Show that for a material with refractive index `mu ge sqrt(2)`~ , light incident at any angle shall be guided along a length perpendicular to the incident face.

Explanation

`rArr` As shown in the figure , consider a light ray `vec (PQ)`, made incident on the surface AB of a denser transparent medium, with angle of incidence i,
After refraction, ray `vec(QR)` is incident at point R on the surface of rarer medium with angle of incidence `pi` . If `pi ge` C then light ray will have total inernal reflection again and again but it will not com out of walls AC and BD. (Where C = cirtical angle of given denser medium w.r.t rarer medium).Thus,
`phit ge C`
`therefore 90^@ - r ge C " "(because phi + r + 90^@)`
`therefore sin (90^@ - r) ge sin C`
`cos r ge (1)/(mu) " " ( because sin C = 1/mu)`
`therefore cos r ge 1`........(1)
`rArr` Applying Snell.s law at point Q,
(1) sin i `= mu sin r`
`therefore sin r = (sin i)/(mu)`.......(2)
`rArr` Now ` cos r sqrt( 1 - sin^2 r)`
`= sqrt(1 - (sin^2 i)/(mu^2))`
` = sqrt(mu^2 - sin^2 i)/(mu)`
`therefore mu cos r = sqrt(mu^2 - sin^2 i)/(mu)`
`therefore cos r = sqrt(mu^2 -sin^2 i).........(3)`
`rArr` From equation (2) and (3),
`sqrt(mu^2 - sin^2i) ge 1`
`therefore mu^2 - sin^2 i ge 1`
`rArr` But maximum value of sin i is 1. Hence if above codition is satisfied for ` i = 90^@` then for all the value of i , above condition will be satisfied,
Hence taking sin `i = sin 90^@ = 1,`
`mu^2 - 1 ge 1`
`therefore mu^2 ge 2`
`therefore mu ge sqrt(2)`
`rArr` Above relation gives required condition.