Power: A Lens's Bending Ability

A shorter focal length bends light more sharply. The power of a lens quantifies this: it is the tangent of the angle by which the lens deflects a ray arriving parallel at unit height (tan⁡δ=h/f\tan\delta = h/f; with h=1h = 1):

P=1f\boxed{P = \frac{1}{f}}

  • SI unit: dioptre (D) =1 m−1= 1\ \text{m}^{-1} — the power of a 1 m focal-length lens. f must be in metres!
  • P positive for converging (convex), negative for diverging (concave) lenses.
  • The optician's language: a prescription of +2.5 D means a convex lens of f=12.5=0.40f = \frac{1}{2.5} = 0.40 m; −4.0 D means a concave lens of f=−25f = -25 cm.

Power of a lens in dioptres and combination of thin lenses in contact

Lenses in Contact: Powers Simply Add

Place two thin lenses (f1,f2f_1, f_2) in contact. The first forms an image at v1v_1 (1v1−1u=1f1\frac{1}{v_1} - \frac{1}{u} = \frac{1}{f_1}), which acts as the object for the second (1v−1v1=1f2\frac{1}{v} - \frac{1}{v_1} = \frac{1}{f_2}). Adding:

1v−1u=1f1+1f2≡1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f_1} + \frac{1}{f_2} \equiv \frac{1}{f}

So for any number of thin lenses in contact:

1f=1f1+1f2+⋯P=P1+P2+⋯m=m1m2⋯\boxed{\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} + \cdots \qquad P = P_1 + P_2 + \cdots \qquad m = m_1 m_2 \cdots}

Powers add algebraically (signs included!) — the reason opticians work in dioptres rather than focal lengths. Combinations let designers tune power and sharpen images (multi-element camera lenses, instrument eyepieces).

[JEE Tip] For lenses separated by distance d (not in contact), process them sequentially: image of lens 1 (shifted by d) becomes the object of lens 2 — exactly NCERT Example 9.8's three-lens chain. A real image that lands beyond the next lens acts as that lens's virtual object (positive u). The chain method never fails; memorised 'd-formulas' often do.

Solved Examples

Example 1: Power of a half-metre lens (NCERT Example 9.7i) [NEET Numerical]

Find the power of a glass lens of focal length 0.5 m.

Solution:

  1. P=1f(in metres)=10.5P = \frac{1}{f(\text{in metres})} = \frac{1}{0.5}.
  2. P=+2P = +2 dioptre — converging.

Example 2: Reading a prescription [NEET Numerical]

What lens does a −4.0 D prescription specify? And +2.5 D?

Solution:

  1. f=1/Pf = 1/P: −4.0-4.0 D gives f=−0.25f = -0.25 m — a concave lens of 25 cm focal length.
  2. +2.5+2.5 D gives f=+0.40f = +0.40 m — a convex lens of 40 cm.
  3. Dioptres are just reciprocal metres with the lens type encoded in the sign.

Example 3: Two lenses in contact [NEET Numerical]

A convex lens (f1=+30f_1 = +30 cm) is in contact with a concave lens (f2=−20f_2 = -20 cm). Find the combination's focal length and power.

Solution:

  1. 1f=130−120=2−360=−160\frac{1}{f} = \frac{1}{30} - \frac{1}{20} = \frac{2-3}{60} = -\frac{1}{60}.
  2. f=−60f = -60 cm: the pair diverges.
  3. P=100−60≈−1.7P = \frac{100}{-60} \approx -1.7 D. (Equivalently P1+P2=3.33−5=−1.67P_1 + P_2 = 3.33 - 5 = -1.67 D ✓.)

Example 4: Power bookkeeping [JEE Numerical]

Three thin lenses of powers +2 D, +3 D and −5 D are in contact. Describe the combination.

Solution:

  1. P=2+3−5=0P = 2 + 3 - 5 = 0 D.
  2. f→∞f \to \infty: the stack behaves as a plane glass sheet — light passes undeviated (in direction).
  3. Zero-power combinations are how instrument makers cancel aberrations while keeping geometry.

Example 5: NCERT's three-lens chain (Example 9.8) [JEE Numerical]

Lenses of f1=+10f_1 = +10 cm, f2=−10f_2 = -10 cm, f3=+30f_3 = +30 cm stand 5 cm apart in sequence. An object sits 30 cm before the first lens. Find the final image.

Solution:

  1. Lens 1: 1v1=110−130⇒v1=+15\frac{1}{v_1} = \frac{1}{10} - \frac{1}{30} \Rightarrow v_1 = +15 cm. This image sits 15−5=1015 - 5 = 10 cm beyond lens 2: a virtual object, u2=+10u_2 = +10 cm.
  2. Lens 2: 1v2=−110+110=0⇒v2→∞\frac{1}{v_2} = -\frac{1}{10} + \frac{1}{10} = 0 \Rightarrow v_2 \to \infty — parallel rays emerge.
  3. Lens 3: parallel input focuses at its focus: v3=+30v_3 = +30 cm.
  4. Final image: 30 cm beyond the third lens. The chain method, step by step.

Example 6: Equivalent power of the eye-plus-spectacle [NEET Numerical]

An eye of power +60 D wears a −2 D spectacle lens (treat as in contact). What is the corrected power and focal length?

Solution:

  1. P=60−2=+58P = 60 - 2 = +58 D.
  2. f=158f = \frac{1}{58} m ≈1.72\approx 1.72 cm.
  3. Spectacles work by adding power — dioptres make eyewear arithmetic trivial.

Example 7: From radii to dioptres [JEE Numerical]

Find the power of a double-convex lens (n = 1.5, radii 10 cm and 15 cm).

Solution:

  1. From Section 6: f=+12f = +12 cm =0.12= 0.12 m.
  2. P=10.12≈+8.3P = \frac{1}{0.12} \approx +8.3 D.
  3. Small focal lengths mean big dioptres — instrument objectives run to tens of D.

Example 8: What does 1 dioptre mean physically?

Interpret P = 1 D via the bending-angle definition.

Solution:

  1. P=tan⁡δP = \tan\delta for a ray arriving parallel at unit height (1 m) from the axis.
  2. A 1 D lens deflects such a ray by tan⁡δ=1/f=1\tan\delta = 1/f = 1 per metre of height — i.e. it converges it to cross the axis 1 m downstream.
  3. Power literally measures deflection per unit ray height — bending ability, not 'strength of glass'.

Example 9: Cancelling a lens [NEET Numerical]

What lens, placed in contact with a +5 D lens, makes the pair behave like a plain sheet? Its focal length?

Solution:

  1. Need P1+P2=0P_1 + P_2 = 0: P2=−5P_2 = -5 D.
  2. f2=−15f_2 = -\frac{1}{5} m =−20= -20 cm — a concave lens.
  3. Equal and opposite powers in contact annul each other — the principle behind achromatic doublets (different materials, cancelling dispersions).

Example 10: Magnification through a chain

Two lenses in contact produce magnifications m1=−2m_1 = -2 and m2=+0.5m_2 = +0.5 in sequence. What is the overall magnification and image character?

Solution:

  1. m=m1m2=−2×0.5=−1m = m_1 m_2 = -2 \times 0.5 = -1.
  2. The final image is inverted, same size as the object.
  3. Multiplying magnifications (with signs) is general for any lens train — microscopes (Section 9) live on this rule.