Wave Behaviour

Reflection is the bouncing back of a wave when it meets a boundary. Refraction is a change of direction on entering a new medium, and diffraction is the spreading of waves around obstacles or through gaps.

Diffraction is the spreading of a wave as it passes through a gap or around an obstacle. It is most noticeable when the gap is about the same size as the wavelength.

Refraction happens because the wave travels at a different speed in the new medium. The change in speed (with frequency fixed by the source) makes the wavefronts bend.

The frequency is set by the source and does not change on crossing a boundary. The speed changes, and since $v = f\lambda$ with $f$ fixed, the wavelength changes too.

Superposition means the displacements add (with sign) at every point. Where they add to a larger value the interference is constructive; where they cancel it is destructive.

A node is a point of permanently zero displacement. Halfway between two nodes is an antinode, where the oscillation is largest.

The fundamental is a single loop: a node at each fixed end and one antinode in the middle. Higher harmonics add more loops (and more antinodes).

Each ray sits $30^\circ$ from the normal, on opposite sides. The angle between them is therefore $30^\circ + 30^\circ = 60^\circ$, i.e. twice the angle of incidence. (Taking just one reflection angle gives $30^\circ$; the $120^\circ$ trap is the deviation from the original direction, $180^\circ - 2i$.)

Reflection at a fixed (rigid) end flips the pulse over – it comes back inverted (a $180^\circ$ phase change). At a free end the pulse reflects the same way up.

The distance between adjacent nodes is half a wavelength, so $\lambda = 2 \times 0.40 = 0.80\text{ m}$. (Forgetting the factor of two gives the $0.40\text{ m}$ trap.)

Frequency stays $500\text{ Hz}$, so $\lambda = \dfrac{v}{f} = \dfrac{170}{500} = 0.34\text{ m}$. The speed halved, so the wavelength halved. ($0.68\text{ m}$ is the original wavelength in air.)

Diffraction is greatest when the wavelength is comparable to, or larger than, the gap width. Waves whose wavelength is far smaller than the gap pass almost straight through with little spreading.

$\sin\theta_2 = \dfrac{n_1\sin\theta_1}{n_2} = \dfrac{(1.00)\sin 30^\circ}{1.50} = 0.333$, so $\theta_2 = 19^\circ$. The ray bends toward the normal on entering the denser medium. (Putting the ratio upside-down gives the $49^\circ$ trap.)

Displacements add with sign: $(+3) + (-2) = +1\text{ cm}$. (Adding the magnitudes as if both were crests gives the $+5\text{ cm}$ trap; this is partial destructive interference.)

A path difference of a half-integer number of wavelengths, $(m + \tfrac{1}{2})\lambda$, puts the waves exactly out of phase, giving destructive interference. A whole number of wavelengths would give constructive interference.

$f_1 = \dfrac{v}{2L} = \dfrac{120}{2(0.60)} = \dfrac{120}{1.2} = 100\text{ Hz}$. (Dividing by $L$ instead of $2L$ gives the $200\text{ Hz}$ trap.)

A whole wavelength of extra path is one full cycle, $360^\circ$. Here the path difference is $\dfrac{1.0}{4.0} = \tfrac{1}{4}$ of a wavelength, so the phase difference is $\tfrac{1}{4} \times 360^\circ = 90^\circ$. (This is partial interference – neither fully constructive nor fully destructive.)

Speed decreases (that is what makes them refract), frequency is fixed by the source so it stays the same, and since $\lambda = \dfrac{v}{f}$ with $f$ constant, the wavelength decreases. Only option A has all three correct.

Each loop is half a wavelength, so three loops span $L = 3 \times \tfrac{\lambda}{2}$. Then $\lambda = \dfrac{2L}{3} = \dfrac{2(1.2)}{3} = 0.80\text{ m}$. (Using $\tfrac{L}{3}$ and forgetting the factor of two gives the $0.40\text{ m}$ trap.)

At the first face the ray refracts toward the normal (to about $19^\circ$ inside the glass). At the second, parallel face the geometry is reversed and it refracts by exactly the same amount, so it emerges at $30^\circ$ – parallel to the original ray, only shifted sideways. ($19^\circ$ is the angle inside the glass, not the emergent angle.)

Path difference $= 8.0 - 6.0 = 2.0\text{ m} = 2\lambda$, a whole number of wavelengths. In-phase sources plus a whole-number path difference means the waves arrive in phase → constructive interference (a loud spot).

Diffraction is significant only when the wavelength is comparable to the size of the obstacle or gap. Audible sound (wavelengths of roughly $0.02$ to $17\text{ m}$) diffracts around everyday objects; visible light (wavelength around $5 \times 10^{-7}\text{ m}$) does not, so it travels in near-straight lines and you cannot see around the corner.

$\sin\theta_2 = \sin\theta_1 \times \dfrac{\lambda_2}{\lambda_1} = \sin 40^\circ \times \dfrac{0.40}{0.60} = 0.643 \times 0.667 = 0.429$, so $\theta_2 = 25^\circ$. The wavelength shrank, so the ray bends toward the normal. (Inverting the ratio gives the $59^\circ$ trap.)

For a string fixed at both ends the harmonics are whole-number multiples: the third harmonic is $3 \times 150 = 450\text{ Hz}$. It has three loops, which means $4$ nodes (the two ends plus two in between). ($300\text{ Hz}$ is the second harmonic, not the third.)

Fundamental: $f_1 = \dfrac{v}{2L} = \dfrac{400}{2(0.50)} = 400\text{ Hz}$. A string fixed at both ends supports all whole-number harmonics, so the next one is $f_2 = 2f_1 = 800\text{ Hz}$. (Only odd harmonics, giving $400$ and $1200\text{ Hz}$, would apply to a pipe closed at one end – not this string.)