Example 1
Find the critical angle from glass n=1.50 to air n=1.00.
sinθc=1.501.00=32, giving θc=41.8∘. Incidence above this value produces total internal reflection.
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Year 11 · SACE Stage 1
Total internal reflection, diffraction and polarisation
Classify light behaviour and calculate critical angles.
Method and worked examples
Total internal reflection requires light travelling from higher to lower refractive index with incidence angle above the critical angle. At the critical angle the transmitted ray is along the boundary: sinθc= nlow/nhigh. Below that angle, some light can transmit. Optical fibres use repeated total internal reflection, with an appropriately higher-index core than cladding.
Diffraction is spreading around edges or through apertures and is strongest when aperture width is comparable to wavelength. It does not require a speed change. Polarisation restricts the direction of electric-field oscillation and demonstrates light’s transverse character. Ideal crossed linear polarisers transmit no light through both; aligned ones transmit the component selected by the first. Keep these mechanisms separate from reflection and refraction in classification questions.
Example 1
Example 2
Which spreads more through the same narrow gap: wavelength 2cm or 10cm?
The 10cm wave has the larger wavelength-to-gap ratio, so it diffracts more strongly under otherwise equal conditions.
Watch out: It must also have incidence angle above the critical angle. At smaller angles a transmitted ray is possible.