Light propagation and reflection

In darkness, a book disappears from view even though it remains on the table. Switch on a lamp: its light reflects from the book into the eye. This route helps us understand shadows, mirrors and where we see an image.

A lamp and the Sun emit their own light; an ordinary book is seen by reflected light. To see an object, light must reach the eye from it: the eye does not send rays to probe the book. A ray is an ideal directed line showing how light propagates. It models a route, not a separate material rod. Many rays reach and leave any one point on an object.

Put an opaque obstacle between lamp and eye: it blocks the direct route. In a uniform transparent medium, geometrical optics describes light with straight rays. This works when obstacles and openings are much larger than the wavelength. At a boundary, direction may change; near very small openings, a straight-ray picture is insufficient. We simplify the wave picture while keeping what matters for the problem.

A small source is nearly a point: behind an opaque object lies a shadow region that direct rays cannot reach. Now imagine enlarging the lamp. From some places behind the obstacle, only part of the source is visible: some light arrives there, forming penumbra. In the umbra, the whole source is hidden. The softness of the shadow edge therefore depends on the source's size and position as well as the object.

Scale the same geometry up to celestial bodies. In a solar eclipse, the Moon lies between the Sun and part of Earth, blocking sunlight. In a lunar eclipse, Earth blocks sunlight travelling to the Moon, which enters Earth's shadow. Moving one body changes which object blocks the light and where the shadow falls. An eclipse needs the right relative alignment, not merely the Moon's presence among the orbits of Earth and Sun.

A ray strikes a mirror. How do we compare its directions before and after? At the point of incidence, draw the normal, perpendicular to the surface. Incidence angle α lies between incident ray and normal; reflection angle β lies between normal and reflected ray. A ray nearly along the mirror makes a small angle with the surface but a large angle with the normal. Without a shared reference line, comparing two numbers can mislead.

Experiments give a simple law: α = β; incident ray, reflected ray and normal lie in one plane. Turn the incident ray farther from the normal, and the reflected ray moves away by the same angle. At α = 0, light returns along its original path. Angles may use degrees or radians, but both start from the normal. Equal incidence and reflection angles do not mean the angle between the two rays equals α.

If a wall reflects light, why does it not show a sharp image like a mirror? Tiny patches of a rough surface have different normals. Each obeys the same reflection law, but reflected directions differ, making the wall visible from many sides. A smooth mirror preserves an ordered ray pattern from each point. Diffuse reflection does not break the law; local surface orientations differ.

Look at one object point in a plane mirror. Its reflected rays diverge in front of the mirror and enter the eye. The eye associates them with a point they seem to have come from; their backward extensions meet behind the mirror. This is a virtual image: actual rays do not pass through that point behind the mirror. 'Virtual' does not mean no light or an invisible image; it distinguishes actual ray geometry from the geometry of extensions.

Constructing two reflected rays shows that the virtual point lies as far behind the mirror as the object is in front, measured perpendicularly. The whole image is upright and the same size. Move the object twice as close: its image also comes twice as close to the mirror plane. A screen behind the mirror cannot catch this image: only our dashed extensions meet there, not light from the object.

A ray strikes at 25° to the mirror surface. Find its reflection angle measured from the normal.

The given 25° is measured from the surface. First find the angle to the perpendicular normal: α = 90° − 25° = 65°.

By α = β, the reflection angle is β = 65°.

Check the diagram: the reflected ray makes 90° − 65° = 25° with the surface on the other side of the normal. The symmetry agrees with the law.

The reflection angle from the normal is 65°. The reflected ray's angle to the surface remains 25°.

We see objects through light entering the eye; the ray model traces its route.

Umbra and penumbra depend on how much of the source an obstacle hides.

The reflection law uses angles from the normal; a virtual image lies where ray extensions meet.

At a transparent boundary, some light can continue in a different direction.