Ray Optics: Exam-Pattern Drill (12)

hard 2 min read

Question

An object is placed 2020 cm in front of a concave mirror of focal length 1515 cm. Find the position, nature, and magnification of the image.

Solution — Step by Step

Following the standard Cartesian sign convention with the pole at origin and incident light from the left: object distance u=20u = -20 cm (object to the left), focal length f=15f = -15 cm (concave mirror, focus on the same side as the object).

1v+1u=1f\frac{1}{v} + \frac{1}{u} = \frac{1}{f} 1v=1f1u=115120=115+120\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-15} - \frac{1}{-20} = -\frac{1}{15} + \frac{1}{20}

Common denominator 6060:

1v=4+360=160    v=60 cm\frac{1}{v} = \frac{-4 + 3}{60} = -\frac{1}{60} \implies v = -60 \text{ cm}
m=vu=6020=3m = -\frac{v}{u} = -\frac{-60}{-20} = -3

The negative magnification means the image is inverted; magnitude m=3|m| = 3 means the image is three times the object size.

Final Answer: Image is at 6060 cm in front of the mirror, real, inverted, and magnified 33 times.

Why This Works

The mirror formula assumes the Cartesian sign convention: distances measured against the incident-light direction are negative. For a concave mirror with the object beyond the focus but inside 2f2f (here u=20u = -20, f=15f = -15, so u<2f=30|u| < |2f| = 30), we expect a real, inverted, magnified image — which matches our answer.

The negative vv value confirms the image is on the same side as the object (in front of the mirror), so it’s a real image. A positive vv would have indicated a virtual image behind the mirror.

Alternative Method

Use the magnification formula m=f/(fu)=15/(15(20))=15/5=3m = f/(f-u) = -15/(-15-(-20)) = -15/5 = -3 directly, skipping the explicit calculation of vv. Useful when the question only asks for magnification.

Forgetting the sign on ff for a concave mirror is the #1 error. Many students plug f=+15f = +15 and get v=+60v = +60, then conclude “virtual image” — completely wrong. Concave ff is negative; convex ff is positive. Drill this until it’s automatic.

JEE and NEET often ask the same problem but stop at a different step — sometimes “find vv”, sometimes “find mm”, sometimes “find himageh_{\text{image}} if hobject=2h_{\text{object}} = 2 cm”. Solve the whole pipeline once; the numbers carry over.

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