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NEET PHYSICSRAY OPTICS AND OPTICAL INSTRUMENTSEasy

Question

An object is mounted on a wall. Its image of equal size is to be obtained on a parallel wall with the help of a convex lens placed between these walls. The lens is kept at distance x in front of the second wall. The required focal length of the lens will be:

A

less than x/4

B

more than x/4 but less than x/2

C

x/2

D

x/4

Step-by-Step Solution

  1. Identify Given Conditions:
  • A real image is formed on a parallel wall (screen), so the image is real and inverted.
  • The image is of equal size to the object, which implies the magnification magnitude ∣m∣=1|m| = 1.
  • The lens is placed at a distance xx from the second wall (where the image is formed). Thus, the image distance v=+xv = +x (using sign convention where light travels L to R).
  1. Apply Lens Concept:
  • For a convex lens, a real image of the same size as the object is formed only when the object is placed at 2f2f and the image is formed at 2f2f on the other side.
  • Therefore, v=2fv = 2f.
  1. Calculation:
  • Since we established v=xv = x, we have x=2fx = 2f.
  • Solving for focal length: f=x/2f = x/2.
  1. Alternative Method (Lens Formula):
  • Magnification m=v/u=−1m = v/u = -1 (for real, inverted, equal size image).
  • So, v=−uv = -u. Given v=xv = x, then u=−xu = -x.
  • Lens formula: 1f=1v−1u\frac{1}{f} = \frac{1}{v} - \frac{1}{u}
  • 1f=1x−1−x=1x+1x=2x\frac{1}{f} = \frac{1}{x} - \frac{1}{-x} = \frac{1}{x} + \frac{1}{x} = \frac{2}{x}
  • f=x/2f = x/2.

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 INSTRUMENTSobjectmountedobtainedparallelconvex

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