10 min readConcept sheetA 3-mark derivation with a ray diagram and 2-mark numericals; also a JEE favourite

Refraction at a spherical surface

A curved surface between two media forms an image given by n₂/v − n₁/u = (n₂ − n₁)/R. It is the building block of the lens maker's formula.

Formulan₂/v − n₁/u = (n₂ − n₁)/R. Light goes from medium n₁ (where the object is) into medium n₂

Signsu, v and R are measured from the pole of the surface. R is positive when the centre of curvature lies on the side the light goes to

AssumptionsPoint object on the principal axis, small aperture and paraxial rays, so sin θ ≈ tan θ ≈ θ

Plane surfaceR = ∞ gives n₂/v = n₁/u: the apparent depth result

The set-up

A spherical refracting surface is part of a sphere that separates two transparent media. Let it be convex towards the object, with centre of curvature C and radius R. The point object O is on the principal axis in medium n₁. The second medium, n₂, is denser (n₂ > n₁). The ray along the axis passes straight through. A ray from O that strikes the surface at N bends towards the normal NC. It meets the axis at I, which is a real image. Let M be the foot of the perpendicular from N to the axis. For paraxial rays, M is almost at the pole, so distances from M equal distances from the pole.

Derivation: n₂/v − n₁/u = (n₂ − n₁)/R

1
Let ∠NOM = α, ∠NCM = β and ∠NIM = γ. For small angles: α ≈ tan α = MN/OM, β ≈ MN/MC and γ ≈ MN/MI.
2
In triangle NOC, the angle of incidence i is an exterior angle: i = α + β.
3
In triangle NCI, β is an exterior angle: β = r + γ, so the angle of refraction r = β − γ.
4
Snell's law for small angles: n₁ sin i = n₂ sin r becomes n₁ i = n₂ r.
5
Substitute: n₁(α + β) = n₂(β − γ), so n₁α + n₂γ = (n₂ − n₁)β.
6
Put in the tangents and cancel MN: n₁/OM + n₂/MI = (n₂ − n₁)/MC.
7
Apply the New Cartesian signs: OM = −u, MI = +v and MC = +R.
8★
So n₁/(−u) + n₂/v = (n₂ − n₁)/R, which gives n₂/v − n₁/u = (n₂ − n₁)/R.

★ key step

How to draw the ray diagram

Draw a horizontal principal axis. Draw an arc bulging towards the left for the surface. Write n₁ on the left and n₂ on the right.
Mark the object O on the axis to the left. Mark the centre of curvature C and the image I on the axis to the right, with C between the surface and I.
Draw a ray from O to a point N on the arc, just above the axis. Draw the normal from C through N as a dotted line, extended beyond N.
Draw the refracted ray from N, bent towards the normal, meeting the axis at I.
Draw NM perpendicular to the axis. Mark α at O, β at C, γ at I, and the angles i and r at N.
Label u = OM, v = MI and R = MC, and put arrows on the rays.

Worked example: images formed by curved surfaces

Problem

(a) A point object in air is 30 cm from a convex glass surface (n = 1.5, R = 10 cm). Find the image. (b) A solid glass sphere (n = 1.5) has a radius of 10 cm. A small air bubble on a diameter is 5 cm from the surface nearest you. Where does it appear? (c) A coin lies 12 cm deep in water (n = 4/3). Use the formula with R = ∞ to find its apparent depth.

Working

(a) n₁ = 1, n₂ = 1.5, u = −30 cm, R = +10 cm (C is on the glass side, where the light goes).
1.5/v = (1.5 − 1)/10 + 1/(−30) = 1/20 − 1/30 = 1/60, so v = 1.5 × 60 = +90 cm.
The image is real, 90 cm inside the glass.
(b) Light goes from glass into air: n₁ = 1.5, n₂ = 1. The bubble is on the incident side, so u = −5 cm.
The surface bulges towards you, so C lies inside the glass, against the light: R = −10 cm.
1/v − 1.5/(−5) = (1 − 1.5)/(−10), so 1/v + 0.3 = 0.05. Then 1/v = −0.25 and v = −4 cm.
The bubble appears 4 cm inside the surface: a virtual image, nearer than the bubble really is.
(c) n₁ = 4/3 (water), n₂ = 1 (air), u = −12 cm, R = ∞: 1/v = (4/3)/(−12) = −1/9, so v = −9 cm.

Answer

(a) Real image 90 cm inside the glass. (b) Virtual image 4 cm inside the surface. (c) The coin appears 9 cm deep

Mirror or refracting surface: where does the real image form?

Spherical mirrorLight turns back after reflection. A real image forms on the same side as the object, so v is negative. A virtual image forms behind the mirror, with v positive.

Refracting surface or lensLight carries on into the second medium. A real image forms on the other side, so v is positive. A virtual image forms on the object's side, with v negative.

In every case the sign tells you only which side the image is on, measured along the incident light.

In real life

A glass paperweight or a round fish bowl makes things inside look bigger, because each curved surface refracts light by this rule. The front of your eye, the cornea, is a curved refracting surface that does most of the eye's focusing.

Exam angle

3 marks: “Derive n₂/v − n₁/u = (n₂ − n₁)/R for refraction at a convex spherical surface separating media n₁ and n₂ (n₂ > n₁), when the image is real.” Ray diagram; α, β and γ as tangents; exterior angles give i = α + β and r = β − γ; Snell's law for small angles, n₁ i = n₂ r; substitute and cancel MN; signs OM = −u, MI = +v, MC = +R; result.

2 marks: “A point object in air is 10 cm from a concave spherical glass surface (n = 1.5) of radius of curvature 20 cm. Find the position of the image.” n₁ = 1, n₂ = 1.5, u = −10 cm, R = −20 cm. 1.5/v = 0.5/(−20) + 1/(−10) = −0.025 − 0.1 = −0.125, so v = −12 cm: a virtual image 12 cm from the surface, on the object's side.

1 marks: “What does the formula n₂/v − n₁/u = (n₂ − n₁)/R become for a plane surface?” With R = ∞, n₂/v = n₁/u, so v = (n₂/n₁)u: the real and apparent depth relation.

Remember it

After over v, minus before over u, equals (after − before) over R: n(after)/v − n(before)/u = [n(after) − n(before)]/R.