Light Worked ExamplesIGCSE Physics 0625

(a) Angles of incidence and reflection are always measured from the normal, which is at $90^{\circ}$ to the mirror. The $28^{\circ}$ is measured from the surface, so subtract it from $90^{\circ}$.

(b) For reflection, the angle of reflection equals the angle of incidence.

(c) The incident ray and the reflected ray lie on either side of the normal, so the angle between them is the sum of the two angles.

(d) The image in a plane mirror is the same distance behind the mirror as the object is in front, on the normal through the object. So the image is $12$ cm behind the mirror, directly opposite O, which puts it $12 + 12 = 24$ cm from the lamp itself.

Any two characteristics: it is virtual (the reflected rays only appear to come from it, so it cannot be caught on a screen), the same size as the object, and upright.

The image is formed on the paper, so light rays really meet there. An image that can be projected onto a screen is real. This rules out C and D, because a virtual image cannot be formed on a screen.

A real image formed by a single converging lens is always inverted. This rules out B.

The window is far away, well beyond $2F$, so its image forms close to the principal focus on the other side of the lens and is diminished (smaller than the object).

Why the distractors tempt: B appeals to anyone who pictures the view "the right way up", and D mixes the magnifying-glass image (virtual, upright) with the size of this one.

(a) The angle of reflection equals the angle of incidence.

(b) The normal at P is vertical, so the reflected ray makes $35^{\circ}$ with the vertical and therefore $90^{\circ} - 35^{\circ} = 55^{\circ}$ with the horizontal. M2 is vertical, so its normal is horizontal. The angle between the ray and the normal to M2 is therefore $55^{\circ}$.

(c) At M2 the angle of reflection is also $55^{\circ}$ from the normal, so the angle with the mirror surface is

(d) Each reflection turns the ray through $180^{\circ} - 2i$. Add the two turns.

A total turn of $180^{\circ}$ means the ray leaving M2 is parallel to the incident ray but travelling in the opposite direction — it is sent back the way it came, displaced sideways.

(a) Refractive index is the ratio of the speed of light in air (or a vacuum) to the speed of light in the material.

(b) Rearrange for the speed in the plastic and substitute.

(c)(i) Use the wave equation $v = f\lambda$ in air.

(c)(ii) The frequency does not change, so the wavelength falls in the same ratio as the speed.

(d) The frequency is set by the source and is unchanged by refraction: $5.0 \times 10^{14}$ Hz. Note that $\lambda_{\text{air}} / \lambda_{\text{plastic}} = 6.0 / 4.0 = 1.5$ — the ratio of wavelengths also equals $n$, but the ratio of frequencies is $1$.

(a) First draw the normal at P, perpendicular to the surface. Each angle to the normal is $90^{\circ}$ minus the angle to the surface.

Check: the ray bends towards the normal on entering the denser glass, so $r$ must be smaller than $i$ — and $33^{\circ} < 52^{\circ}$.

(b) Use the refractive index equation.

(a) Use the critical-angle equation and rearrange for $\sin c$.

Take the inverse sine.

(b) Ray A travels from resin into air. In $n = \sin i / \sin r$ the angle in air goes on top, so the angle in air is the unknown here.

Take the inverse sine. The ray bends away from the normal, as expected on entering a less dense medium.

(c) Marking points: light in B is travelling from the resin (where it is slower, the denser medium) towards air (less dense); its angle of incidence, $50^{\circ}$, is greater than the critical angle of $40.2^{\circ}$; so no light refracts out — all of it is totally internally reflected back into the resin, at an angle of reflection of $50^{\circ}$.

Total internal reflection needs two conditions: the light must be travelling in the denser medium towards a less dense one (glass to air), and the angle of incidence must be greater than the critical angle.

C satisfies both: glass to air, and $50^{\circ} > 42^{\circ}$. All the light is reflected back into the glass.

A is wrong because light going from air into glass bends towards the normal and always enters the glass — there is no critical angle in that direction. B is wrong because $30^{\circ}$ is less than the critical angle, so most of the light refracts out (only a weak partial reflection occurs). D describes what happens below the critical angle, not above it.

