(a) Using the relation for refraction at a curved spherical surface, derive the expression for lens maker's formula.

(b) Three lenses $L_1$ , $L_2$ and $L_3$ , each of focal length 40 cm, are placed coaxially. The distance between $L_1$ and $L_2$ and between $L_2$ and $L_3$ are 120 cm and 20 cm respectively. An object is kept at a distance of 80 cm to the left of lens $L_1$ . Find the distance of the final image formed from the object.

OR

(a) Draw a ray diagram to show the image formation by a concave mirror when the object is kept between its focus and the centre of curvature. Using this diagram, derive the mirror formula.

(b) A concave mirror produces a two times magnified virtual image of an object kept 10 cm in front of it. Calculate the focal length of the mirror.

Ray Optics and Optical Instruments

Lens Maker's Formula and Multi-Lens Image Formation

Official Answer

This 5-mark answer derives the lens maker's formula and then tracks an object through three coaxial lenses.

Part (a) — Lens maker's formula

A thin lens is treated as two refracting spherical surfaces. For refraction at a single spherical surface separating media n₁ and n₂:

$n_2/v - n_1/u = (n_2 - n_1)/R$

Surface 1 (radius R₁), object in air ( $n_1 = 1$ ) forms an intermediate image at v₁: $n_2/v_1 - 1/u = (n_2 - 1)/R_1$ .
Surface 2 (radius R₂), this image acts as object for refraction from glass (n₂) back to air: $1/v - n_2/v_1 = (1 - n_2)/R_2$ .

Adding the two equations, the v₁ terms cancel:

$1/v - 1/u = (n_2 - 1)(1/R_1 - 1/R_2)$

Comparing with $1/v - 1/u = 1/f$ gives the lens maker's formula:

$1/f = (n - 1)(1/R_1 - 1/R_2)$ (with $n = n_2/n_1$ ).

Part (b) — Three-lens system (each $f = +40\text{ cm}$ )

L₁: $u = -80\text{ cm} \to 1/v = 1/40 + 1/(-80) = 1/80 \to v = +80\text{ cm}$ (80 cm right of L₁).
$L_1L_2 = 120\text{ cm}$ , so this image is $120 - 80 = 40\text{ cm}$ left of $L_2$ → $u = -40\text{ cm}$ for L₂.
L₂: $1/v = 1/40 + 1/(-40) = 0 \to v = \infty$ (parallel rays leave L₂).
L₃: parallel incident rays → image at focus → $v = +40\text{ cm}$ (40 cm right of L₃).

Taking $L_1$ at origin: $L_2$ at $+120$ , $L_3$ at $+140$ , final image at $+180\text{ cm}$ . Object at $-80\text{ cm}$ .

Distance of final image from object = $180 - (-80) = 260\text{ cm}$ .

OR alternative (concave mirror): For a $2\times$ magnified virtual image, $m = +2 \to v = -2u = +20\text{ cm}$ ( $u = -10\text{ cm}$ ). Mirror formula $1/v + 1/u = 1/f \to 1/20 - 1/10 = -1/20$ → $f = -20\text{ cm}$ (focal length 20 cm).

lens maker's formularefraction at spherical surfaceradii of curvaturecoaxial lensesimage at infinitysign conventionfinal image distancefocal length

Marking Scheme

11 mark: set up refraction at each surface ( $n_2/v_1 - 1/u = (n_2-1)/R_1$ and $1/v - n_2/v_1 = (1-n_2)/R_2$ ).
21 mark: add and simplify to $1/f = (n-1)(1/R_1 - 1/R_2)$ .
31 mark: L₁ gives $v = +80\text{ cm}$ ; identify object for L₂ as $u = -40\text{ cm}$ .
41 mark: L₂ gives image at infinity; L₃ forms image 40 cm to its right.
51 mark: add distances to get final image 260 cm from the object.
6OR: (a) 2 marks ray diagram + mirror formula derivation, 1 mark sign convention; (b) $m=+2 \to v=+20\text{ cm}$ (1 mark), $f = -20\text{ cm}$ (1 mark).

