Past exam of the physics 2 paper of the University of Oxford / 2021 resit / 9

The Huygens-Fresnel principle treats every point of an advancing wavefront as a source of secondary spherical waves; their forward envelope constructs the next wavefront. Consider two points where an oblique plane wavefront meets a plane interface a distance $AB$ apart. After its first point reaches $A$, let time $t$ elapse until the second reaches $B$. In medium 1, the incident travel distance is $v_{1}t=AB\sin i$. Meanwhile, a secondary wave from $A$ has travelled distance $v_{2}t$ in medium 2. The tangent from $B$ to that secondary wavefront gives $v_{2}t=AB\sin r$. ThereforeHere $i,r$ are the angle of incidence and angle of refraction, measured from the interface normal vector, and $n_j=c/v_j$ are the refractive indices at the common frequency. This is Snell's law.

Let $r_1,r_2$ be the internal angle of refraction values measured from the two face normals, and let $i,e$ be the external incidence and emergence angles. Geometry of an optical prism gives $r_1+r_2=A$ and total deviation $\delta =(i-r_1)+(e-r_2)=i+e-A$.

For an isosceles optical prism whose internal beam is parallel to its base, reflection symmetry gives $r_1=r_2=A/2$. Applying Snell's law at both faces, with air refractive index approximated by one, gives $\sin i=n\sin (A/2)=\sin e$. For transmitted forward rays, $i=e$, so $i=(\delta +A)/2$. HenceThis is the symmetric minimum deviation geometry at a single wavelength. Transmission requires $n\sin (A/2)\le 1$; otherwise the proposed symmetric path cannot enter and leave through the two faces.

Because optical dispersion makes the refractive index depend on wavelength, no single angle of incidence makes every component of white light travel parallel to the base. Choose the supplied central visible wavelength $535\mathrm{nm}$, with $n_0=1.86$, as the design component. Snell's law givesKeep this angle of incidence fixed for all colours. The other components haveSubstitution givesAll three emerge: the largest value of $n\sin r_2$ is about $0.9323<1$. The angular spread between the extreme supplied wavelengths isThe respective deviations are $72.988^{\circ }$, $63.376^{\circ }$ and $60.398^{\circ }$. Applying the symmetric formula independently to the two extreme refractive indices would instead give $72.221^{\circ }-60.344^{\circ }=11.877^{\circ }$; that compares differently oriented optical prisms, not the emerging components of one fixed white-light beam. It can be a rough approximation but misses the stated fixed-incidence geometry. The supplied $370\mathrm{nm}$ value is near ultraviolet, so the computed spread refers to the entire supplied band, not a universally defined visible-white spectrum.

A simple converging optical lens bends different wavelengths through different angles because of optical dispersion. For a fixed shape in air, the lensmaker's equation gives $1/f\propto n(\lambda )-1$: the larger short-wavelength refractive index gives a shorter focal length. Different colours therefore form sharp images at different axial positions, causing longitudinal chromatic aberration, blur and coloured fringes in a white-light image. Off the optical axis, colour-dependent transverse magnification can also give lateral chromatic aberration. An achromatic doublet combines suitable optical glasses and curvatures to reduce this wavelength dependence.