(a) The line through the centre of the lens, at right angles to it, is the principal axis.

(b) The focal length is the distance from the centre of the lens to F: $5$ small squares, so $5 \times 0.5 = 2.5$ cm.

(c) Ray 1: from the top of O draw a ray parallel to the principal axis to the centre line of the lens, then through F on the far side. Taking the lens centre as the origin, this ray goes from $(0,\ 1.5)$ through $(2.5,\ 0)$, so it falls $1.5$ cm for every $2.5$ cm across.

Ray 2: from the top of O draw a straight line through the centre of the lens, undeviated. It goes from $(-4.0,\ 1.5)$ through $(0,\ 0)$.

Extend both rays until they cross. Check with the two slopes: at $x = 6.7$ cm both are at $y = -2.5$ cm.

So the image is about $6.7$ cm from the lens on the opposite side from O (just over $13$ small squares), and its top is $2.5$ cm below the axis: height $2.5$ cm.

(d) The image is enlarged ($2.5$ cm compared with $1.5$ cm), inverted (below the axis) and real (the rays actually meet, so it could be caught on a screen). This is expected: O is between F and $2F$.

(a) The object must be between the lens and the principal focus F — closer to the lens than one focal length. Here O is $2.4$ cm from the lens and F is $4.0$ cm away.

(b) From the diagram the image is virtual (dashed), upright (above the axis, the same way up as O) and enlarged ($2.5$ cm tall against $1.0$ cm).

(c) After the lens the two rays diverge — they never actually meet. The image is found only by extrapolating the rays backwards (dashed lines). No light actually reaches the position of I, so nothing would appear on a screen placed there.

(d) The image is on the same side of the lens as the object and further from the lens: $6.0$ cm (12 small squares) compared with $2.4$ cm for O.

(a) Dispersion — the glass has a slightly different refractive index for each frequency, so each colour is refracted by a different amount.

(b) Violet has the larger refractive index, so it is bent (deviated) more. Y is further from the original direction, so Y is violet and X is red.

(c) Rearrange $n = \dfrac{\text{speed in air}}{\text{speed in glass}}$ for each colour.

Violet light travels more slowly in glass, which is why it is refracted more.

(d) The visible spectrum in order of increasing frequency is red, orange, yellow, green, blue, indigo, violet. So: red, orange, green, violet.

(e) Monochromatic light is light of a single frequency.

(a) The points lie close to a straight line through the origin, so $\sin i$ is directly proportional to $\sin r$. That means $\sin i / \sin r$ is a constant — the refractive index.

(b) Use a large triangle on the line, not two of the plotted points. From the origin to the point where $\sin r = 0.60$, the line reads $\sin i = 0.90$.

(c) The gradient is $\sin i / \sin r$, so $n = 1.5$ (no unit).

(d) A single reading carries the full error of one angle measurement: $\sin 28^{\circ} / \sin 19^{\circ} = 1.44$, whereas the six individual values range from $1.44$ to $1.51$. The best-fit line averages out these random errors across all the readings (and would reveal an anomalous one), so its gradient is more reliable.

(e) Any two: place the pins as far apart as possible on each ray; keep the pins vertical and line up their bases when viewing; use thin pins and a sharp pencil for thin lines; use a long ruler to draw each ray in one stroke.

(a) Apply $n = \sin i / \sin r$ at the end face.

(b) The normal at the end face runs along the fibre, and the normal at Q is at right angles to it. The ray and these two normals form a right-angled triangle, so $r$ and $\theta$ add up to $90^{\circ}$.

(c) Use the critical-angle equation.

(d) At Q the light is in glass travelling towards air (denser to less dense), and $\theta = 58.4^{\circ}$ is greater than the critical angle $43.2^{\circ}$. So the ray is totally internally reflected, at $58.4^{\circ}$. The fibre sides are parallel, so it meets the opposite side at the same $58.4^{\circ}$ and is totally internally reflected again, and so on — the light zig-zags along the fibre with no loss through the sides.

(e) Glass is transparent to visible light and some infrared, and visible light and short-wavelength infrared can carry high rates of data.