>Start by writing down what you know with its sign attached, before touching any algebra. The method below works for lenses and mirrors alike.<\/p>\n\n<ol>\n<li><strong>Sign your knowns.<\/strong> Object distance positive. Focal length positive for converging, negative for diverging. For a mirror, get f from f = R\/2 first.<\/li>\n<li><strong>Rearrange for the unknown.<\/strong> Solving for image distance gives d<sub>i<\/sub> = 1 \/ (1\/f \u2212 1\/d<sub>o<\/sub>).<\/li>\n<li><strong>Compute the reciprocals separately.<\/strong> Work out 1\/f and 1\/d<sub>o<\/sub> as decimals, subtract, then invert at the very end.<\/li>\n<li><strong>Read the signs back.<\/strong> Positive d<sub>i<\/sub> means real; negative means virtual. Then find m = \u2212d<sub>i<\/sub>\/d<sub>o<\/sub>: negative m means inverted, and a magnitude above 1 means enlarged.<\/li>\n<\/ol>\n\n<p>Step three is where most arithmetic slips happen, because it is tempting to invert too early. If you would rather have the algebra checked for you while you learn the pattern, our <a href=\"https:\/\/physicsfundamentalsinfo.com\/calculators\/lens-mirror\">Lens &amp; Mirror Calculator<\/a> solves for whichever quantity you leave blank and shows the working line by line.<\/p>\n\n<p>The simulator below is the fastest way to build intuition. Drag the object closer and watch the image distance swing to infinity and come back on the other side as a negative number.<\/p>\n\n<div class=\"pf-sim-slot\"><div class=\"pf-sim-slot-header\"><span class=\"icon-dot\"><\/span><span class=\"label\">Lens Equation Lab<\/span><\/div><div class=\"pf-sim-slot-body\"><style>.pf-sim-frame{width:100%;border:none;height:600px}@media(max-width:760px){.pf-sim-frame{height:1000px}}<\/style><iframe src=\"\/labs\/lens-equation.html?embed=1\" class=\"pf-sim-frame\" loading=\"lazy\"><\/iframe><\/div><\/div>\n\n<h2>Where the Lens Equation Comes From<\/h2>\n\n<p>The equation is not a rule handed down from nowhere. It falls out of two similar triangles and, underneath that, out of Snell&#8217;s law applied twice.<\/p>\n\n<p>Look back at the ray diagram. The undeviated ray through the centre of the lens creates a pair of similar triangles either side, which gives h<sub>i<\/sub>\/h<sub>o<\/sub> = d<sub>i<\/sub>\/d<sub>o<\/sub> \u2014 that is the magnification relation.<\/p>\n\n<p>The ray that arrives parallel to the axis and leaves through the far focal point makes a second similar pair, this time involving f. Set the two ratios equal, tidy the algebra, and 1\/f = 1\/d<sub>o<\/sub> + 1\/d<sub>i<\/sub> is what is left.<\/p>\n\n<p>Go one level deeper and the lens is simply two refracting surfaces in series. Light bends entering the glass and bends again leaving it, each bend governed by <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/waves\/snells-law-explained\/\">Snell&#8217;s law<\/a> \u2014 the same <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/waves\/reflection-and-refraction\/\">reflection and refraction<\/a> behaviour you see at any boundary. OpenStax works through this two-surface derivation in full in <a href=\"https:\/\/openstax.org\/books\/university-physics-volume-3\/pages\/2-4-thin-lenses\" target=\"_blank\" rel=\"noopener\">University Physics Volume 3, section 2.4<\/a>.<\/p>\n\n<p>That derivation also produces the lensmaker&#8217;s equation, 1\/f = (n \u2212 1)(1\/R<sub>1<\/sub> \u2212 1\/R<sub>2<\/sub>), which builds the focal length from the glass&#8217;s <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/waves\/refractive-index\/\">refractive index<\/a> and the curvature of its two faces. In short: geometry sets where the image goes, and the material sets f.<\/p>\n\n<h2>What the Equation Predicts as the Object Moves Closer<\/h2>\n\n<p>Feed a converging lens a steadily shrinking object distance and the image runs through a complete life cycle. This single table answers most exam questions about image type without any calculation at all.