Back to Directory
NEET PHYSICSEasy

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.
Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started