Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Section A

For the same massless spring of spring constant $k$, Hooke's law gives the vertical simple harmonic motion about the displaced equilibrium position:Gravity shifts the equilibrium extension but does not change this frequency. Use the measured sample means in the ratio estimate. For small measurement errors, logarithmic differentiation givesAssuming the two frequency estimates have independent errors, their variances add. Writing $ϵ=\delta \nu /\overline{\nu }$ for their common fractional standard uncertainty, monomial uncertainty propagation yieldsThis is a first-order uncertainty formula, rather than an exact identity for an arbitrary noisy ratio. It assumes that errors in the two measured frequencies have zero covariance.

There is a statistical distinction in the wording: the standard deviation of individual repetitions is not the standard error of their sample mean. If the supplied $\sigma ({\nu }_i)$ describes individual trials, the frequency means have errors $\sigma ({\nu }_i)/\sqrt{n}$, and the ratio formed from those means instead satisfiesThus the printed formula describes either a single-pair ratio or a convention in which $\delta \nu$ already denotes the uncertainty of each averaged frequency. Under either consistent interpretation, independent repeated measurements reduce statistical error in proportion to $n^{-1/2}$. Halving it requires $4n$ trials in total for each mass, or $3n$ additional trials. Repetition alone does not reduce a fixed systematic error.

The spacecraft clock measures proper time $\tau$, while a clock at rest with Earth measures the launch-to-emission interval $t$. Neglect Earth's subsequent motion and gravity, as in the intended special relativity model. The Lorentz factor and time dilation giveIn Earth's inertial frame, the spacecraft's displacement at emission is $d=vt$. ThereforeHere $\tau =43200\mathrm{s}$ and the speed of light is $c=2.99792458\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$. The distance is evaluated at the emission event, using simultaneous Earth-frame endpoints, as required by relativity of simultaneity; the return signal has not yet reached Earth. A distance assigned on the spacecraft's simultaneous time slice is different: Earth has moved away at $c/3$ for $12\mathrm{h}$ in that frame, giving $c\tau /3=4$ light-hours. Specifying the reference frame is essential to the numerical answer.

Use dimensions $\mathsf{M},\mathsf{L},\mathsf{T}$ for mass, length and time. The reduced Planck constant, gravitational constant and speed of light have dimensionsFor $l_p={\hslash }^{\alpha }{G}^{\beta }{c}^{\gamma }$ to have the dimensions of a length, dimensional analysis requiresThe first equation gives $\beta =\alpha$; the third gives $\gamma =-3\alpha$. Substitution in the second gives $2\alpha =1$. HenceWith $\hslash =1.054571817\times 10^{-34}\mathrm{J}\mathrm{s}$, $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$ and $c=2.99792458\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$, the Planck length isThe proposed monomial fixes the dimensionless prefactor to one; dimensional analysis on its own cannot select an arbitrary dimensionless multiplier.

First intensity minimum. In Fraunhofer diffraction, the path difference between adjacent slits is $d\sin \theta$, so the relative wave phase is $\delta =2\pi d\sin \theta /{\lambda }_1$. Summing the $N$ equal complex amplitudes as a geometric series gives the diffraction grating interference factorThe central maximum has $\delta =0$. Its nearest zero occurs at $N\delta /2=\pi$, where the denominator is nonzero. ThusThere is a symmetric first zero on each side of the incident direction. The narrow slits are treated as having a sufficiently broad single-slit diffraction envelope.

Wavelength resolution and the missing order specification. Adopt the Rayleigh criterion: the principal maximum of one wavelength lies at the adjacent minimum of the other in the same nonzero diffraction order $m$. For ${\lambda }_2>{\lambda }_1$ and $m>0$, those directions obeyConsequently the ideal resolving power of a diffraction grating is ${\lambda }_1/\Delta \lambda =mN$, givingThe usual first-order reading therefore gives $64\mathrm{nm}$. Zero diffraction order cannot separate the two wavelengths, because both central maxima coincide.

The paper does not specify which diffraction order is observed, so $64\mathrm{nm}$ is not an order-independent minimum. In the ideal infinitely narrow-slit model, the largest usable order for this positive wavelength separation must satisfy $(m+1/N){\lambda }_1\le d$. Since $d/{\lambda }_1=125$, this permits $m\le 124$, giving the formal all-order resultThat pair meets the criterion at $\theta \simeq 83.1^{\circ }$. Order $125$ of ${\lambda }_1$ is at grazing incidence and the corresponding longer-wavelength peak cannot propagate. In a real device, finite slit width, weak high-order intensity and screen acceptance may prevent such an extreme order from being used; those data are absent. Thus the specified-order expression, and $64\mathrm{nm}$ when first order is intended, are the defensible resolution answers under the stated criterion.

For a photon, the photon energy formula and the vacuum relation $\nu =c/\lambda$ give $E_{\gamma }=hc/\lambda$. For an Electron, the de Broglie wavelength gives the same momentum magnitude $p=h/\lambda$, but the nonrelativistic kinetic energy is $K_e=p^2/(2m_e)$. Using $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $m_e=9.10938\times 10^{-31}\mathrm{kg}$ and $1\mathrm{eV}=1.602176634\times 10^{-19}\mathrm{J}$,Thus, to the precision of the wavelength,The electron answer is its kinetic energy, as required for the diffraction experiment; its relativistic energy would additionally include the approximately $511\mathrm{keV}$ rest energy. Since $K_e\ll m_e{c}^2$, the nonrelativistic approximation is well justified. Equal wavelengths imply equal momenta, not equal energies for particles with different energy–momentum relations.

Put the charged sheet at $z=0$ and the grounded plates at $z=\pm d/2$. By symmetry the electric field points away from the positive sheet with equal magnitude in the two gaps. A Gaussian pillbox crossing the sheet givesThe plate boundary conditions fix their electric potentials to zero. For the charged sheet between grounded parallel plates, the central potential isWith vacuum permittivity ${\epsilon }_0=8.85419\times 10^{-12}\mathrm{F}{\mathrm{m}}^{-1}$, $V_s\simeq 28.24\mathrm{V}$. An Electron has electric charge $-e$, so moving from a grounded plate to this positive potential decreases its potential energy by $e{V}_s$. Conservation of energy then givesThe unrounded value is $3.15\times 10^6\mathrm{m}{\mathrm{s}}^{-1}$, about $0.0105c$. Either plate gives the same speed. The ideal calculation neglects the electron's disturbance of the prescribed field and image-force effects at its release point, consistently with treating it as a test particle.