science-2018:P.W.2
science-2018:P.W.2 Light phenomena
Learning goal: Represent the propagation of light with ray diagrams, apply the law of reflection to plane mirrors and to the placement of mirrors in a light path, use Snell's law and n = c/v to solve refraction problems and to find an index of refraction from data, and draw ray diagrams and use the mirror and lens equation to locate and describe images formed by plane, concave and convex mirrors and by converging and diverging lenses.
Rays: the propagation of light. Light is a wave, but in a uniform medium its energy travels in straight lines, so its path can be drawn as rays: arrows showing the direction of travel, perpendicular to the wavefronts. A ray diagram is a geometric model; it does not show the wave itself, only where the wave goes. Two rules govern most of this lesson. First, the law of reflection: because light is a wave, when it reflects from a surface the angle of incidence equals the angle of reflection, both measured from the normal, the line perpendicular to the surface at the point where the ray hits. Second, Snell's law for refraction at the boundary between two transparent media: n₁ sin θ₁ = n₂ sin θ₂, where n is the index of refraction of each medium and θ is the angle between the ray and the normal in that medium.
Index of refraction. The index of refraction of a material is n = c/v, the speed of light in a vacuum (c = 3.00 × 10⁸ m/s) divided by its speed in the material. Light is slower in optically denser transparent media such as water and glass, so n is greater than 1: about 1.00 for air, 1.33 for water and about 1.5 for common glass. Light entering a slower medium bends toward the normal; light entering a faster medium bends away from it; a ray along the normal (θ₁ = 0) passes straight through with no bending, although its speed and wavelength still change. Worked model: a ray in air strikes water at 40° to the normal. sin θ₂ = (1.00)(sin 40°)/1.33 = 0.6428/1.33 = 0.4833, so θ₂ = 28.9°. The refracted ray is closer to the normal than the incident ray, as expected for a slower medium. Its speed in water is v = c/n = 3.00 × 10⁸/1.33 = 2.26 × 10⁸ m/s. Snell's law also runs the other way: measure the two angles and you can find n. If a ray enters a glass block at 50° and refracts at 30.7°, n = sin 50°/sin 30.7° = 0.766/0.511 = 1.50.
Images formed by refraction at a flat surface. A ray diagram also explains why a pool looks shallower than it is. Rays leaving a coin on the bottom bend away from the normal as they pass from water into air, so when they are traced straight back they appear to come from a point higher up than the coin: a virtual image, upright and the same size, but nearer the surface. For viewing close to straight down, Snell's law with small angles gives apparent depth = real depth × (n of the viewer's medium ÷ n of the object's medium). Worked model: a coin 0.40 m deep in water seen from directly above appears 0.40 × 1.00/1.33 = 0.30 m deep. Viewed at a slant, the rays bend more and the image appears shallower still, so the simple ratio holds only near the normal. The same reasoning, run in reverse, is why a fish looking up sees objects above the water as higher than they are.
Plane mirrors and the light path. A plane mirror forms an image as far behind the mirror as the object is in front, upright, the same size, and virtual: the rays only appear to come from it, so it cannot be caught on a screen. To steer a beam around obstacles, place each mirror so that the required change of direction is split equally about the mirror's normal: a beam that must turn through 90° needs a mirror at 45° to it, and in general a mirror turns a beam through twice the angle between the beam and the mirror surface. Drawing the path to scale, predicting the hit point, testing it and correcting the diagram is exactly the design cycle used in a laser maze.
Curved mirrors and lenses: three principal rays. A concave mirror and a converging lens both bring parallel rays to a focal point F at a distance f (the focal length); a convex mirror and a diverging lens spread parallel rays so that they appear to come from a focal point behind the mirror or on the incoming side of the lens. To locate an image, draw any two of three principal rays from the top of the object: a ray parallel to the axis, which passes through F after reflection or refraction (or appears to come from F for the diverging types); a ray through F (or heading toward the far F for the diverging types), which leaves parallel to the axis; and for a lens a ray through the center, undeviated, or for a mirror a ray to the vertex, reflected symmetrically. Where the rays cross is a real image, which can be projected on a screen and is inverted; where they only appear to diverge from is a virtual image, upright.
The mirror and lens equation. For a thin lens, or a curved mirror, with rays that stay close to the axis, and with the object distance d_o, the image distance d_i and the focal length f, 1/d_o + 1/d_i = 1/f, and the magnification is m = −d_i/d_o. Sign conventions: f is positive for concave mirrors and converging lenses, negative for convex mirrors and diverging lenses; d_i is positive for a real image (in front of a mirror, on the far side of a lens) and negative for a virtual image; m is negative for inverted images and positive for upright ones, with |m| > 1 meaning enlarged. Worked model: a converging lens with f = 10 cm. Object at 30 cm: 1/d_i = 1/10 − 1/30 = 2/30, so d_i = 15 cm, m = −0.5: real, inverted, half size. Object at 20 cm (twice the focal length): d_i = 20 cm, m = −1: real, inverted, same size. Object at 5 cm (inside the focal length): 1/d_i = 1/10 − 1/5 = −1/10, d_i = −10 cm, m = +2: virtual, upright, enlarged, which is how a magnifying glass works. A diverging lens or a convex mirror always gives a virtual, upright, reduced image for a real object, whatever the distance.
