Refraction and lenses

A spoon in a glass appears bent at the water–air boundary. If light can change direction between media, can we choose surfaces that bring rays together to form an image?

Light from the submerged spoon travels through water and then air. Its speed differs in these media, and at oblique incidence its direction usually changes too: refraction occurs. The eye traces arriving directions backwards as straight lines, so the submerged part appears displaced. At normal incidence, symmetry leaves the direction unchanged even though speed changes. A speed change therefore need not produce a turn in every case.

Entering water from air, a ray bends towards the normal; on emerging, a transmitted ray bends away from it. Why is a link with speed reasonable? At oblique entry, one part of a light wave reaches the slower medium first, changing the orientation of a surface of equal wave phase and hence the propagation direction. A medium with lower light speed is called optically denser. This describes light in the medium, not a simple comparison of the masses of a kilogram of water and glass.

One plane boundary changes a ray's direction once. A lens has two refracting surfaces: the transparent body redirects rays on entry and exit. Curvature gives different parts of the beam different normals. In the thin-lens model, thickness is small compared with the main distances in the problem, and ray constructions concern rays near the principal axis. This replaces two-stage refraction with simple rules.

Shine a beam parallel to the principal axis onto a converging lens. In this model, outgoing rays meet at one point, the principal focus; a screen there can receive a bright spot. The principal axis passes through the optical centre. Send a parallel beam from the other side and a second principal focus appears. This does not mean light from every object always gathers at a focus: the condition here is an incoming beam parallel to the axis.

A diverging lens acts differently on the same parallel beam: outgoing rays spread apart. Extend them backwards and the extensions meet at the principal virtual focus in front of the lens. It marks where outgoing rays seem to originate; the light emerging from the lens does not gather there. As with a mirror, distinguish actual intersection from apparent intersection: ‘virtual’ describes the construction, not an absence of light.

For a thin lens, the magnitude of focal length is the distance from optical centre to principal focus, measured in metres. Calculations assign a sign: F > 0 for a converging lens and F < 0 for a diverging lens. A negative F is not a negative length measured with a ruler; the sign encodes the lens action. A nearer focus means the lens changes ray directions more strongly at equal distances from the axis.

It is useful to describe lens action with a quantity that grows as focal length decreases: optical power D = 1/F, with F necessarily in metres. The unit is the dioptre: 1 дптр = 1 м⁻¹. Halving |F| doubles |D|; D and F have the same sign. A very distant focus gives D close to zero and only a small change in ray direction. Substituting a literal F = 0 is invalid: it does not describe an ordinary thin lens.

Finding an image does not require drawing every ray. From the object's tip, draw one parallel to the principal axis: its later path follows from the focus definition. After a converging lens it passes through the far principal focus. After a diverging lens, its backward extension passes through the near focus. Keep track of actual light paths versus auxiliary extensions.

Draw a second useful ray from the same tip through the optical centre. In an ideal thin lens it does not change direction; for a real thin lens this is an approximation. If the two outgoing rays actually meet, light from the tip gathers there. This is a real image, which a screen can receive. If only backward extensions meet, the image is virtual, as in the mirror example.

Place the object beyond 2F in front of a converging lens and draw both rays from its tip. They meet across the axis beyond the lens, between F and 2F: the image is inverted and smaller. As the object recedes, the image approaches the focus and shrinks. With the object at 2F, symmetry gives an equal-sized image at 2F on the other side. These properties follow from ray intersections, not just the lens name.

Move the object between F and 2F. The outgoing rays now meet beyond 2F: the real image remains inverted but becomes enlarged. As the object approaches F, the screen must be moved farther away. With the object exactly at the focus, rays from one point emerge parallel and do not meet at a finite distance. This limiting case explains why we cannot keep looking for an ordinary screen image near the lens.

What if the object is closer than F? Rays from its tip diverge after the converging lens. Backward extensions meet on the object side, giving an upright, enlarged, virtual image: the action of a magnifying glass. The lens is still converging: it reduces the divergence but cannot make the beam converge for this object position. A lens name therefore does not guarantee a real image in every situation.

For a real object in front of a diverging lens, outgoing rays diverge and their extensions give an image between the lens and near focus. It is virtual, upright and smaller. As the object approaches the lens, the image also approaches it and becomes closer in size to the object, while remaining smaller in this model. A screen behind the lens cannot receive it: the eye sees the arriving light's direction, not rays actually meeting at the apparent point.

Check optical power for a converging lens: F = 20 см = 0,20 м. Then D = 1/0,20 = +5 дптр. Incorrectly dividing one by 20 gives a reciprocal centimetre, which cannot be called dioptres without conversion. A diverging lens with the same focal-length magnitude would have −5 дптр. Check that a nearer focus means a larger |D| and that the sign matches the lens type.

A thin converging lens has F = 10 см. An object is 5 cm in front of it. Can its image be obtained on a screen behind the lens?

First compare position with the focus: 5 см < 10 см. The object is inside the converging lens's focal length.

Draw a ray parallel to the axis and a ray through the centre. Beyond the lens they diverge; only their backward extensions meet on the object side.

The image is therefore upright, enlarged and virtual. A screen behind the lens cannot receive it because the real rays do not converge there, although the eye can see it through the lens.

No. The image is virtual, upright and enlarged. The lens acts as a magnifying glass and does not form a screen image behind it.

Refraction changes direction at a boundary for oblique passage; a lens redirects rays at two surfaces.

D = 1/F describes lens action using reciprocal focal length in metres, with consistent signs.

Image type depends on actual rays and their extensions, and on the object's position relative to the focus.

Later, precise refraction laws and the lens equation will add distance calculations to these constructions.