How mirrors and lenses bend and focus light predictably enough to build a camera, a microscope, and your own eye — treating light as simple straight-line rays.
This chapter uses ray optics — treating light as straight lines, ignoring its wave nature (that comes back in the next chapter). A single, consistent sign convention makes every mirror and lens problem solvable with the same two formulas.
Light bends when it crosses a boundary between media of different optical density, because its speed changes while its frequency stays fixed.
Going from a denser to a rarer medium, the refracted ray bends further from the normal as the angle of incidence increases. Past a certain angle — the critical angle — the refracted ray would need to bend more than 90°, which is impossible; instead, all the light reflects back into the denser medium.
Applying the spherical-surface refraction equation twice — once at each face of the lens — gives a formula for focal length purely in terms of the lens's shape and material:
A ray passing through a prism bends twice — once at each face — and the total deviation depends on the angle of incidence, reaching a minimum value at one specific angle.
Refractive index depends slightly on wavelength — violet light bends more than red. A prism therefore splits white light into its constituent colours (VIBGYOR), spread out by angle since each wavelength deviates by a different amount.
The eye's lens adjusts its focal length (accommodation) to focus objects at different distances onto the retina. The near point (closest comfortable focus) is conventionally taken as 25 cm; the far point of a normal eye is at infinity.
An object is placed 30 cm in front of a concave mirror of focal length 10 cm. Find the image position and magnification.
Solution: u = −30 cm, f = −10 cm. Using 1/v = 1/f − 1/u = −1/10 − (−1/30) = −3/30 + 1/30 = −2/30, so v = −15 cm.
m = −v/u = −(−15)/(−30) = −0.5. The image is real, inverted, and half the object's size, forming 15 cm in front of the mirror.
A convex lens of focal length 20 cm forms an image of an object placed 30 cm from the lens. Find the image position and magnification.
Solution: u = −30 cm, f = +20 cm. Using 1/v = 1/f + 1/u = 1/20 − 1/30 = (3−2)/60 = 1/60, so v = +60 cm.
m = v/u = 60/(−30) = −2. The image is real, inverted, twice the object's size, formed 60 cm on the far side of the lens.
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