Misconception: the image is on the mirror's surface. The rays reaching your eye from a plane mirror diverge from a point behind the mirror, which is why you must refocus your eyes when you look from the frame to your reflection. A second misconception is that covering half a lens removes half the image; it only makes the whole image dimmer, because every part of the lens sends rays to every part of the image.
Investigation: focal length and images of a converging lens. Materials: a converging lens (a magnifying glass), a lens holder, a white card, a meter stick, a cool LED lamp behind a card with an arrow-shaped hole cut in it as the object (the card is held in its own stand, clear of the lamp). Steps: point the lens at a distant window and move the card until the window image is sharp; the lens-to-card distance is f (parallel rays from far away focus at F). Then place the lamp at 3f, 2f, 1.5f and 0.5f from the lens, find the sharp image on the card where one exists, and record d_o, d_i, whether the image is inverted, and its size relative to the object. Check each pair against 1/d_o + 1/d_i = 1/f. A result that supports the model: measured d_i values within about 5 % of prediction (a suggested classroom criterion; compare it with the uncertainty of your own distance measurements), inversion for all real images, and no image on the card for the object inside f (look through the lens instead to see the upright virtual image). Safety: never use the Sun as the distant object; a lens concentrates sunlight enough to burn paper and eyes. What the ray model cannot show: color fringes and blur at the edges of the image, which come from glass bending different colors by slightly different amounts and from the lens's imperfect shape.
Worked example 1 — Snell's law into glass. A ray in air meets a glass surface (n = 1.52) at 35° to the normal. sin θ₂ = sin 35°/1.52 = 0.5736/1.52 = 0.3774, θ₂ = 22.2°. The ray bends toward the normal because glass is the slower medium; its speed inside is 3.00 × 10⁸/1.52 = 1.97 × 10⁸ m/s.
Worked example 2 — out of water. A ray in water (n = 1.33) reaches the surface at 30° to the normal. sin θ₂ = 1.33 sin 30°/1.00 = 0.665, θ₂ = 41.7° in air, bending away from the normal on entering the faster medium. (At larger angles the sine would exceed 1, meaning no refracted ray emerges; that case is beyond this lesson.)
Worked example 3 — finding n from data. A student shines a narrow beam into a semicircular block and measures pairs of angles: 20° in air gives 13.2° inside; 40° gives 25.4°; 60° gives 35.3°. n = sin 20°/sin 13.2° = 0.342/0.228 = 1.50; sin 40°/sin 25.4° = 0.643/0.429 = 1.50; sin 60°/sin 35.3° = 0.866/0.578 = 1.50. The constant ratio is the evidence that Snell's law holds and that the block has n ≈ 1.50.
Worked example 4 — curved mirrors. Concave mirror, f = 12 cm, object 36 cm away: 1/d_i = 1/12 − 1/36 = 2/36, d_i = 18 cm in front of the mirror, m = −0.5: real, inverted, half size. Convex mirror, f = −12 cm, object 24 cm away: 1/d_i = −1/12 − 1/24 = −3/24, d_i = −8.0 cm (behind the mirror), m = +1/3: virtual, upright, one third the size, the wide field of a security mirror.
Worked example 5 — placing a mirror. A laser beam traveling due east must be turned to travel due north to hit a target. The beam must turn through 90°, so incidence and reflection angles are 45° each, and the mirror's surface must lie at 45° to the beam (its normal at 45° to both the east and north directions). Marking the hit point, drawing the normal and checking that both angles are 45° confirms the placement before the test.
Representation task: ray diagrams to scale. For a converging lens with f = 6.0 cm, draw three ray diagrams on graph paper with the object at 18 cm, 12 cm and 4.0 cm, using two principal rays each. Label the image as real or virtual, upright or inverted, and enlarged or reduced, and measure d_i from each diagram; then calculate d_i from the lens equation (9.0 cm, 12 cm and −12 cm) and compare. Add a fourth diagram for a convex mirror with f = −6.0 cm and the object at 12 cm (image at −4.0 cm, upright, m = +1/3).
Independent performance: determine an index of refraction. Plan and carry out an investigation to find the index of refraction of water or a glass block using a narrow beam from a ray box or a slit over a lamp, a protractor and a semicircular container. Success criteria: a labeled ray diagram of the set-up showing the normal, the incident and refracted rays and both angles; at least five incident angles between 10° and 60° with measured refracted angles; n calculated for each pair and averaged, with the spread stated as an uncertainty; a comparison with the accepted value for the material if known (1.33 for water) and a percent difference; a paragraph on which angle measurements contributed most to the uncertainty and how you reduced it; and a statement that a laser or bright beam was never directed toward anyone's eyes.
