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JEE Advanced 2006
Exam
JEE Advanced
Year
2006
Subject
Physics
Chapter
Ray Optics

JEE Advanced 2006 · Physics · Ray Optics

The graph between object distance u and image distance v for a lens...

The graph between object distance uu and image distance vv for a lens is given below. The focal length of the lens is The graph between object distance u and image distance v for a lens is given below. The focal length of the lens is

Options

  1. A5±0.15 \pm 0.1
  2. B5±0.055 \pm 0.05
  3. C0.5±0.10.5 \pm 0.1
  4. D0.5±0.050.5 \pm 0.05

Correct answer

B. 5±0.055 \pm 0.05

Step-by-step solution

From the lens formula: 1f=1v−1u\frac{1}{f}=\frac{1}{v}-\frac{1}{u} we have, 1f=110−1−10\frac{1}{f}=\frac{1}{10}-\frac{1}{-10} or f=+5f=+5 Further, Δu=0.1\Delta u=0.1 and Δv=0.1\Delta v=0.1 (from the graph) Now, differentiating the lens formula we have, Δff2=Δvv2+Δuu2 or Δ\frac{\Delta f}{f^2}=\frac{\Delta v}{v^2}+\frac{\Delta u}{u^2} \text{ or } \Deltaf=\left(\frac{\Delta v}{v^2}+\frac{\Delta u}{u^2}\right)f2Substitutingthevalueswehave,f^2 Substituting the values we have,\Deltaf=(0.1102+0.1102)(f=\left(\frac{0.1}{10^2}+\frac{0.1}{10^2}\right)(5)^2 \Rightarrow f ±Δ\pm \Deltaf=5 ±\pm0.05
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