Intensity, refraction and polarisation flashcards

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Intensity
Power per unit area, on an area perpendicular to the direction of travel of the wave. Unit W m⁻².
Refractive index
Speed of light in a vacuum ÷ speed of light in the material, n = c/v. No unit; at least 1.
Critical angle
The angle of incidence in the denser medium for which the angle of refraction is 90°.
Total internal reflection
All of the light is reflected at a boundary: happens when light travels towards a lower refractive index at an angle of incidence greater than the critical angle.
Plane-polarised wave
A transverse wave whose oscillations are in one plane only, the plane containing the direction of travel.
Unpolarised light
Light whose oscillations occur in all directions perpendicular to its direction of travel, e.g. from the Sun or a filament lamp.
Pulse-echo technique
A short pulse is sent out; the time for its echo to return from a reflector gives the reflector's distance, d = vt/2.
Intensity from power and area
I = P/A (Yes)
Intensity at distance r from a point source
I = P/4πr² (the power spreads over a sphere of area 4πr²). (No — learn it)
Snell's law
n₁ sin θ₁ = n₂ sin θ₂, angles measured from the normal. (Yes)
Refractive index from speeds
n = c/v (Yes)
Critical angle, material to air
sin C = 1/n (Yes)
Critical angle between two materials (n₁ > n₂)
sin C = n₂/n₁ (put θ₂ = 90° in Snell's law). (No — derive it)
Pulse-echo distance
d = vt/2, because t is the time there and back. (No — derive it)
Shape of an intensity–distance graph for a point source
A 1/r² curve: doubling r quarters I; trebling r gives I/9.
Graph of sin i against sin r for a glass block
Straight line through the origin; gradient = refractive index n.
Wavefronts entering a slower medium at an angle
Wavefronts get closer together (shorter wavelength) and the direction of travel bends towards the normal.
Semicircular block: ray at the flat face below, at and above C
Below C: refracts out plus weak reflection. At C: refracted ray along the surface. Above C: total internal reflection.
Echo trace (A-scan)
Signal against time: one spike per reflecting boundary; the delay gives the depth, d = vt/2; deeper echoes are weaker.
What happens to a wave at a boundary between two media?
Part is reflected and part transmitted. Frequency stays the same; speed and wavelength change.
Why does an ultrasound scan need gel on the skin?
An air gap would reflect almost all of the ultrasound; the gel removes the air so most is transmitted into the body.
How does an optical fibre guide light?
Light in the core hits the core–cladding boundary at more than the critical angle and is totally internally reflected along the fibre.
Why does an optical fibre have cladding?
It protects the core surface from scratches, stops light crossing between touching fibres, and keeps the boundary condition the same along the fibre.
Why can sound not be polarised?
It is longitudinal: the oscillations are along the direction of travel, so there is no plane to select.
How do polarising sunglasses reduce glare?
Light reflected from water or roads is partly polarised in a horizontal plane; filters with a vertical axis absorb much of it.
Stress analysis with polarised light
A transparent plastic model between crossed polarisers shows coloured fringes; crowded fringes mark where stress is concentrated.
Two things that limit the detail in a pulse-echo image
Wavelength (features smaller than about λ are not resolved) and pulse duration (boundaries closer than vτ/2 give overlapping echoes).
Refractive index practical: apparatus and method
Ray box with narrow slit, glass block on paper, pencil crosses to mark rays, protractor to measure i and r from the normal; repeat for 6+ angles.
A source radiates 18 W equally in all directions. Intensity at 3.0 m?
I = 18 ÷ (4π × 3.0²) = 0.16 W m⁻² (0.159).
Light in air hits glass (n = 1.50) at 45°. Angle of refraction?
sin r = sin 45° ÷ 1.50 = 0.471, so r = 28.1°.
Speed of light in water, n = 1.33?
v = c/n = 3.00 × 10⁸ ÷ 1.33 = 2.26 × 10⁸ m s⁻¹.
Critical angle for a plastic of refractive index 1.45 (into air)?
sin C = 1/1.45 = 0.690, so C = 43.6°.
A sonar echo returns after 0.60 s; sound travels at 1500 m s⁻¹ in water. Distance?
d = vt/2 = 1500 × 0.60 ÷ 2 = 450 m.
The angle given is measured from the surface
Subtract from 90° first: Snell's law and the critical angle use angles to the normal.
Proving total internal reflection happens
State both conditions: travelling towards lower n, and angle of incidence > C — and quote your value of C.
Pulse-echo answer twice too big
The time was not halved: the pulse goes there and back, so d = vt/2.
Area for a source radiating in all directions
4πr² (sphere), not πr² (circle) — otherwise the intensity is four times too big.
Does light change frequency in glass?
No. Frequency is fixed by the source; speed and wavelength both fall by the factor n.

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