reflection refraction and total internal reflection laws

At a smooth interface between two media, an incoming wave is partly reflected and partly transmitted, and the directions of the resulting rays follow from applying boundary conditions to the wave field on both sides of the surface. Along the interface, the tangential component of the wave vector must be continuous; this phase-matching condition yields the law of reflection $\theta_i=\theta_r$ and Snell's law $n_1\sin\theta_1=n_2\sin\theta_2$, where angles are measured from the normal. The refractive index $n=c/v$ measures the factor by which the wave slows in a medium, so the wavelength shortens to $\lambda_2=\lambda_1 n_1/n_2$ while the frequency stays fixed. Because $|\sin\theta_2|\le 1$, a wave traveling from a denser to a less dense medium ( $n_1>n_2$ ) has no transmitted direction when $\theta_1$ exceeds the critical angle $\theta_c=\arcsin(n_2/n_1)$; all incident energy is then reflected back into the first medium, a phenomenon called total internal reflection.

This picture is consistent with the wave propagation and interference ideas you already met: reflection and refraction are coherent superpositions of scattered wavelets whose phases add constructively only in the directions that satisfy the boundary condition. For angles beyond $\theta_c$, the transmitted field becomes an evanescent wave that decays exponentially away from the interface and carries no net energy across it. Total internal reflection underlies optical fibers, prisms, and many sensing devices.

A ray of light strikes a flat mirror. The angle between the incident ray and the mirror surface is $35^\circ$. Find (a) the angle of reflection and (b) the angle between the incident and reflected rays.

Light traveling in water ($n_1=1.33$) strikes a water–glass boundary at an angle of incidence $40.0^\circ$. The glass has $n_2=1.52$. Find the angle of refraction. If the vacuum wavelength is $600\,\mathrm{nm}$, find the speed and wavelength of the light in the glass.

An optical fiber has a core with $n_1=1.50$ and a cladding with $n_2=1.45$. (a) Find the critical angle at the core–cladding boundary. (b) A ray inside the core hits the boundary at $80^\circ$ to the normal. Is it guided by total internal reflection? (c) What happens if the incidence angle drops to $70^\circ$?

Use the PhET Bending Light simulation (or an equivalent ray-optics simulator). Set the top medium to air ($n=1.00$) and the bottom to water ($n=1.33$). Vary the angle of incidence from $10^\circ$ to $80^\circ$ in steps of $10^\circ$; for each angle record $\theta_1$, $\theta_2$, and compute $\sin\theta_1$ and $\sin\theta_2$. Plot $\sin\theta_2$ versus $\sin\theta_1$ and verify that the slope is $n_1/n_2$ within reading error. Then reverse the media (water above, air below) and slowly increase $\theta_1$ until the refracted ray disappears; record the critical angle and compare it with $\arcsin(1/1.33)\approx 48.8^\circ$. Do not use a real laser pointer: direct eye exposure to even a low-power beam can cause injury, and the simulation gives the same angles without risk.

To make the geometry of reflection, refraction, and total internal reflection visible: identify the normal and the incident, reflected, and refracted rays, and see that beyond the critical angle the refracted ray vanishes.

A two-panel labeled sketch. Panel A: a horizontal boundary between medium 1 (top, $n_1$) and medium 2 (bottom, $n_2$); a dashed vertical normal; an incident ray making angle $\theta_1$ with the normal; a reflected ray at $\theta_1'=\theta_1$; a refracted ray bent toward the normal at $\theta_2<\theta_1$ for $n_2>n_1$. Panel B: the same boundary with light going from denser medium 1 to rarer medium 2; $\theta_1$ is drawn larger than the critical angle $\theta_c$, the refracted ray is absent, and only the reflected ray remains, with the label 'total internal reflection'.

A swimmer lies 1.5 m below the surface and looks straight up. The water has $n_1 = 1.33$ and the air above it $n_2 = 1.00$. The visible sky is squeezed into a circle of half-angle $\theta_c = \arcsin(1.00/1.33) = 48.8^\circ$; beyond that circle the surface acts as a mirror and reflects the pool floor and the swimmer's own legs.

Snell's law $n_1\sin\theta_1 = n_2\sin\theta_2$ is solved for transmission into air: once $\theta_1 > \theta_c$ no real angle $\theta_2$ satisfies it ($\sin\theta_2 > 1$), the transmitted wave disappears and only the reflected ray remains — total internal reflection.

A 12 km fibre cable carries a 1550 nm signal. The core has $n_1 = 1.48$ and the cladding $n_2 = 1.46$, so the critical angle at the core wall is $\theta_c = \arcsin(1.46/1.48) = 80.6^\circ$. A ray launched into the end face inside the acceptance cone strikes the wall at more than $80.6^\circ$ to the normal and is guided along the whole 12 km, losing only about 0.2 dB/km to absorption.

The law of reflection fixes the angle at every bounce off the wall, while the total internal reflection condition $\sin\theta_c = n_2/n_1$ keeps the ray inside the core instead of leaking into the cladding.

A straight straw stands in a glass of water 8 cm deep. Light from a point on the straw reaches the surface at $\theta_1 = 45^\circ$ to the normal inside the water ($n_1 = 1.33$) and leaves into air ($n_2 = 1.00$) at $\theta_2 = \arcsin(1.33\sin 45^\circ) = 70.2^\circ$, bent strongly away from the normal. The eye continues the rays straight back and places that point at about $8/1.33 = 6$ cm, so the straw looks broken and shifted.

Snell's law $n_1\sin\theta_1 = n_2\sin\theta_2$: on entering the optically rarer medium the ray bends away from the normal, so the exit angle exceeds the angle of incidence.