Past exam of the physics 2 paper of the University of Oxford / 2018 / 10

The Huygens-Fresnel principle constructs a later wavefront as the envelope of secondary wavelets emitted from the earlier one. At a planar interface, take a time interval during which one point on the incoming front travels distance $v_1\Delta t$ before reaching the interface, while the point that reached it first generates a wavelet of radius $v_2\Delta t$. The two right triangles share the interface segment, giving $\sin i/\sin r=v_1/v_2$. Since the refractive index is $n=c/v$, this yields Snell's law:The angles are measured from the interface normal. This geometric construction assumes homogeneous media and uses the propagation speeds at the given frequency.

Let $r=\arcsin (\sin \theta /n)$ be the angle inside the prism relative to the inward normal of face A. For the entry position shown, below the midpoint of A, and $0<\theta <90^{\circ }$, the transmitted ray first reaches face B. Its incidence angle there is $i_B=|\alpha -r|$. If $n\sin i_B<1$, a transmitted ray leaves B according to Snell's law. If $n\sin i_B>1$, it undergoes total internal reflection at B and then reaches C. At the equality, transmission is grazing.

On the reflected route through this equilateral prism, the incidence angle at C is $r$, so $n\sin r=\sin \theta <1$ permits emergence from C. The one-reflection path through an equilateral prism is drawn alongside the direct-transmission route.

Real interfaces also produce partial Fresnel reflection below the critical angle; these additional weaker rays are omitted from the principal-path sketch. Which face a ray first reaches also depends on its entrance position, so the route statement uses the entrance geometry illustrated in the source.

For $n=1.5$, the critical angle is $i_c=\arcsin (1/1.5)\simeq 41.8103^{\circ }$. Here $0\le r<i_c<\alpha =60^{\circ }$, so the condition at B is $\alpha -r>i_c$. At the threshold,Thus the requested positive-angle interval for strict total internal reflection isAt the upper equality the transmitted ray is grazing; normal incidence $\theta =0$ also gives total internal reflection at B if the endpoint is included. The distinction follows the question's positive-angle convention.

Take the positive horizontal direction along B. The initial direction is $(\alpha -90^{\circ })-\theta$. Reflection at the horizontal face B reverses the vertical direction. At C, symmetry of the equilateral geometry gives internal incidence $r$, so Snell's law makes the emerging angle to C's normal equal to $\theta$. Its direction is therefore $(90^{\circ }-\alpha )+\theta$. Subtracting the incoming direction givesThis is the smaller angle between the incident and emergent propagation directions in the requested reflection interval. The cancellation of $n$ occurs because the entry and exit refraction angles have equal magnitudes; refractive index still determines whether the reflected route exists and where it exits.

Assume normal dispersion, so shorter visible wavelengths have larger refractive indices. At fixed external angle $\theta$, the relevant criticality parameter iswhich increases with $n$ on this route. Thus shorter-wavelength blue light is totally reflected at B and reaches C, while longer-wavelength red light can transmit through B; green lies at the critical threshold. The prism separates the routes by wavelength near the green cutoff.

The totally reflected rays emerging from C have the same external deviation $180^{\circ }-2\alpha +2\theta$ despite their different refractive indices; their exit positions can differ. Red light also has a partially reflected component by the Fresnel equations, so C is not a mathematically perfect colour cutoff. The observation concerns total versus partial reflection, with the transmitted B rays spread by wavelength. Anomalous dispersion, if present, would require the actual $n(\lambda )$ instead of the normal ordering assumed here.