Refraction at a spherical surface
One curved surface
A lens has two curved surfaces. To understand it, first take just one: a curved boundary between two media, like the end of a glass rod rounded into part of a sphere.
The medium on the left has index n₁, the medium on the right n₂. The surface has its pole at M and its centre of curvature at C, a distance R away.
A point object O sits on the axis. One ray goes along the axis and passes straight through. Another hits the surface at N, close to the axis.
At N, the normal is the radius NC. The ray refracts towards this normal and crosses the axis at I. Every paraxial ray from O crosses at the same point, so I is the image of O.
Because the angles are small, each angle equals its tangent. The angle at O is MN over OM, the angle at C is MN over MC, and the angle at I is MN over MI.
From the triangle NOC, the angle of incidence i is the angle at O plus the angle at C. From the triangle NCI, the angle of refraction r is the angle at C minus the angle at I.
Snell’s law for small angles says n₁ i = n₂ r. Substituting and cancelling MN gives n₁ over OM, plus n₂ over MI, equals n₂ minus n₁, over MC.
Now the signs: OM = −u, MI = +v and MC = +R. This gives the result: n₂/v − n₁/u = (n₂ − n₁)/R.
It holds for any spherical surface, convex or concave, if we use the signs. R is positive when the centre of curvature is on the side the light goes to, and negative when it is on the side the light comes from.
Images through a curved surface
Example. A point source in air is 100 centimetres from a convex glass surface of radius 20 centimetres. The glass has n = 1.5. Where is the image?
Signs: u = −100 centimetres and R = +20 centimetres, because the centre of curvature is inside the glass, on the far side.
Substitute: 1.5 over v, plus 1 over 100, equals 0.5 over 20. So 1.5 over v equals 0.025 minus 0.010, which is 0.015.
v = 1.5 ÷ 0.015 = +100 centimetres. A real image forms 100 centimetres inside the glass, on the far side of the surface.
If the object were very far away, u would go to infinity and the image would form at the second focus: f₂ = n₂R over (n₂ − n₁) = 1.5 × 20 ÷ 0.5 = 60 centimetres.
Example two: a fish swims at the very centre of a round fish bowl of radius 15 centimetres. Light from the fish goes from water, n = 1.33, into air.
The light travels outwards, and the centre of the bowl is behind the surface, on the side the light comes from. So u = −15 and R = −15 centimetres.
1 over v equals −0.33 over −15, plus 1.33 over −15. That is 0.022 minus 0.089, which is −0.067. So v = −15 centimetres: the fish appears exactly where it is.
That makes sense: every ray from the centre travels along a radius and hits the surface along the normal, so it is not bent. But the fish looks bigger: m = n₁v over n₂u = 1.33.
Try it: a curved glass surface
- This is Example one: u = −100 cm, R = +20 cm, n₂ = 1.5. The image is real, at v = +100 cm.
- Move the object closer. When it passes F₁ (40 cm from the surface), the image becomes virtual.
- The blue ray goes through the centre of curvature C. It meets the surface along the normal, so it does not bend.
- Make R negative (surface concave towards the object). Now the surface spreads the rays out.
Where is the image?
A point object in air is 80 cm from a convex spherical glass surface (n = 1.5, R = 20 cm). Where is the image?
- 120 cm inside the glass
- 80 cm inside the glass
- 120 cm in front of the surface, in air
- 60 cm inside the glass
Answer
A. n₂/v − n₁/u = (n₂ − n₁)/R with u = −80 cm, R = +20 cm: 1.5/v + 1/80 = 0.5/20 = 0.025. So 1.5/v = 0.025 − 0.0125 = 0.0125, and v = +120 cm: real, inside the glass. (80 cm comes from forgetting the n₂ in n₂/v; 60 cm is the focal length f₂, the image of a very distant object.)
The fish at the centre
A small fish is at the centre of a spherical fish bowl of radius 15 cm filled with water (n = 1.33). Seen from outside, where does the fish appear, and how big?
- At the centre, 1.33 times bigger
- At the centre, the same size
- 11.3 cm from the glass, the same size
- 20 cm from the glass, 1.33 times smaller
Answer
A. With u = −15 cm, R = −15 cm, n₁ = 1.33, n₂ = 1: 1/v = (1 − 1.33)/(−15) + 1.33/(−15) = 0.022 − 0.089 = −0.067, so v = −15 cm: at the centre. m = n₁v/(n₂u) = 1.33 × (−15)/(1 × (−15)) = 1.33. (11.3 cm is the flat-surface answer 15/1.33; a curved surface behaves differently.)
Summary
- Refraction at a spherical surface:
n₂/v − n₁/u = (n₂ − n₁)/R - Focal lengths:
f₂ = n₂R/(n₂ − n₁), f₁ = −n₁R/(n₂ − n₁) - Magnification:
m = n₁v / (n₂u)
- Derived for paraxial rays using small angles and n₁ i = n₂ r.
- R is positive if the centre of curvature is on the side the light goes to (outgoing side), negative otherwise.
- A ray through the centre of curvature meets the surface along the normal and is not bent.
- A lens is two such surfaces one after the other: this result leads straight to the lens maker’s formula.