<\/p>\n\n<div class=\"pf-table-scroll\" style=\"display:block;width:100%;max-width:100%;overflow-x:auto;-webkit-overflow-scrolling:touch;margin:1.5em 0;\">\n<table style=\"width:100%;border-collapse:collapse;\">\n<thead>\n<tr style=\"background:#142139;color:#FAF6EE;\">\n<th style=\"padding:10px;text-align:left;border:1px solid #D9CFB8;\">Object position<\/th>\n<th style=\"padding:10px;text-align:left;border:1px solid #D9CFB8;\">Image distance<\/th>\n<th style=\"padding:10px;text-align:left;border:1px solid #D9CFB8;\">Image type<\/th>\n<th style=\"padding:10px;text-align:left;border:1px solid #D9CFB8;\">Size<\/th>\n<th style=\"padding:10px;text-align:left;border:1px solid #D9CFB8;\">Everyday case<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Very far away<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">d<sub>i<\/sub> is close to f<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Real, inverted<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Tiny<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Telescope, phone camera<\/td>\n<\/tr>\n<tr style=\"background:#F5F2EA;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Beyond 2f<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Between f and 2f<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Real, inverted<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Diminished<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Ordinary photography<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Exactly at 2f<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">d<sub>i<\/sub> = 2f<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Real, inverted<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Same size<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">1:1 macro copying<\/td>\n<\/tr>\n<tr style=\"background:#F5F2EA;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Between f and 2f<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Beyond 2f<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Real, inverted<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Enlarged<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Projector, enlarger<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Exactly at f<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Infinite<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">No image forms<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">\u2014<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Lighthouse, collimator<\/td>\n<\/tr>\n<tr style=\"background:#F5F2EA;\">\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Closer than f<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Negative<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Virtual, upright<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Enlarged<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Magnifying glass<\/td>\n<\/tr>\n<tr>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Diverging lens, any distance<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Always negative<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Virtual, upright<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Diminished<\/td>\n<td style=\"padding:10px;border:1px solid #D9CFB8;\">Door peephole, myopia lens<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n\n<p>That row where the image distance runs to infinity is the magnifying-glass flip you saw at the start. Cross the point where d<sub>o<\/sub> equals f and the answer changes sign \u2014 real becomes virtual, inverted becomes upright.<\/p>\n\n<h2>Real-World Examples of the Lens Equation<\/h2>\n\n<h3>Your phone camera barely moves its lens<\/h3>\n\n<p>A phone camera has a focal length of roughly 5 mm. For a selfie at 50 cm, the equation puts the sensor at 5.05 mm behind the lens; for a mountain 5 km away, at 5.005 mm.<\/p>\n\n<p>The entire focus range of the camera is about forty microns of travel \u2014 thinner than a sheet of paper. That is why phone autofocus is a whisper-quiet voice coil rather than a visible barrel twist.