Lens and mirror formula problems — step by step approach with sign convention

Question

An object is placed 30 cm from a concave mirror of focal length 20 cm. Find the image position, nature, and magnification. How does the sign convention work?

(CBSE 10/12 + JEE Main + NEET)

Solution — Step by Step

Apply the New Cartesian Sign Convention

Rule	Convention
All distances measured from the pole/optical centre	Origin is at the mirror/lens
Direction of incident light is positive (+ve)	Left to right is positive
Distances opposite to incident light are negative	Towards the left of mirror = negative
Heights above principal axis are positive	Below = negative

For a concave mirror: $f = -20$ cm (focus is in front of mirror, negative direction).

Object distance: $u = -30$ cm (object is always in front, negative).

Use the mirror formula

Mirror Formula

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

$\frac{1}{v} + \frac{1}{-30} = \frac{1}{-20}$

$\frac{1}{v} = -\frac{1}{20} + \frac{1}{30} = \frac{-3 + 2}{60} = -\frac{1}{60}$

$v = \mathbf{-60 \text{ cm}}$

Negative $v$ means the image is on the same side as the object — in front of the mirror. This is a real image.

Calculate magnification

Magnification

$m = -\frac{v}{u} = -\frac{-60}{-30} = -2$

$|m| = 2$ means the image is twice the size of the object.

Negative $m$ means the image is inverted.

Summarise the image properties

Position: 60 cm in front of the mirror (between C and infinity)
Nature: Real and inverted
Size: Magnified (twice the object size)

This makes sense: object at $C$ gives image at $C$ (same size). Object between $C$ and $F$ gives image beyond $C$ (magnified). Our object at 30 cm is between $C = 40$ cm and $F = 20$ cm.

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Why This Works

The mirror formula is derived from geometry — similar triangles formed by the incident ray, reflected ray, and the principal axis. The sign convention converts a geometric construction into algebraic equations where the signs automatically tell us the nature of the image (real vs virtual, erect vs inverted).

The magnification formula $m = -v/u$ encodes both size and orientation in one number. If $|m| > 1$ , the image is enlarged. If $|m| < 1$ , it is diminished. The sign gives the orientation.

Alternative Method

Instead of the formula, use ray diagrams to predict the image qualitatively. Draw two rays from the object:

Ray parallel to principal axis reflects through the focus
Ray through the centre of curvature reflects back on itself

Where these intersect is the image. This gives the position, nature, and size visually — then use the formula for exact numbers.

💡 Expert Tip

For NEET and JEE, memorise this table: for a concave mirror, object beyond C gives a real, inverted, diminished image between F and C. Object at C gives image at C (same size). Object between C and F gives a real, inverted, magnified image beyond C. Object at F gives image at infinity. Object between F and pole gives a virtual, erect, magnified image behind the mirror.

Common Mistake

⚠️ Common Mistake

The most frequent error: forgetting the sign convention and using positive values for everything. With $u = 30$ , $f = 20$ : $\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{1}{60}$ , giving $v = 60$ (wrong sign). The correct approach uses $u = -30$ , $f = -20$ , which gives $v = -60$ . The negative signs are not optional — they carry the physics.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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