Light: reflection, refraction and lenses
Physics · WAEC and JAMB · SS2 and SS3
Geometrical optics is the most drawing-heavy topic in the syllabus and a ray diagram earns marks on its own. WAEC reliably sets a refractive index or critical angle calculation and a mirror or lens formula calculation, plus objectives on image characteristics.
What you need to know
- The laws of reflection state that the incident ray, the reflected ray and the normal at the point of incidence all lie in the same plane, and that the angle of incidence equals the angle of reflection.
- A plane mirror forms an image that is virtual, erect, the same size as the object, laterally inverted and as far behind the mirror as the object is in front. When the mirror rotates through an angle, the reflected ray turns through twice that angle.
- The number of images formed by two plane mirrors inclined at angle theta is n = (360/theta) - 1. Two mirrors at 60 degrees therefore give five images, which is the principle of the kaleidoscope.
- A concave mirror converges light and can form real or virtual images depending on object position; a convex mirror always diverges light and always gives a virtual, erect, diminished image. That is why convex mirrors are used as driving mirrors and in shops: they give a wide field of view.
- For any mirror or lens, 1/f = 1/u + 1/v using the real-is-positive convention, where f = r/2 for a spherical mirror. Magnification m = v/u = image height / object height.
- Concave mirror image positions to know: object beyond C gives a real, inverted, diminished image between F and C; object at C gives a real, inverted, same-size image at C; object between F and C gives a real, inverted, magnified image beyond C; object at F gives an image at infinity; object between F and the pole gives a virtual, erect, magnified image behind the mirror, which is the shaving or make-up mirror.
- Refraction is the change in direction of light as it passes from one medium to another because its speed changes. Light bends towards the normal entering a denser medium and away from the normal entering a less dense medium; a ray along the normal passes straight through undeviated.
- Snell's law states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given pair of media: n = sin i / sin r. The refractive index of glass is about 1.5 and that of water about 1.33.
- Refractive index can also be written as n = speed of light in vacuum / speed in the medium, or as real depth / apparent depth. The apparent-depth relation explains why a swimming pool looks shallower than it is and why a coin in a bowl of water appears raised.
- Total internal reflection happens when light travels from a denser to a less dense medium and the angle of incidence exceeds the critical angle. The critical angle c is given by sin c = 1/n, so for glass of n = 1.5, c is about 42 degrees.
- Total internal reflection explains the shimmering mirage on a hot road, the sparkle of a diamond, the working of optical fibres in telecommunications and endoscopes, and the use of right-angled prisms in periscopes and binoculars.
- A converging (convex) lens is thicker at the centre and brings parallel rays to a real focus; a diverging (concave) lens is thinner at the centre and spreads parallel rays so they appear to come from a virtual focus. Power of a lens P = 1/f with f in metres, measured in dioptres.
- A converging lens gives a real inverted image for any object beyond F, and a virtual, erect, magnified image when the object is inside F, which is the simple magnifying glass. A diverging lens always gives a virtual, erect, diminished image.
- Defects of vision: short sight (myopia) means distant objects focus in front of the retina and is corrected with a diverging lens; long sight (hypermetropia) means near objects focus behind the retina and is corrected with a converging lens. In the camera the image is real, inverted and diminished on the film or sensor, and focusing is done by moving the lens, while the eye focuses by changing the thickness of its lens.
Key terms
- Real image
- An image formed by the actual intersection of light rays, which can be caught on a screen.
- Virtual image
- An image formed where light rays only appear to intersect, which cannot be caught on a screen.
- Principal focus
- The point on the principal axis to which rays parallel and close to the axis converge, or from which they appear to diverge, after reflection or refraction.
- Refractive index
- The ratio of the sine of the angle of incidence in vacuum or air to the sine of the angle of refraction in the medium.
- Critical angle
- The angle of incidence in the denser medium for which the angle of refraction in the less dense medium is 90 degrees.
- Total internal reflection
- The complete reflection of light back into a denser medium when the angle of incidence exceeds the critical angle.
- Power of a lens
- The reciprocal of the focal length in metres, measured in dioptres.
Formulae
angle of incidence = angle of reflectionnumber of images in inclined mirrors: n = (360/theta) - 1mirror and lens formula: 1/f = 1/u + 1/vf = r/2 for a spherical mirrormagnification m = v/u = image height / object heightSnell's law: n = sin(i)/sin(r)n = speed of light in air / speed of light in mediumn = real depth / apparent depthsin(c) = 1/n for the critical anglepower of a lens P = 1/f, with f in metres
Worked examples
A ray of light travelling in air strikes the surface of a glass block at an angle of incidence of 60 degrees. If the refractive index of the glass is 1.5, calculate the angle of refraction and the critical angle for the glass-air boundary.
- Snell's law: n = sin i / sin r, so sin r = sin i / n.
- sin 60 = 0.8660, so sin r = 0.8660 / 1.5 = 0.5774.
- r = sin^-1(0.5774) = 35.3 degrees.
- For the critical angle, sin c = 1/n = 1/1.5 = 0.6667.
- c = sin^-1(0.6667) = 41.8 degrees.
Answer: Angle of refraction = 35.3 degrees; critical angle = 41.8 degrees
An object is placed 15 cm from a converging lens of focal length 10 cm. Find the position, nature and magnification of the image.
- Use 1/f = 1/u + 1/v with f = 10 cm and u = 15 cm, taking real distances as positive.
- 1/v = 1/f - 1/u = 1/10 - 1/15.
- Common denominator 30: 1/v = 3/30 - 2/30 = 1/30.
- v = 30 cm, positive, so the image is real and formed on the opposite side of the lens, 30 cm from it.
- Magnification m = v/u = 30/15 = 2.
- The object lies between F and 2F, so the image is real, inverted and twice the size of the object, formed beyond 2F.
Answer: Image is 30 cm from the lens on the far side; real, inverted and magnified, with magnification 2
The mistake to avoid
Candidates invert the lens formula and write 1/v = 1/u - 1/f, or forget that for a virtual image or a diverging lens the value of v or f is negative in the real-is-positive convention. Make the term you want the subject carefully, and remember the final step: after getting 1/v you must take the reciprocal, not leave 1/30 as the answer. Many lose the reciprocal mark exactly there.
In the exam
Draw the ray diagram even when the question only asks for a calculation; it costs two minutes and tells you at once whether the image should be real or virtual, so you can catch a sign error. Learn the six concave-mirror object positions as a table of image nature, position and size. In refraction calculations, work to at least four decimal places in the sine values before taking the inverse, or your angle will be a degree off.