A transparent light guide has refractive index 1.54 relative to air. Light inside it reaches a flat guide–air boundary. (a) Calculate the critical angle using sin c = 1/n. (b) Predict what happens for incidence angles of 30° and 60°, both measured from the normal. (c) Explain why a ray entering the guide from air cannot undergo total internal reflection at that first boundary.

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Critical angle 40.5°; 30° gives transmission with partial reflection; 60° gives total internal reflection; entry is from lower to higher refractive index. At the critical angle the transmitted ray would run along the boundary at 90° to the normal. Thus sin c = 1/1.54 and c = 40.5°, which must be found with the calculator in degree mode. The 30° incidence angle is below 40.5°, so a refracted ray enters the air, bending away from the normal. Some light can also be reflected, but the reflection is not total. The 60° incidence angle is greater than the critical angle for this guide. Since the ray travels from higher to lower refractive index, both conditions are satisfied and it is totally internally reflected. A ray entering from air travels from lower to higher refractive index. It therefore fails one of the necessary conditions for total internal reflection, regardless of its entry angle. Always compare angles measured from the normal, not from the surface.

Total internal reflection in a light guide 8

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