A ray enters glass from air at 50.0° to the normal and refracts at 30.0°. (a) Calculate refractive index. (b) Calculate critical angle. (c) Explain the parallel but displaced emerging ray. (d) Describe a ray-pin experiment, precautions and a graph method for refractive index.

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Refractive index 1.53; critical angle 40.8°. Opposite refractions at parallel faces restore direction while finite thickness creates lateral shift. For air to glass, n = sin i/sin r = sin 50.0°/sin 30.0° = 1.532, so n = 1.53. At a glass–air critical angle, refracted angle is 90° and sin c = 1/n. Thus c = sin⁻¹(1/1.532) = 40.8°. Parallel faces have parallel normals. Refraction toward the normal at entry is reversed by refraction away at exit, restoring direction; travel through finite thickness leaves lateral displacement. Trace the block, place two widely separated incident pins, align two image pins without parallax, remove the block and draw rays and normals accurately. Use thin vertical pins. Repeat several incidence angles and plot sin i vertically against sin r horizontally. A best-fit gradient through the origin estimates n; repeats and large pin separation reduce uncertainty.

Refraction through a glass block

A ray enters glass from air at 50.0° to the normal and refracts at 30.0°. (a) Calculate refractive index. (b) Calculate critical angle. (c) Explain the parallel but displaced emerging ray. (d) Describe a ray-pin experiment, precautions and a graph method for refractive index.

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