<\/p>\n\n<h3>The eye does the opposite<\/h3>\n\n<p>Your retina cannot move, so d<sub>i<\/sub> is locked at roughly 2 cm. To keep the equation balanced as objects come closer, the eye changes the only quantity it can \u2014 the ciliary muscle squeezes the lens and shortens f.<\/p>\n\n<p>That process is called accommodation, and it stiffens with age. Reading glasses simply add the focal power the lens can no longer supply.<\/p>\n\n<h3>Cinema projectors sit just outside the focal point<\/h3>\n\n<p>Put the film a fraction beyond f and the image lands far away and hugely enlarged \u2014 row four of the table. Because the image is real and inverted, the film runs through the gate upside down so the picture arrives the right way up.<\/p>\n\n<h3>Car wing mirrors are honest liars<\/h3>\n\n<p>A convex mirror has a negative focal length, so d<sub>i<\/sub> is negative for every possible object position. The image is always virtual, upright and shrunken \u2014 which widens the field of view but makes cars look further away than they are.<\/p>\n\n<p>That is the entire physics behind the etched warning on the glass.<\/p>\n\n<h3>Telescopes park the sensor at the focal plane<\/h3>\n\n<p>For a star, d<sub>o<\/sub> is effectively infinite, so the 1\/d<sub>o<\/sub> term vanishes and d<sub>i<\/sub> = f exactly. Every telescope is built around that simplification: the detector sits at the focal plane and nothing needs refocusing between targets.<\/p>\n\n<h2>Common Misconceptions About the Lens Equation<\/h2>\n\n<h3>&#8220;So f equals d<sub>o<\/sub> plus d<sub>i<\/sub>&#8220;<\/h3>\n\n<p>No \u2014 reciprocals do not add that way. With d<sub>o<\/sub> = 30 cm and d<sub>i<\/sub> = 15 cm, the correct focal length is 10 cm, not 45 cm.<\/p>\n\n<p>Always add the reciprocals first, then invert once at the end. Inverting term by term is the single most common error in this topic.<\/p>\n\n<h3>&#8220;A negative image distance means I made a mistake&#8221;<\/h3>\n\n<p>It means you found a virtual image, which is a legitimate physical answer. Every magnifying glass, every peephole and every wing mirror produces one.<\/p>\n\n<p>Report it with the minus sign intact and say what it means: image on the same side as the object, upright, not projectable onto a screen.<\/p>\n\n<h3>&#8220;Covering half the lens cuts the image in half&#8221;<\/h3>\n\n<p>It does not. Every point on the object sends light through the <em>whole<\/em> lens, and all of those rays reconverge at the same image point.<\/p>\n\n<p>Block half and you remove half the rays, so the image gets dimmer \u2014 but it stays complete. Try it with a magnifying glass and a card; the effect is genuinely surprising the first time.<\/p>\n\n<h3>&#8220;The equation always works&#8221;<\/h3>\n\n<p>It is an approximation with two conditions attached. The lens must be thin compared with its radii of curvature, and the rays must stay close to the axis (paraxial).<\/p>\n\n<p>Break either and real lenses show aberrations. Because refractive index varies with wavelength, f itself differs slightly for red and blue light \u2014 the same <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/waves\/dispersion-of-light\/\">dispersion of light<\/a> that makes a prism work also produces the coloured fringing that camera designers spend fortunes correcting.<\/p>\n\n<h2>How the Lens Equation Relates to Mirrors, Refraction and Lens Power<\/h2>\n\n<p>Think of this equation as the hub of geometric optics, with three spokes running off it.<\/p>\n\n<ul>\n<li><strong>To mirrors:<\/strong> the same formula, with f = R\/2 converting curvature into focal length. Which sign R takes depends on whether the mirror is concave or convex.<\/li>\n<li><strong>To refraction:<\/strong> the lensmaker&#8217;s equation supplies f from the glass&#8217;s refractive index and the two surface curvatures, so Snell&#8217;s law sits one level below.<\/li>\n<li><strong>To optical power:<\/strong> P = 1\/f in dioptres. A prescription of minus 2.5 D is a diverging lens with f = \u22120.40 m, and closely spaced lens powers simply add.