Reflection and refraction of light

A spoon under water appears displaced because light from it changes direction at the surface. Let us trace the actual ray path and find the angles at which light can leave water or glass.

A ray from a point under water reaches the surface. To compare its directions on either side, first draw the normal: a line perpendicular to the surface at the point of incidence. Measure the angle of incidence α, reflection β and refraction γ from this line. An angle to the surface and its corresponding angle to the normal add up to 90°. The same ray direction can therefore be described by two different numbers. Measuring from the wrong line spoils a calculation even when the formula is right.

Some light may return to the original medium. For reflection, \alpha=\beta; the incident ray, reflected ray and normal lie in one plane. Draw equal angles on opposite sides of the normal, with the reflected arrow pointing away from the surface. At normal incidence, the light returns along the same line. Increasing α by a certain angle increases β by the same amount, while the angle to the surface decreases.

A transparent material changes the speed at which light propagates. Its absolute refractive index n_i=c/v_i compares that speed with the vacuum value. Here n₁ describes the incident medium and n₂ the receiving medium. This ratio has no units. If n₂ is greater than n₁, v₂ is less than v₁. The indices apply to a chosen frequency and can differ for different colours. A larger refractive index does not necessarily mean a larger mass density.

An oblique wavefront crosses the boundary in stages: one part has already changed speed while another is still in the original medium. The front turns, and the ray refracts according to n_1\sin\alpha=n_2\sin\gamma. For n₂>n₁ it bends towards the normal; for n₂<n₁ it bends away, provided a transmitted ray exists. At α=0 the whole front crosses at once. Its direction is unchanged even though its speed changes.

Sometimes it is easier to compare the two media directly using the relative index n=n_2/n_1=v_1/v_2. This is not automatically equal to the absolute index n₂. At oblique incidence, \sin\alpha/\sin\gamma=n. Swapping the media inverts the relative index: the larger and smaller angles exchange roles. At normal incidence the sine ratio is 0/0, so use the original equation n₁ sin α=n₂ sin γ instead of this ratio.

At a stationary boundary, the wave frequency stays the same. Otherwise the oscillations on either side would continually slip out of step in time. From v=λν, \lambda_1/\lambda_2=v_1/v_2=n_2/n_1. In a medium with a larger absolute index, light travels more slowly and has a shorter wavelength. Express both wavelengths in the same units when comparing them. A wavelength change during refraction is not a change in the source frequency.

The law often gives the sine of the unknown angle first. Use the inverse operation to find the angle itself: arcsin(0,5)=30°=π/6 radians. A half-turn of π radians equals 180°. Set the calculator to the same angular units as the numbers you use. Before pressing any buttons, check the limit: the sine of a physical angle cannot exceed 1. A result above this limit may mean that no ordinary refracted ray exists, rather than that the arithmetic is wrong.

Consider light leaving a medium with n₁>n₂, and gradually increase the angle of incidence. The refracted ray moves away from the normal more rapidly until it runs along the boundary: γ=90°. The corresponding α₀ is the critical angle. Starting in the medium with the smaller index gives no such limit: the ray bends towards the normal instead. A large angle alone therefore does not guarantee total internal reflection.

Setting sin γ=1 gives \sin\alpha_0=n_2/n_1. For α>α₀ there is no propagating refracted ray in the second medium: total internal reflection occurs. If n₂≈1, we can write 1/n₁ using the absolute index of the original medium. But for the relative index defined above, n=n₂/n₁, we have sin α₀=n, not 1/n. Checking the meaning of the symbols prevents an impossible sine greater than one before we even look for the angle.

Light travels from glass with n₁=1,5 into air with n₂≈1. Find the critical angle and determine what happens at an incidence angle of 50°.

Light leaves glass with n₁ = 1,5 and enters air with n₂ ≈ 1. Since n₁ > n₂, a critical angle exists.

sin α₀ = n₂/n₁ = 1/1,5 ≈ 0,667. This is an allowed sine value between zero and one.

In degree mode, α₀ = arcsin(0,667) ≈ 41,8°.

The given angle, 50°, exceeds 41.8°. A transmitted ray would require an impossible sine greater than one, so total internal reflection occurs.

The critical angle is α₀ ≈ 41,8°. At an incidence angle of 50°, total internal reflection occurs: light is heading towards the lower-index medium and the incidence angle exceeds the critical angle.

Measure angles from the normal. A refractive index compares light speeds, not mass densities.

At a stationary boundary, frequency is unchanged. A change in speed changes the wavelength and, at oblique incidence, the direction.

Total internal reflection requires n₁>n₂ and α>α₀. Do not confuse absolute and relative refractive indices.

Next we will bring rays together with a lens and explore situations that need a wave model of light.