Mirror formula: 1/f = 1/v + 1/u. u = -30, f = -15. Solve for v. Real, inverted image at -30 cm.

Object 30 cm in Front of Concave Mirror, f = 15 cm — Find Image

Question

An object is placed 30 cm in front of a concave mirror whose focal length is 15 cm. Find the position, nature, and size of the image formed.

Solution — Step by Step

Set up the sign convention

We use the New Cartesian Sign Convention: the pole of the mirror is the origin, and distances are measured along the principal axis. Distances in the direction of incident light (towards the mirror) are negative.

So, u = −30 cm (object is in front of the mirror) and f = −15 cm (concave mirror, focal length is negative).

Write the mirror formula

The mirror formula connects object distance, image distance, and focal length:

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

We know f and u — we solve for v.

Substitute and solve for v

$\frac{1}{-15} = \frac{1}{v} + \frac{1}{-30}$

$\frac{1}{v} = \frac{1}{-15} + \frac{1}{30}$

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

$\boxed{v = -30 \text{ cm}}$

Find the magnification

$m = -\frac{v}{u} = -\frac{-30}{-30} = -1$

Magnification is −1, which tells us the image is the same size as the object and inverted.

Interpret the result

Position: 30 cm in front of the mirror (same side as the object, v is negative)
Nature: Real (v is negative for a mirror) and inverted (m is negative)
Size: Same as the object (|m| = 1)

The image forms at the centre of curvature (C), which makes complete sense — the object was also placed at C.

Why This Works

When an object sits exactly at the centre of curvature of a concave mirror, every ray from it strikes the mirror and reflects straight back through C. This is because rays through C are normal to the mirror surface — they hit at 0° and return along the same path.

The result is a real, inverted image of the same size, formed at C itself. This is one of the special cases worth memorising for boards — object at C gives image at C, size unchanged.

This specific case (u = 2f) is a reference point. If u > 2f, the image shrinks and moves between F and C. If u < 2f (but beyond F), the image grows and moves beyond C. The u = 2f case is the crossover.

Alternative Method — Using the Relation u = 2f

We can skip the formula entirely once we recognise the setup.

For a concave mirror, when u = 2f:

Image distance: v = 2f
Magnification: m = −1

Here, f = 15 cm, so 2f = 30 cm. The object is at 30 cm, which equals 2f. So directly: v = −30 cm, m = −1.

💡 Expert Tip

Memorise the three special cases for concave mirrors:

Object at infinity → image at F (point-sized)
Object at C (u = 2f) → image at C, same size
Object at F → image at infinity

These appear as direct 1-mark questions in CBSE Class 10 boards almost every year.

Common Mistake

⚠️ Common Mistake

The most common error here is using f = +15 cm for a concave mirror. Students forget that in the New Cartesian system, the focal length of a concave mirror is negative. If you plug in f = +15, you get v = +30 cm — which would imply a virtual image behind the mirror. That's physically wrong for an object placed beyond F. Always assign signs before substituting.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the Relation u = 2f

Common Mistake

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