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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Geometrical Optics

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Geometrical Optics

To account for the effect of the lens thickness t , four...

To account for the effect of the lens thickness tt, four modifications to the lens formula are proposed. Which of these is most likely correct?

Options

  1. A1v−t−1u+t=1f\dfrac{1}{v - t} - \dfrac{1}{u + t} = \dfrac{1}{f}
  2. B1v−1u=tuf\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{t}{uf}
  3. Ctv2−1u=1f\dfrac{t}{v^2} - \dfrac{1}{u} = \dfrac{1}{f}
  4. D1v−1u+tuv=tf\dfrac{1}{v} - \dfrac{1}{u} + \dfrac{t}{uv} = \dfrac{t}{f}

Correct answer

A. 1v−t−1u+t=1f\dfrac{1}{v - t} - \dfrac{1}{u + t} = \dfrac{1}{f}

Step-by-step solution

For a modified lens formula to be correct, it must reduce to the standard thin lens formula when the thickness of the lens approaches zero (t→0t \to 0).

The standard thin lens formula is given by:
1v−1u=1f\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}

Substituting t=0t = 0 into the first expression gives 1v−0−1u+0=1v−1u=1f\dfrac{1}{v - 0} - \dfrac{1}{u + 0} = \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}, which matches the standard formula.

Substituting t=0t = 0 into the second expression gives 1v−1u=0\dfrac{1}{v} - \dfrac{1}{u} = 0, which is incorrect.

Substituting t=0t = 0 into the third expression gives −1u=1f-\dfrac{1}{u} = \dfrac{1}{f}, which is incorrect.

Substituting t=0t = 0 into the fourth expression gives 1v−1u=0\dfrac{1}{v} - \dfrac{1}{u} = 0, which is incorrect.

Additionally, the fourth expression is dimensionally incorrect as the right-hand side tf\dfrac{t}{f} is dimensionless, while the left-hand side has dimensions of [L]−1[L]^{-1}.

Therefore, only the first modification is physically consistent.
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