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NEET PHYSICSRAY OPTICS AND OPTICAL INSTRUMENTSEasy

Question

An object is placed on the principal axis of a concave mirror at a distance of 1.5f1.5f (ff is the focal length). The image will be at:

A

-3f

B

1.5f

C

-1.5f

D

3f

Step-by-Step Solution

  1. Identify Formula: Use the mirror formula: 1v+1u=1f\frac{1}{v} + \frac{1}{u} = \frac{1}{f}.
  2. Sign Convention: According to the Cartesian sign convention for a concave mirror:
  • Object distance (uu) is taken as negative: u=1.5fu = -1.5f.
  • Focal length (ff) is taken as negative for a concave mirror: Focal length = f-f.
  1. Calculation: Substitute the values into the formula: 1v+11.5f=1f\frac{1}{v} + \frac{1}{-1.5f} = \frac{1}{-f} 1v=11.5f1f\frac{1}{v} = \frac{1}{1.5f} - \frac{1}{f} 1v=1015f1f=23f33f\frac{1}{v} = \frac{10}{15f} - \frac{1}{f} = \frac{2}{3f} - \frac{3}{3f} 1v=233f=13f\frac{1}{v} = \frac{2 - 3}{3f} = \frac{-1}{3f} v=3fv = -3f
  2. Conclusion: The image is formed at a distance of 3f3f from the mirror. The negative sign indicates that the image is formed on the same side as the object (real image).

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from RAY OPTICS AND OPTICAL INSTRUMENTS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSRAY OPTICS AND OPTICAL INSTRUMENTSobjectplacedprincipalconcavemirror

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