<\/li>\n<\/ul>\n\n<p>Once the equation is fluent, the rest of the subject unpacks quickly. If you want the device-by-device picture rather than the algebra, the guides to <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/waves\/concave-convex-lens\/\">concave and convex lenses<\/a> and <a href=\"https:\/\/physicsfundamentalsinfo.com\/blog\/waves\/concave-convex-mirror\/\">concave and convex mirrors<\/a> cover the ray diagrams case by case.<\/p>\n\n<figure style=\"margin:32px auto;max-width:640px;text-align:center;\"><img decoding=\"async\" src=\"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-content\/uploads\/2026\/08\/Convex_lens_magnifying_glass_and_upside-down_image.jpg\" alt=\"Lens equation in practice: a converging lens forming a real inverted image\" loading=\"lazy\" style=\"width:100%;height:auto;border-radius:4px;\" width=\"1920\" height=\"1103\"><figcaption style=\"font-size:13px;color:#1F2E47;font-style:italic;margin-top:8px;\">With the object beyond the focal point, the lens equation returns a positive image distance &mdash; and the image arrives inverted.<\/figcaption><\/figure>\n\n<h2>Worked Problems<\/h2>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 1<\/div><div class=\"pf-problem-question\">An object is placed 36 cm from a converging lens of focal length 12 cm. Find the image distance and the magnification.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Sign the knowns. Converging lens, so f = +12 cm; real object, so d<sub>o<\/sub> = +36 cm. Use 1\/f = 1\/d<sub>o<\/sub> + 1\/d<sub>i<\/sub>.<\/p>\n<p>Step 2: Rearrange and substitute. 1\/d<sub>i<\/sub> = 1\/f \u2212 1\/d<sub>o<\/sub> = 1\/12 \u2212 1\/36 cm<sup>\u22121<\/sup>.<\/p>\n<p>Step 3: Common denominator: 3\/36 \u2212 1\/36 = 2\/36 = 1\/18 cm<sup>\u22121<\/sup>, so d<sub>i<\/sub> = +18 cm.<\/p>\n<p>Step 4: Magnification. m = \u2212d<sub>i<\/sub>\/d<sub>o<\/sub> = \u221218\/36 = \u22120.50. Positive d<sub>i<\/sub> means real; negative m means inverted.<\/p>\n<p><strong>Answer: d<sub>i<\/sub> = +18 cm, m = \u22120.50 \u2014 a real, inverted image at half size.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 2<\/div><div class=\"pf-problem-question\">A magnifying glass has a focal length of 10 cm. A stamp is held 6.0 cm from it. Where is the image and how big is it?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: f = +10 cm, d<sub>o<\/sub> = +6.0 cm. The object is inside the focal length, so expect a virtual image.<\/p>\n<p>Step 2: 1\/d<sub>i<\/sub> = 1\/10 \u2212 1\/6.0 = 3\/30 \u2212 5\/30 = \u22122\/30 = \u22121\/15 cm<sup>\u22121<\/sup>.<\/p>\n<p>Step 3: d<sub>i<\/sub> = \u221215 cm. The minus sign places the image 15 cm from the lens on the <em>same<\/em> side as the stamp.<\/p>\n<p>Step 4: m = \u2212d<sub>i<\/sub>\/d<sub>o<\/sub> = \u2212(\u221215)\/6.0 = +2.5.<\/p>\n<p><strong>Answer: d<sub>i<\/sub> = \u221215 cm, m = +2.5 \u2014 a virtual, upright image 2.5 times larger.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 3<\/div><div class=\"pf-problem-question\">A 4.0 cm tall candle stands 30 cm in front of a diverging lens of focal length 20 cm. Find the image distance and image height.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Diverging lens, so f = \u221220 cm. d<sub>o<\/sub> = +30 cm, h<sub>o<\/sub> = 4.0 cm.<\/p>\n<p>Step 2: 1\/d<sub>i<\/sub> = 1\/(\u221220) \u2212 1\/30 = \u22123\/60 \u2212 2\/60 = \u22125\/60 = \u22121\/12 cm<sup>\u22121<\/sup>.<\/p>\n<p>Step 3: d<sub>i<\/sub> = \u221212 cm, so the image is virtual and 12 cm from the lens on the object&#8217;s side.<\/p>\n<p>Step 4: m = \u2212(\u221212)\/30 = +0.40, so h<sub>i<\/sub> = 0.40 \u00d7 4.0 cm = 1.6 cm.<\/p>\n<p><strong>Answer: d<sub>i<\/sub> = \u221212 cm, h<sub>i<\/sub> = 1.6 cm upright \u2014 virtual, upright and diminished, as a diverging lens always gives.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 4<\/div><div class=\"pf-problem-question\">A concave mirror has a radius of curvature of 40 cm. An object sits 30 cm in front of it. Locate the image and describe it.