Mirror Formula 1/v + 1/u = 1/f — Sign Convention and Example

Question

A concave mirror has a focal length of 15 cm. An object is placed 20 cm in front of the mirror. Find the position of the image. Is the image real or virtual?

Solution — Step by Step

Why This Works

The mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ comes directly from the geometry of reflection — it holds for all spherical mirrors as long as we follow sign convention consistently. The "New Cartesian" system simply picks the principal axis as the x-axis with the pole as origin, so every coordinate has a definite sign.

For a concave mirror, $f$ is always negative because the focus lies in front of the mirror (same side as the incoming light). For a convex mirror, $f$ is always positive because the focus is behind the mirror. This is the one rule you must never mix up in CBSE board exams — it accounts for a large share of sign-convention errors.

When $v$ comes out negative, the image is real (concave mirror) and can be caught on a screen. When $v$ comes out positive for a concave mirror, the image is virtual and behind the mirror.

Alternative Method — Using the Magnification Check

After finding $v = -60$ cm, we can cross-verify using linear magnification:

$m = -\frac{v}{u} = -\frac{(-60)}{(-20)} = -3$

The magnification is $-3$, meaning the image is 3 times larger than the object and inverted (negative $m$). This is consistent with the object being placed between $f$ and $C$ — wait, actually here the object at 20 cm is between $f = 15$ cm and $C = 30$ cm, so a real, magnified, inverted image beyond $C$ is exactly what we expect. The geometry and formula agree. Always do this sanity check in exams.

Common Mistake

A related slip: computing $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ but then forgetting to carry the negative signs through the arithmetic. Write each fraction with its sign explicitly before combining — don't do it mentally.