Question

Using Huygen's wave theory, show that (for refraction of light):

`sin i/sin r = "constant"`

where terms have their usual meaning. You must draw a neat and labelled diagram.

Answer 1

Let SS' represent the surface separating media 1 and 2 of refractive indices n₁ and n₂, respectively. Let c₁ and c₂ be the velocities of light in the two media. The second medium is optically denser than the first, hence c₁ > c₂.

APB is the incident wavefront. By the time the disturbance at B reaches C, the secondary wavelet from A has travelled a distance of AD = c₂t, where t is the time it took the waves to travel the distance of BC. Therefore, BC= c₁ t and AD= c₂t.· With A as the centre and radius AD(= c₂t), we draw a sphere and a tangent CD to the sphere is also drawn from the point C. Thus, CD represents the refracted plane wavefront. To prove that CD is the common wavefront, it is enough to show that in the time the disturbance travels from B to C or A to D, the disturbance at P reaches L. With M as the centre, we draw a sphere such that CD happens to be the tangent of the sphere.

From the similar triangles ACD and MCL,

`("AD")/("ML") = ("AC")/("MC")`  ...(i)

Similarly, from similar triangles (ACE and MCN),

`("AE")/("MN") = ("AC")/("MC")`  ...(ii)

From (i) and (ii),

`("AD")/("ML") = ("AE")/("MN")`

or `("AE")/("AD") = ("MN")/("ML")`

∵ AE = BC = c₁t and AD = c₂t, we can write,

`("AE")/("AD") = (c_1t)/(c_2t) = ("MN")/("ML")`

or `("BC")/("AD") = ("MN")/("ML") = (c_1)/(c_2)`  ...(iii)

The radius of the secondary wavefront for point A is therefore AD, while the radius of the secondary wavefront for point M is ML.

Let r and i represent the angles of refraction and incidence, respectively.

From triangle ABC and ACD,

`sini/sinr = ("BC"/"AC")/("AD"/"AC") = ("BC")/("AD") = (c_1t)/(c_2t) = c_1/c_2  = ""_1n_2`

₁n₂ is constant and is called the refractive index of the second medium with respect to the first medium. So, `sini/sinr` = constant