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Unit IX · Ray optics and optical instruments
10 min readConcept sheet1-mark MCQ and assertion-reason, 2-mark numericals on power, and a 3-mark derivation for thin lenses in contact; also a NEET favourite

Magnification, power of a lens and lenses in contact

A lens magnifies by m = v/u. Its power is P = 1/f, in dioptres. For thin lenses in contact, powers add and magnifications multiply.

Magnification of a lensm = h′/h = v/u. Negative m: real, inverted image. Positive m: virtual, erect image
Power of a lensP = 1/f, with f in metres. SI unit: dioptre (D), where 1 D = 1 m⁻¹. Converging lens: P positive; diverging lens: P negative
Thin lenses in contact1/f = 1/f₁ + 1/f₂ + ...; P = P₁ + P₂ + ...; m = m₁ × m₂ × ...

Magnification and power

The linear magnification of a lens is m = h′/h = v/u. Watch the sign: a mirror has m = −v/u, but a lens has m = +v/u. The power of a lens measures how strongly it converges or diverges light. It is defined as P = 1/f, with f in metres. Its SI unit is the dioptre (D): a lens of focal length 1 m has a power of 1 D. A convex lens has positive power, and a concave lens has negative power. A shorter focal length means a stronger lens and a larger power. Opticians write spectacle prescriptions in dioptres.

Worked example: thin lens formula and magnification

Problem

(a) A 3 cm tall object is 30 cm from a convex lens of focal length 20 cm. Find the position, size and nature of the image. (b) Repeat for a concave lens of focal length 20 cm. (c) Find the power of each lens.

Working
  1. (a) u = −30 cm, f = +20 cm. 1/v = 1/f + 1/u = 1/20 − 1/30 = 1/60, so v = +60 cm.
  2. m = v/u = 60/(−30) = −2, so h′ = −2 × 3 cm = −6 cm.
  3. The image is real, inverted and 6 cm tall, 60 cm from the lens on the far side.
  4. (b) u = −30 cm, f = −20 cm. 1/v = −1/20 − 1/30 = −5/60 = −1/12, so v = −12 cm.
  5. m = v/u = (−12)/(−30) = +0.4, so h′ = +1.2 cm.
  6. The image is virtual, erect and smaller, 12 cm from the lens on the object's side.
  7. (c) Convex lens: P = 1/(+0.20 m) = +5 D. Concave lens: P = 1/(−0.20 m) = −5 D.
Answer

(a) v = +60 cm; 6 cm tall, real and inverted. (b) v = −12 cm; 1.2 cm tall, virtual and erect. (c) +5 D and −5 D

Derivation: two thin lenses in contact

Two thin lenses A and B, of focal lengths f₁ and f₂, touch each other on a common axis. A point object O is at distance u.

  1. 1
    Lens A alone forms an image I₁ at distance v₁: 1/v₁ − 1/u = 1/f₁. ...(1)
  2. 2
    I₁ acts as the object for lens B. The lenses are thin and touching, so I₁ is also at distance v₁ from B.
  3. 3
    Lens B forms the final image I at distance v: 1/v − 1/v₁ = 1/f₂. ...(2)
  4. 4
    Add (1) and (2): 1/v − 1/u = 1/f₁ + 1/f₂.
  5. 5
    A single equivalent lens of focal length f would form the same image: 1/v − 1/u = 1/f.
  6. 6★
    So, 1/f = 1/f₁ + 1/f₂, and in powers, P = P₁ + P₂. For several lenses, 1/f = 1/f₁ + 1/f₂ + 1/f₃ + ...
  7. 7
    Magnification: m₁ = v₁/u and m₂ = v/v₁, so the total m = m₁ × m₂ = v/u.

★ key step

How to draw the diagram for lenses in contact

  • Draw a principal axis and two thin convex lenses, A and B, touching each other.
  • Mark the object O on the axis on the left.
  • Draw a ray from O to lens A. After A, continue it as a dotted line to I₁, the image that A alone would form.
  • At lens B, bend the ray more steeply so that it meets the axis at I, closer to the lenses than I₁.
  • Label u (from O to the lenses), v₁ (to I₁), v (to I), f₁ and f₂. Put arrows on the rays.

Worked example: lenses in contact

Problem

(a) A convex lens of focal length 20 cm and a concave lens of focal length 40 cm are in contact. Find the focal length and power of the combination. (b) Lenses of power +4 D and −6 D are in contact. Is the combination converging or diverging?

Working
  1. (a) f₁ = +20 cm and f₂ = −40 cm. 1/f = 1/20 − 1/40 = (2 − 1)/40 = 1/40, so f = +40 cm.
  2. P = 1/(0.40 m) = +2.5 D. Check: P₁ + P₂ = +5 D − 2.5 D = +2.5 D.
  3. f is positive, so the pair acts as a converging lens.
  4. (b) P = +4 D − 6 D = −2 D, so f = 1/(−2 D) = −0.5 m = −50 cm.
  5. f is negative, so the combination is diverging.
Answer

(a) f = +40 cm, P = +2.5 D, converging. (b) P = −2 D, f = −50 cm, diverging

Activity: refractive index of a liquid using a convex lens and a plane mirror

The liquid between the lens and the mirror forms a plano-concave liquid lens in contact with the glass lens.

  1. 1
    Put a plane mirror on the base of a stand and a convex lens on it. Move a needle above the lens until the needle and its inverted image coincide with no parallax.
  2. 2
    Light now returns along its own path from the plane mirror, so the needle is at the focus. Measure f₁, the focal length of the glass lens.
  3. 3
    Put a few drops of the liquid on the mirror and replace the lens. Repeat to find F, the focal length of the glass lens and liquid lens together.
  4. 4
    Find the liquid lens from 1/f₂ = 1/F − 1/f₁. It comes out negative, because F is longer than f₁.
  5. 5
    Measure the radius R of the lens face touching the liquid with a spherometer. For the plano-concave liquid lens, 1/|f₂| = (n − 1)/R, so n = 1 + R/|f₂|.
  6. 6
    Result (example): f₁ = 20 cm, F = 30 cm and R = 20 cm give 1/f₂ = 1/30 − 1/20 = −1/60. So f₂ = −60 cm and n = 1 + 20/60 ≈ 1.33, the value for water.
In real life

An eye-test report gives lens power in dioptres: −1.5 D is a concave lens for short sight, and +2.0 D a convex lens for reading. Phone cameras combine several lenses to get the power and sharpness they need.

Exam angle

3 marks: “Derive the expression for the equivalent focal length of two thin lenses in contact. Write the power of the combination.” Lens A: 1/v₁ − 1/u = 1/f₁. I₁ is the object for B, also at v₁: 1/v − 1/v₁ = 1/f₂. Adding: 1/v − 1/u = 1/f₁ + 1/f₂ = 1/f. Power P = P₁ + P₂.

2 marks: “Two thin lenses of power +10 D and −5 D are placed in contact. Find the focal length of the combination and say whether it converges or diverges light.” P = +10 D − 5 D = +5 D. f = 1/5 m = 20 cm. Positive, so it is converging.

1 marks: “Define 1 dioptre.” 1 dioptre is the power of a lens of focal length 1 metre.

Remember it

Powers add, magnifications multiply.