Hint

Add the two single-surface refraction equations (air→glass, glass→air) to get $1/f = (n-1)(1/R_1-1/R_2)$ . In (b) trace lens by lens; note $L_2$ 's object sits at its focus, and finally add all distances plus the 80 cm to reach the object.

Quick Oral Answer

A thin lens is two spherical surfaces in series; adding their refraction equations gives $1/f = (n-1)(1/R_1-1/R_2)$ . For the three-lens numerical, $L_1$ images at $+80\text{ cm}$ , L₂ (object at its focus) sends rays to infinity, and L₃ focuses them 40 cm behind it — 260 cm from the original object.

Analysis & Explanation

Concept. Part (a) shows that a thin lens is nothing but two spherical refracting surfaces in series; summing their refraction equations produces the lens maker's formula. Part (b) is a classic multi-lens ray-tracing problem where the image of one lens becomes the object of the next.

The elegant middle step. The striking feature of part (b) is that L₂ receives its object at exactly one focal length ( $u = -40\text{ cm} = -f$ ), so it throws the image to infinity and emits a parallel beam. L₃ then focuses that parallel beam precisely at its focal point. Recognising “object at focus → image at infinity → next lens images at its focus” saves a lot of arithmetic.

Exam traps.

Sign convention: with the real-is-positive-to-the-right convention, a real object has u negative; keep track of which lens the distance is measured from.
The final question asks for distance from the object, not from L₃. Students often stop at 40 cm; the correct answer adds up all separations and the initial 80 cm to give 260 cm.

Real-world link. Multi-lens design is exactly how camera lenses, microscopes and telescopes control magnification and aberration; the lens maker's formula lets designers pick glass (n) and curvatures (R₁, R₂) to hit a target focal length.

OR part. A virtual, magnified, erect image from a concave mirror means the object is inside the focus; the positive magnification (+2) fixes the image position before the mirror formula gives $f = -20\text{ cm}$ .

Common Mistakes

1Reporting the final image distance from L₃ (40 cm) instead of from the object (260 cm) as the question demands.
2Sign-convention errors: forgetting that the intermediate image 80 cm right of L₁ becomes an object 40 cm to the LEFT of L₂ ( $u = -40\text{ cm}$ ).
3In the derivation, forgetting to reverse n₁ and n₂ at the second surface (glass → air), which flips the sign of the $(n-1)$ term.

Interesting Facts

The lens maker's formula predicts that a biconvex lens becomes diverging if immersed in a medium denser than its glass — because $(n_{lens}/n_{medium} - 1)$ turns negative.

In part (b), L₂ sitting exactly one focal length from its object turns the beam parallel — the same trick used inside collimators and in the eyepiece of a telescope in normal adjustment.

Camera 'zoom' lenses contain 10–20 individual elements; each interface is designed with the very refraction-at-a-surface relation used to derive this formula.

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Frequently Asked Questions

How is the lens maker's formula derived from refraction at a spherical surface?

A thin lens is modelled as two refracting spherical surfaces. Apply $n_2/v - n_1/u = (n_2-n_1)/R$ at the first surface (air to glass) and again at the second surface (glass to air), using the first image as the object for the second. Adding the two equations cancels the intermediate image distance and yields $1/v - 1/u = (n-1)(1/R_1 - 1/R_2)$ , i.e. $1/f = (n-1)(1/R_1 - 1/R_2)$ .

Why does lens L₂ form its image at infinity in part (b)?

The image from L₁ lands 40 cm to the left of L₂, so the object distance for L₂ is $u = -40\text{ cm}$ , which equals minus its focal length. When an object sits exactly at the focus of a converging lens, the refracted rays emerge parallel, so the image is at infinity. These parallel rays are then focused by L₃ at its own focal point, 40 cm behind it.

Why is the final answer 260 cm and not 40 cm?

40 cm is only the distance of the final image from lens L₃. The question asks for the distance from the object. Measuring from L₁ at origin: L₂ is at 120 cm, L₃ at 140 cm, and the final image at 180 cm; the object is at −80 cm. So the object-to-image distance is $180 - (-80) = 260\text{ cm}$ .