Magnification by lenses is a core Ray Optics & Optical Instruments concept in Physics. This guide explains what it is, walks through a fully worked example, and lists the key equations you need — with a short quiz to test yourself.

Key equations and worked example

A convex lens has focal length f = 10 cm. An object stands 30 cm away. 1/v = 1/f − 1/u = 1/10 − 1/30 = 2/30 → v = 15 cm. The image forms 15 cm beyond the lens. Magnification m = −v/u = −15/30 = −0.5: the image is real, inverted, and half the object's size. Move the object inside f (say 5 cm) and the image becomes virtual, upright and magnified — a magnifying glass.

• <code>Lens formula: 1/f = 1/v + 1/u</code>
• <code>Magnification m = −v/u = (image height)/(object height)</code>
• <code>Power P = 1/f (dioptres, f in metres)</code>
• <code>Two thin lenses in contact: 1/F = 1/f₁ + 1/f₂</code>

Magnification by lenses in detail

Magnification by lenses is one of the central ideas in Ray Optics &amp; Optical Instruments, and it appears in Physics curricula under Reflection and refraction. It is worth learning deeply because it connects to so many other topics in this section.

A convex lens is thicker at the centre, so parallel rays refract toward the principal axis and meet at the focal point. Image formation follows the lens formula 1/f = 1/v + 1/u (with the sign convention used here, all distances measured as positive magnitudes for a real image). Objects beyond 2f give diminished real images; between f and 2f give magnified real images; inside f give virtual magnified images.

For exams, the pattern is predictable: first a definition or statement of the result, then a direct numerical application of one of the equations above, then a "why" question — why the formula takes that form, or what changes when a variable is doubled or halved. The worked example and quiz below cover exactly that progression.

Quick self-check:

• Q: Where must the object be for a convex lens to act as a magnifying glass?<br />A: Inside the focal length (u &lt; f) — the image is then virtual, upright and enlarged.
• Q: An object is at 2f from a convex lens. Where is the image?<br />A: Also at 2f on the other side, real, inverted, and the same size (m = −1).
• Q: What does a negative magnification signify?<br />A: The image is inverted relative to the object (real image case).
• Q: How is the power of a lens related to its focal length?<br />A: P = 1/f with f in metres — shorter focal length means greater converging power, in dioptres.