Q.Explain refraction of light on the basis of wave theory. Hence prove the laws of refraction. Two coherent sources of light having intensity ratio 81 : 1 produce interference fringes. Calculate the ratio of intensities at the maxima and minima in the interference pattern. OR State Brewster's law and show that when light is incident at polarizing angle the reflected and refracted rays are mutually perpendicular to each other. Monochromatic light of wavelength 4300 Å falls on a slit of width 'a'. For what value of 'a' the first maximum falls at 30°?

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