Question

Problem 2 Show that for an index ofrefraction k that D(a) is minimized when sin α- 4 Use this to obtain p- arcsin to obtain ß arcsinand fil in the following table. 4-k2 3k2 a in degrees B in degrees R(a) in degrees Color Red Orange Yellow Green Bluc Indigo Violet 1.331 1.332 1.333 1.335 1337 1.340 1.344 The values of R(a) you obtained in the table accounts for the various color bands that appear in the rainbow as sunlight hits droplets at various angles Snlight The primary rainbow is created by light entering the upper portion of the raindrop If light is bright enough then a secondary rainbow is created from light entering the lower portion of the raindrop as in the diagram below D(a)

Answer 1

Given that sin alpha = squareroot 4 - k^2/3 alpha = sin^-1(squareroot 4 - k^2/3) beta = sin^-1(squareroot 4 - k^2/3k^2 R(alpha) = D(alpha) - pi D(alpha) = 2pi + 2 alpha - 4 beta \r\n given rightarrow D(alpha) = (alpha - beta) + (360 degrees - 4 beta) + (d - beta) = 360 degrees + 2 alpha - 6 beta D(alpha = 2pi + 2 alpha - 6 beta To find the value of alpha we consider scells Law. sin alpha = k sin beta and D(alpha) = 2 alpha - 6 beta + 2pi Differentiate D w.r.t to alpha. partial differential D/partial differential alpha = 2 - 6 partial differential beta/partial differential alpha = 0 implies partial differential beta/partial differential alpha = 1/3 sin alpha = k sin beta [Differentiate w.r.t alpha] cos alpha = k cos beta d beta/d alpha cos alpha = k/3 cos beta [sin^2 theta cos^2 theta = 1 cos^2 theta = 1 - sin^2 theta sin beta = sin alpha/k