Q.Derive the laws of refraction of light using Huygens' principle.

Applying Huygens' principle at a refracting surface and using simple triangle geometry directly yields sin(i)/sin(r) = v1/v2, i.e. Snell's law.

Let XY be a plane interface separating medium 1 (wave speed v1) from medium 2 (wave speed v2). Let a plane wavefront AB be incident on XY, touching it first at A, with angle of incidence i (the angle between the incident wavefront's normal, i.e. the ray, and the normal to XY).

Let the wavefront take time t to travel from B to a point C on the surface, so BC = v1 t.

During this same time t, the secondary wavelet from A (now already in medium 2) spreads out with radius AD = v2 t.

The new refracted wavefront is the common tangent CD to these secondary wavelets, making angle of refraction r with the normal to XY.

In right triangle ABC (right angle at B): sin(i) = BC / AC = v1 t / AC

In right triangle ADC (right angle at D): sin(r) = AD / AC = v2 t / AC

Dividing:

sin(i) / sin(r) = v1 / v2

Since absolute refractive index n = c/v, we have v1 = c/n1 and v2 = c/n2, so v1/v2 = n2/n1. Substituting:

sin(i)/sin(r) = n2/n1

n1 sin(i) = n2 sin(r)

which is Snell's law of refraction.

By constructing the secondary wavelets from a plane wavefront as it crosses a boundary into a medium of different speed, simple triangle geometry reproduces Snell's law of refraction.

Consider a plane wavefront AB incident on a plane interface XY separating medium 1 (wave speed $v_1$) from medium 2 (wave speed $v_2$), with angle of incidence $i$. By Huygens' principle, every point on the wavefront acts as a source of secondary wavelets.

Let the time taken for the wavelet from B to reach the interface at C be $t$: $BC = v_1 t$. During the same time $t$, the wavelet from A (which reached the interface first) spreads into medium 2 with radius $AD = v_2 t$. The new wavefront in medium 2 is the tangent CD from this secondary wavelet.

From the geometry, in right triangle ABC: $\sin i = \dfrac{BC}{AC} = \dfrac{v_1 t}{AC}$

In right triangle ADC: $\sin r = \dfrac{AD}{AC} = \dfrac{v_2 t}{AC}$

Dividing:

$\dfrac{\sin i}{\sin r} = \dfrac{v_1}{v_2}$

Since refractive index $n = c/v$, this ratio $v_1/v_2 = n_2/n_1$, giving:

$n_1 \sin i = n_2 \sin r$ — Snell's law.

(Also, since A, D, C, B all lie in the same plane, the incident ray, refracted ray, and normal are shown to lie in one plane — the first law of refraction.)

By constructing the refracted wavefront geometrically from Huygens' principle and comparing the two right triangles formed at the interface, the ratio sin i / sin r reduces to a constant, giving Snell's law.

Consider a plane wavefront AB incident on a plane refracting surface XY (separating medium 1, speed v1, from medium 2, speed v2), making angle of incidence i with the normal.

By Huygens' principle, every point on the wavefront AB acts as a source of secondary wavelets. Let the wavefront take time t to travel from B to C in medium 1 (so BC = v1·t), while the secondary wavelet from A spreads into medium 2, covering a distance AD = v2·t in the same time t. The new refracted wavefront is the tangent CD to this secondary wavelet, making angle of refraction r with the surface.

From right triangle ABC: $\sin i = \dfrac{BC}{AC} = \dfrac{v_1 t}{AC}$

From right triangle ADC: $\sin r = \dfrac{AD}{AC} = \dfrac{v_2 t}{AC}$

Dividing:

$\dfrac{\sin i}{\sin r} = \dfrac{v_1}{v_2}$

Since $v_1$ and $v_2$ are constants for the two given media, this ratio is a constant, denoted $n_{21}$ (refractive index of medium 2 with respect to medium 1):

$\dfrac{\sin i}{\sin r} = n_{21} = \text{constant}$

This is Snell's law of refraction, derived purely from the wave picture of light.

Comparing wavefronts of an incident and refracted plane wave via Huygens' construction gives $\sin i/\sin r = v_1/v_2$, a constant — the law of refraction.

Let a plane wavefront AB be incident on a plane interface XY separating a rarer medium (speed $v_1$) from a denser medium (speed $v_2 < v_1$), the wavefront making angle of incidence $i$ with the interface's normal. By Huygens' principle, every point on a wavefront is a source of secondary wavelets; the new wavefront is the common tangent (envelope) to these wavelets.

Let the time taken for the disturbance from B (on the interface) to reach C (on the interface, at the far end of the incident wavefront's footprint) be $\tau$, so $BC = v_1\tau$. In the same time $\tau$, the secondary wavelet from A (which reaches the interface first) spreads out into the denser medium with radius $AD = v_2\tau$. The refracted wavefront is the tangent CD drawn from C to this wavelet.

From the geometry, in triangle ABC: $\sin i = \dfrac{BC}{AC} = \dfrac{v_1\tau}{AC}$.

In triangle ACD: $\sin r = \dfrac{AD}{AC} = \dfrac{v_2\tau}{AC}$.

Dividing:

$\dfrac{\sin i}{\sin r} = \dfrac{v_1}{v_2}$

Since $v_1$ and $v_2$ are constants for a given pair of media, this ratio is a constant, $^1n_2$ — this is Snell's law: $\dfrac{\sin i}{\sin r} = {}^1n_2 = \text{constant}$. Also, since $i$ (in the incident medium) and $r$ (refracted ray) lie in the plane of incidence, along with the normal, this proves the incident ray, refracted ray and normal are coplanar (the second law of refraction). Since the wave is going from a rarer to a denser medium ($v_1>v_2$), $r<i$ — the ray bends towards the normal.

(Ray diagram: draw interface XY horizontal; incident wavefront AB tilted at angle i to XY with A touching XY; normal at A drawn vertical; refracted wavefront CD tilted at angle r to XY on the other side, with the arc of radius $v_2\tau$ centred at A tangent to CD.)

Part (a): A wavefront is a surface of constant phase; Huygens' construction gives Snell's law $\dfrac{\sin i}{\sin r}=\dfrac{v_1}{v_2}=\dfrac{n_2}{n_1}$, with the ray bending away from the normal on going from a denser to a rarer medium.

Part (b): Applying the refraction-at-a-surface relation at both faces of a thin biconvex lens and adding yields the thin-lens formula $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$, with $\dfrac{1}{f}=(\mu-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$.