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

easy CBSE CBSE 2024 Board Exam 4 min read

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

All distances are measured from the pole of the mirror. Distances in the direction of incident light (towards the mirror) are negative; distances opposite to incident light are also negative for a concave mirror’s focus.

For our problem: object distance u=20u = -20 cm (object is in front of the mirror), focal length f=15f = -15 cm (concave mirror, focus is in front).

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

Substitute the known values:

1v+120=115\frac{1}{v} + \frac{1}{-20} = \frac{1}{-15}
1v=115120=115+120\frac{1}{v} = \frac{1}{-15} - \frac{1}{-20} = -\frac{1}{15} + \frac{1}{20}

Finding LCM of 15 and 20 is 60:

1v=460+360=160\frac{1}{v} = \frac{-4}{60} + \frac{3}{60} = \frac{-1}{60}
v=60 cmv = -60 \text{ cm}

The negative sign tells us the image is formed in front of the mirror — on the same side as the object.

v=60v = -60 cm → image is 60 cm in front of the mirror. Since vv is negative (real side of concave mirror), the image is real and inverted.

Final Answer: Image is formed 60 cm in front of the concave mirror. It is real and inverted.


Why This Works

The mirror formula 1v+1u=1f\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, ff is always negative because the focus lies in front of the mirror (same side as the incoming light). For a convex mirror, ff 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 vv comes out negative, the image is real (concave mirror) and can be caught on a screen. When vv comes out positive for a concave mirror, the image is virtual and behind the mirror.


Alternative Method — Using the Magnification Check

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

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

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

Quick memory aid for sign convention: “Object is always negative for mirrors.” Since we place objects in front of mirrors, uu is always negative. If your uu comes out positive, you’ve made a sign error.


Common Mistake

The most common error in CBSE 2024 and previous years: students write f=+15f = +15 cm for a concave mirror. They remember “concave converges” and think ff should be positive. Wrong — in New Cartesian convention, concave mirror has ff negative because the focus is in front of the mirror (negative x-direction). Convex mirror has ff positive. If you flip these, your image distance will come out with the wrong sign and the entire interpretation collapses.

A related slip: computing 1v=1f1u\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.

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