Q.(A) The following questions has choice :

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 …

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:

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

…

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:

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

…

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:

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

…

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 $\dfrac{1}{f}=(\mu-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$.