Ray Optics: Step-by-Step Worked Examples (10)

easy • 2 min read
Tags Ray Optics

Question

An object is placed 30 cm30\ \text{cm} in front of a concave mirror of focal length 20 cm20\ \text{cm}. Find the image distance, magnification, and nature of the image.

Solution — Step by Step

Using the standard Cartesian convention (light travels left to right, all distances measured from the pole): for a concave mirror, f=−20 cmf = -20\ \text{cm}. The object is in front, so u=−30 cmu = -30\ \text{cm}.

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

Substituting:

1v=1−20−1−30=−120+130=−160\frac{1}{v} = \frac{1}{-20} - \frac{1}{-30} = -\frac{1}{20} + \frac{1}{30} = -\frac{1}{60}

So v=−60 cmv = -60\ \text{cm}.

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

Negative vv means the image is on the same side as the object — real. ∣m∣=2|m| = 2 means inverted and magnified twice.

Final answer: v=−60 cmv = -60\ \text{cm}, m=−2m = -2, image is real, inverted, and twice as large.

Why This Works

Sign convention is the entire game in ray optics. Once you fix it (we use Cartesian: distances measured from pole, in the direction of incident light positive), every problem becomes substitution.

The mirror formula 1v+1u=1f\tfrac{1}{v} + \tfrac{1}{u} = \tfrac{1}{f} works for both concave and convex if signs are honest. Don’t memorise separate formulas.

Alternative Method

Use the magnification formula m=ff−u=−20−20−(−30)=−2010=−2m = \tfrac{f}{f-u} = \tfrac{-20}{-20-(-30)} = \tfrac{-20}{10} = -2. Faster for “find magnification only” MCQs.

When u=2fu = 2f for a concave mirror, image is at v=2fv = 2f with m=−1m = -1. Memorise this benchmark — many NEET PYQs use u=2fu = 2f as the test condition.

Common Mistake

Using f=+20f = +20 for a concave mirror. Concave mirrors converge light, but in Cartesian convention, the focus lies in the direction of incident light reversed, so ff is negative. Pluging f=+20f = +20 gives v=+12 cmv = +12\ \text{cm} (virtual, wrong nature).

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