Q9
1 markMCQSection A

A concave lens of focal length 10 cm is cut into two identical plano-concave lenses. The focal length of each lens will be

Ray Optics and Optical Instruments
Lens Maker's Formula — Cutting a Lens

Options

(A)20 cm
(B)30 cm
(C)40 cm
(D)5 cm
Official Answer

Correct option: (A) 20 cm


A symmetric biconcave lens (radii R each) has:


  • 1/f=(n−1)(1/R1−1/R2)=(n−1)(−1/R−1/R)=−2(n−1)/R1/f = (n-1)(1/R_1 - 1/R_2) = (n-1)(-1/R - 1/R) = -2(n-1)/R.
  • Cutting perpendicular to the principal axis gives two plano-concave lenses (one flat surface, one concave of radius R): 1/f′=(n−1)(−1/R)=−(n−1)/R1/f' = (n-1)(-1/R) = -(n-1)/R.
  • Therefore f′=2f=2×10=20 cmf' = 2f = 2 \times 10 = 20\text{ cm} in magnitude.
plano-concave lenslens maker's formulafocal length doublescutting a lenspower halvedbiconcave lens20 cmrefraction spherical surface

Marking Scheme

  • 11 mark: correct option (A) 20 cm.
  • 2Key reasoning: removing one curved surface halves the power, doubling the focal length: f′=2ff' = 2f.

Hint

Each cut piece keeps only one curved surface, so its power is halved — what happens to focal length?

Quick Oral Answer

When a symmetric biconcave lens is cut across its middle, each half keeps only one curved surface, so its power is halved and its focal length doubles from 10 cm to 20 cm.

Analysis & Explanation

This applies the lens maker's formula to a lens that is cut, a favourite CBSE conceptual MCQ.


Concept: For a symmetric biconcave lens the two surfaces contribute equally. Using the lens maker's formula with R1=−RR_1 = -R and R2=+RR_2 = +R:

  • 1/f=(n−1)(1/R1−1/R2)=−2(n−1)/R1/f = (n-1)(1/R_1 - 1/R_2) = -2(n-1)/R.

Cutting the lens through a plane containing the principal axis... no — cutting it perpendicular to the principal axis (across the middle) removes one curved surface from each half, leaving a plane face and a single concave face:

  • 1/f′=(n−1)(1/∞−1/R)=−(n−1)/R=12(1/f)1/f' = (n-1)(1/\infty - 1/R) = -(n-1)/R = \frac{1}{2}(1/f).
  • Hence f′=2f=20 cmf' = 2f = 20\text{ cm}.

Why (A) is correct: each half has only one refracting curved surface, so its power is halved and its focal length doubles.


Why the others are wrong:

  • (D) 5 cm assumes the focal length halves — the opposite of what removing a surface does.
  • (B) 30 cm and (C) 40 cm correspond to tripling/quadrupling, which has no physical basis here.

Exam trap: Contrast this with cutting a lens along the principal axis (into two half-lenses), which leaves the focal length unchanged.

Common Mistakes

  1. 1Assuming the focal length halves to 5 cm — cutting off one curved surface reduces power, so focal length increases, not decreases.
  2. 2Confusing a cut perpendicular to the axis (plano-concave, f doubles) with a cut along the axis (half lens, f unchanged).
  3. 3Ignoring the sign/geometry and treating a symmetric lens surface contribution as unequal.

Interesting Facts

Because power P=1/fP = 1/f is additive, joining the two plano-concave halves back-to-back reproduces the original 10 cm lens: 20 cm and 20 cm in contact give 10 cm.

The same halving-of-power idea explains why a plano-convex lens has exactly twice the focal length of an equivalent symmetric biconvex lens of the same glass and radius.

Spotted a mistake or something unclear?

Tell us — we fix reported answers fast.

Frequently Asked Questions

Why does the focal length double when the lens is cut into plano-concave halves?

The lens maker's formula sums the contributions of both surfaces. A symmetric biconcave lens has two equal curved surfaces; each plano-concave half retains only one, so its power (1/f1/f) is exactly halved and its focal length doubles.

What if the lens were cut along the principal axis instead?

Cutting along the principal axis produces two half-lenses that each keep both curved surfaces, so the focal length stays the same (10 cm); only the aperture is reduced.