Ray Optics: Application Problems (11)

Question

An object is placed $20\,\text{cm}$ in front of a concave mirror of focal length $15\,\text{cm}$. Find the position, nature, and magnification of the image.

Solution — Step by Step

The image is at $60\,\text{cm}$ in front of the mirror, real, inverted, and magnified $3\times$.

Why This Works

The concave mirror with the object between $f$ and $C$ (the centre of curvature at $30\,\text{cm}$) always produces a real, inverted, magnified image beyond $C$. We’ve placed the object at $20\,\text{cm}$, which is between $f = 15$ and $C = 30$, so the textbook prediction matches.

The negative sign of $v$ confirms the image is on the same side as the object (in front of the mirror), which means it is real. The negative magnification confirms inversion.

Alternative Method

Use the magnification formula directly: $m = f/(f-u) = -15/(-15-(-20)) = -15/5 = -3$. Then $v = -mu = -(-3)(-20) = -60\,\text{cm}$. Same answer in two lines.

Common Mistake

Sign errors. Students drop the minus on $u$ and get $v = +60\,\text{cm}$, then call the image virtual. Always write all three quantities ($u$, $v$, $f$) with their signs before substituting. NEET 2024 had a four-mark trap on exactly this.