<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Convert curvature to focal length. f = R\/2 = 40\/2 = +20 cm, positive because the mirror is concave.<\/p>\n<p>Step 2: Same equation as a lens. 1\/d<sub>i<\/sub> = 1\/20 \u2212 1\/30 cm<sup>\u22121<\/sup>.<\/p>\n<p>Step 3: 3\/60 \u2212 2\/60 = 1\/60 cm<sup>\u22121<\/sup>, so d<sub>i<\/sub> = +60 cm, in front of the mirror.<\/p>\n<p>Step 4: m = \u221260\/30 = \u22122.0. The object lies between f and 2f, so an enlarged real image is exactly what the table predicts.<\/p>\n<p><strong>Answer: d<sub>i<\/sub> = +60 cm, m = \u22122.0 \u2014 real, inverted and twice the size.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 5<\/div><div class=\"pf-problem-question\">A car wing mirror is convex with a radius of curvature of 3.0 m. A cyclist is 5.0 m behind the car. Where does the image appear and at what scale?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Convex mirror, so f = \u2212R\/2 = \u22121.5 m. d<sub>o<\/sub> = +5.0 m.<\/p>\n<p>Step 2: 1\/d<sub>i<\/sub> = 1\/(\u22121.5) \u2212 1\/5.0 = \u22120.6667 \u2212 0.2000 = \u22120.8667 m<sup>\u22121<\/sup>.<\/p>\n<p>Step 3: d<sub>i<\/sub> = \u22121.15 m \u2014 virtual, about 1.2 m behind the mirror surface.<\/p>\n<p>Step 4: m = \u2212(\u22121.15)\/5.0 = +0.23.<\/p>\n<p><strong>Answer: d<sub>i<\/sub> = \u22121.15 m, m = +0.23 \u2014 upright and 23% of full size, which is precisely why objects are closer than they appear.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 6<\/div><div class=\"pf-problem-question\">A projector must throw a sharp image onto a screen 3.0 m away while the slide sits 0.60 m from the lens. What focal length is needed, and what is the magnification?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: Both distances are known and the image is real, so d<sub>o<\/sub> = +0.60 m and d<sub>i<\/sub> = +3.0 m. Solve for f.<\/p>\n<p>Step 2: 1\/f = 1\/d<sub>o<\/sub> + 1\/d<sub>i<\/sub> = 1\/0.60 + 1\/3.0 = 1.6667 + 0.3333 m<sup>\u22121<\/sup>.<\/p>\n<p>Step 3: 1\/f = 2.000 m<sup>\u22121<\/sup>, so f = 0.50 m = 50 cm.<\/p>\n<p>Step 4: m = \u22123.0\/0.60 = \u22125.0, so the picture is five times the slide and inverted \u2014 hence the film is loaded upside down.<\/p>\n<p><strong>Answer: f = 0.50 m, m = \u22125.0.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<div class=\"pf-problem\"><div class=\"pf-problem-num\">Problem 7<\/div><div class=\"pf-problem-question\">A converging lens of focal length 10 cm must project a real image magnified exactly 4 times onto a screen. How far from the lens should the object be, and how far away is the screen?<\/div><details><summary>Show Solution<\/summary><div class=\"pf-problem-solution\">\n<p><strong>Solution:<\/strong><\/p>\n<p>Step 1: A real image is inverted, so m = \u22124.0. From m = \u2212d<sub>i<\/sub>\/d<sub>o<\/sub> this gives d<sub>i<\/sub> = 4.0 d<sub>o<\/sub>.<\/p>\n<p>Step 2: Substitute into the lens equation. 1\/10 = 1\/d<sub>o<\/sub> + 1\/(4.0 d<sub>o<\/sub>).<\/p>\n<p>Step 3: Combine the right side over a common denominator: 1\/10 = (4 + 1)\/(4 d<sub>o<\/sub>) = 5\/(4 d<sub>o<\/sub>).<\/p>\n<p>Step 4: Cross-multiply. 4 d<sub>o<\/sub> = 50, so d<sub>o<\/sub> = 12.5 cm and d<sub>i<\/sub> = 4.0 \u00d7 12.5 = 50 cm.<\/p>\n<p>Step 5: Check. 1\/12.5 + 1\/50 = 0.0800 + 0.0200 = 0.100 = 1\/10 cm<sup>\u22121<\/sup>. Correct.<\/p>\n<p><strong>Answer: d<sub>o<\/sub> = 12.5 cm and d<sub>i<\/sub> = 50 cm, so lens to screen is 50 cm and the total object-to-screen span is 62.5 cm.<\/strong><\/p>\n<\/div><\/details><\/div>\n\n<h2>Frequently Asked Questions<\/h2>\n\n<details class=\"pf-faq-item\"><summary>What is the lens equation?<\/summary><div class=\"pf-faq-item-answer\">\nThe lens equation is 1\/f = 1\/d<sub>o<\/sub> + 1\/d<sub>i<\/sub>, connecting focal length, object distance and image distance for a thin lens or spherical mirror. Knowing any two lets you calculate the third. The sign of the image distance reveals the image type: positive for real, negative for virtual.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Is the lens equation the same as the mirror equation?