Waves and Optics: cram pack

Formula sheets, the key method steps for every topic and 4 practice problems, with full solutions at the back.

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1Formulas and methods

Springs & pendulums

Mass on a spring
T = 2π·√(m/k)
Pendulum
T = 2π·√(L/g)
Angular frequency
ω = √(k/m) = 2π/T
Maximum speed and acceleration
vmax = A·ω, amax = A·ω²
Energy
E = ½·k·A²

Waves, sound & optics

Wave speed
v = fλ, T = 1/f
String
v = √(T/μ), μ = m/L
Sound level
β = 10·log10(I/I0), I0 = 10⁻¹² W/m²
Point source
I = P/(4πr²)
Standing waves
string or open pipe fn = n·v/(2L); closed pipe fn = n·v/(4L), n odd
Doppler effect
f′ = f·(v ± vo)/(v ∓ vs)
Refractive index
n = c/v, λ = λ0/n
Snell’s law
n1sin θ1 = n2sin θ2
Critical angle
sin θc = n2/n1
Optical fibre
NA = sin θa = √(n1² − n2²)
Lenses and mirrors
1/f = 1/do + 1/di, m = −di/do, mirror f = R/2
Double slit
d·sin θ = mλ, ym = mλL/d
Grating
d = 1/N, d·sin θ = mλ, mmax ≤ d/λ
Single slit
a·sin θ = mλ (dark), w = 2λL/a

Method

  1. Identify the kind of problem: wave speed, sound level, standing wave, Doppler, refraction, lens or mirror, or interference.
  2. Write the relation first: v = fλ, β = 10·log(I/I₀), fₙ = n·v/(2L) or n·v/(4L), f′ = f(v ± vo)/(v ∓ vs), n₁sin θ₁ = n₂sin θ₂, 1/f = 1/d₀ + 1/dᵢ, d·sin θ = mλ.
  3. Convert units: nm, mm and cm to m, and check whether angles are measured from the normal.
  4. Check signs and plausibility: a virtual image has dᵢ < 0, an approaching source sounds higher, a closed pipe has only odd harmonics.

2Practice problems

1.Springs & pendulumsMedium
L = ?

How long must a simple pendulum be to have a period of 1.3 s? Use g = 9.81 m/s². Find L.

T = 1.3 sg = 9.81 m/s²
L = m
2.Springs & pendulumsHard
1.5 kgk = 300 N/mA = 17 cm

A 1.5 kg mass on a spring with k = 300 N/m is pulled 17 cm from equilibrium and released. Find the total energy E of the oscillation.

m = 1.5 kgk = 300 N/mA = 17 cm
E = J
3.Waves, sound & opticsMedium
38°90°n1​ = ?air, n2​ = 1.00

Light inside a transparent block is totally reflected at the boundary with air (n = 1.00) for every angle of incidence above 38°. Find the refractive index n1 of the block.

θc = 38°n2 = 1.00
n1 =
4.Waves, sound & opticsHard
f1​ = 15 cmf2​ = 20 cm25 cm50 cm

Two thin lenses share an axis, 50 cm apart: lens 1 with f₁ = 15 cm and lens 2 with f₂ = 20 cm. An object stands 25 cm in front of lens 1. Find the position di2 of the final image measured from lens 2 (positive beyond lens 2, real-is-positive convention).

f1 = 15 cmf2 = 20 cms = 50 cmdo = 25 cm
di2 = cm

3Solutions

1.Springs & pendulums

Answer: L = 0.4199 m

Simple harmonic motion: a mass m on a spring of stiffness k oscillates with angular frequency ω = √(k/m) and period T = 2π/ω = 2π·√(m/k). A simple pendulum of length L has T = 2π·√(L/g). With amplitude A, the largest speed is A·ω, the largest acceleration A·ω² and the energy ½·k·A².

Solve for the length

L = g·(T/2π)² = 9.81·(1.3/2π)² = 0.4199 m

Answer

L = 0.4199 m
2.Springs & pendulums

Answer: E = 4.335 J

Simple harmonic motion: a mass m on a spring of stiffness k oscillates with angular frequency ω = √(k/m) and period T = 2π/ω = 2π·√(m/k). A simple pendulum of length L has T = 2π·√(L/g). With amplitude A, the largest speed is A·ω, the largest acceleration A·ω² and the energy ½·k·A².

Angular frequency

ω = √(300/1.5) = 14.14 rad/s

Energy

E = ½·k·A² = ½·300·0.17² = 4.335 J

Answer

E = 4.335 J
3.Waves, sound & optics

Answer: n1 = 1.624

Geometric optics: at a boundary, n₁·sin θ₁ = n₂·sin θ₂ (Snell's law), with angles measured from the normal and n = c/v. Going into a slower medium the ray bends towards the normal. From a denser medium (n₁ > n₂) beyond the critical angle θc = arcsin(n₂/n₁) all light is reflected: total internal reflection. Thin lenses and spherical mirrors obey 1/f = 1/do + 1/di with the magnification m = −di/do. Sign convention: f > 0 for a converging lens and a concave mirror (f = R/2), f < 0 for a diverging lens and a convex mirror; di > 0 for a real image, di < 0 for a virtual one, and m < 0 for an inverted image.

Refractive index of the block

n₁ = n₂/sin θc = 1.00/0.6157 = 1.624

Answer

n1 = 1.624
4.Waves, sound & optics

Answer: di2 = −33.33 cm

Geometric optics: at a boundary, n₁·sin θ₁ = n₂·sin θ₂ (Snell's law), with angles measured from the normal and n = c/v. Going into a slower medium the ray bends towards the normal. From a denser medium (n₁ > n₂) beyond the critical angle θc = arcsin(n₂/n₁) all light is reflected: total internal reflection. Thin lenses and spherical mirrors obey 1/f = 1/do + 1/di with the magnification m = −di/do. Sign convention: f > 0 for a converging lens and a concave mirror (f = R/2), f < 0 for a diverging lens and a convex mirror; di > 0 for a real image, di < 0 for a virtual one, and m < 0 for an inverted image.

Image of lens 1

1/di1 = 1/15 − 1/25 → di1 = 37.5 cm

Object for lens 2

do2 = 50 − 37.5 = 12.5 cm

Final image

1/di2 = 1/20 − 1/12.5 → di2 = −33.33 cm

Answer

di2 = −33.33 cm

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