An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.
A ray of light in air strikes a flat glass surface (n = 1.50) at 48° to the normal. Find the angle of refraction and state which way the ray bends and why.
Snell's law: (1.00) sin 48° = 1.50 sin θ₂, so sin θ₂ = 0.743/1.50 = 0.495 and θ₂ = 29.7°. The ray bends toward the normal because light travels more slowly in glass than in air (n = c/v = 1.50 means v = 2.00 × 10⁸ m/s), and a wave entering a slower medium at an angle turns toward the normal.
Light travels through a plastic at 1.90 × 10⁸ m/s. Calculate the plastic's index of refraction. A ray inside the plastic strikes the plastic-air boundary at 25° to the normal; find the angle of the refracted ray in air.
n = c/v = (3.00 × 10⁸)/(1.90 × 10⁸) = 1.58. Leaving into air: 1.58 sin 25° = 1.00 sin θ₂, so sin θ₂ = 1.58 × 0.4226 = 0.668 and θ₂ = 41.9°. The ray bends away from the normal because air is the faster medium.
A student measures an incident angle of 55° in air and a refracted angle of 33° in a liquid. What is the liquid's index of refraction, and what single further measurement would best check the result?
n = sin 55°/sin 33° = 0.819/0.545 = 1.50. The best check is a second pair of angles at a different incidence, for example 30°, which should give sin 30°/n = 0.333 and a refracted angle near 19.5°; if the ratio sin θ₁/sin θ₂ is the same for both pairs, Snell's law is confirmed and the value is trustworthy. Measuring small angles less precisely is the main source of error, so a mid-range angle is better than a very small one.
State the law of reflection and use it to explain where a plane mirror must be placed to turn a horizontal laser beam through 60°.
The angle of incidence equals the angle of reflection, both measured from the normal to the mirror. Let θ be the angle of incidence. A beam that hits a mirror head-on (θ = 0) is sent straight back, a turn of 180°; in general the beam's direction turns through 180° − 2θ. For a 60° turn, 180° − 2θ = 60°, so θ = 60°: the incident and reflected rays each make 60° with the normal (and 120° with each other, measured at the mirror). The mirror surface, perpendicular to the normal, therefore makes 90° − 60° = 30° with the incoming beam; check with the rule that a mirror turns a beam through twice the angle between the beam and the mirror surface: 2 × 30° = 60°. Tilting the mirror 30° the other way turns the beam 60° to the other side.
A converging lens has f = 15 cm. Locate and describe the image of an object placed 45 cm away, and then one placed 10 cm away.
At 45 cm: 1/d_i = 1/15 − 1/45 = 2/45, d_i = 22.5 cm on the far side; m = −22.5/45 = −0.5: real, inverted, half size, projectable on a screen. At 10 cm (inside the focal length): 1/d_i = 1/15 − 1/10 = −1/30, d_i = −30 cm, meaning virtual and on the same side as the object; m = −(−30)/10 = +3: upright and three times larger, seen by looking through the lens, as with a magnifying glass.
Compare the images formed by a converging lens and a diverging lens of the same focal-length magnitude for an object 30 cm away when |f| = 10 cm.
Converging (f = +10 cm): 1/d_i = 1/10 − 1/30 = 2/30, d_i = 15 cm, m = −0.5: real, inverted, reduced, on the far side. Diverging (f = −10 cm): 1/d_i = −1/10 − 1/30 = −4/30, d_i = −7.5 cm, m = +0.25: virtual, upright, reduced, on the object's side. A diverging lens never forms a real image of a real object; a converging lens does whenever the object is outside its focal length.
Describe how you would experimentally determine the focal length of a converging lens, and explain why the method works in terms of ray diagrams.
Point the lens at a distant object such as a far window and move a card behind the lens until the image is sharp; the lens-to-card distance is the focal length. Rays from a very distant object arrive nearly parallel to the axis, and by definition a converging lens brings parallel rays to the focal point, so the sharp image sits at F. A check: place an object at 2f and confirm a same-size inverted image at 2f on the other side, as 1/d_o + 1/d_i = 1/f predicts. Never use the Sun as the distant object.
A swimming pool is 2.0 m deep. Using a ray diagram and n = 1.33 for water, explain why the bottom looks shallower when viewed from directly above, find its apparent depth, and say whether the image is real or virtual.
Rays from the bottom travel up through the water and bend away from the normal as they enter the air, because air is the faster medium. Traced straight back, the bent rays seem to come from a point above the true bottom, so the bottom looks raised. For viewing near the normal, apparent depth = 2.0 × 1.00/1.33 = 1.5 m. The image is virtual, because the rays only appear to come from it and it cannot be projected on a screen; it is upright and the same size. Viewed at a slant, the bottom looks shallower still, which is one reason pools are more dangerous to jump into than they look.