<\/summary><div class=\"pf-faq-item-answer\">\nYes, in the real-is-positive convention they are literally the same equation. A thin lens and a spherical mirror both obey 1\/f = 1\/d<sub>o<\/sub> + 1\/d<sub>i<\/sub>, and both use m = \u2212d<sub>i<\/sub>\/d<sub>o<\/sub>. The only extra step for a mirror is finding the focal length from its curvature using f = R\/2.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What does a negative image distance mean?<\/summary><div class=\"pf-faq-item-answer\">\nA negative image distance means the image is virtual. Light does not actually converge there, so the image cannot be caught on a screen, and it sits on the same side as the object. Virtual images are always upright relative to the object. Magnifying glasses, diverging lenses and convex mirrors all produce them.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Why do some textbooks write 1\/v minus 1\/u = 1\/f instead?<\/summary><div class=\"pf-faq-item-answer\">\nThat is the Cartesian sign convention, common in UK and Indian syllabuses. It treats distances as signed coordinates measured along the direction light travels, so a real object gets a negative distance, u = \u2212d<sub>o<\/sub>. Substitute that in and the two formulas become identical. Both give the same physical answer; just never mix them within one problem.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>How do you calculate magnification from the lens equation?<\/summary><div class=\"pf-faq-item-answer\">\nUse m = \u2212d<sub>i<\/sub>\/d<sub>o<\/sub>, which also equals h<sub>i<\/sub>\/h<sub>o<\/sub>. Find the image distance from the lens equation first, then divide and flip the sign. A negative m means an inverted image, a positive m means upright, and a magnitude above 1 means enlarged.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>What happens when the object is exactly at the focal point?<\/summary><div class=\"pf-faq-item-answer\">\nNo image forms. Setting d<sub>o<\/sub> = f gives 1\/d<sub>i<\/sub> = 0, so the image distance is infinite. Rays leave the lens perfectly parallel instead of converging anywhere. This is not a failure of the equation but a useful configuration, and it is how lighthouses and collimators produce a straight beam.\n<\/div><\/details>\n\n<details class=\"pf-faq-item\"><summary>Does the lens equation work for thick lenses?<\/summary><div class=\"pf-faq-item-answer\">\nNot directly. The formula assumes a thin lens, meaning its thickness is much smaller than the radii of curvature of its surfaces, and it assumes paraxial rays close to the optical axis. Thick lenses need the distances measured from principal planes instead. Wide-angle rays introduce spherical aberration that the simple equation cannot predict.\n<\/div><\/details>\n","protected":false},"excerpt":{"rendered":"<p>The lens equation, 1\/f = 1\/do + 1\/di, locates the image formed by any thin lens or spherical mirror. This guide covers the three sign rules, both textbook conventions, an interactive simulator and seven worked problems.<\/p>\n","protected":false},"author":1,"featured_media":749,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[4],"tags":[],"class_list":["post-748","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-waves"],"_links":{"self":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/748","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/comments?post=748"}],"version-history":[{"count":7,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/748\/revisions"}],"predecessor-version":[{"id":1839,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/posts\/748\/revisions\/1839"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media\/749"}],"wp:attachment":[{"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/media?parent=748"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/categories?post=748"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicsfundamentalsinfo.com\/blog\/wp-json\/wp\/v2\/tags?post=748"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}