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

Any ray entering at an angle I shall be guided along AC if the angle ray makes with the face AC `(phi)` is greater than the critical angle as per the principle of total internal reflection `phi+r 90^(@)` , therefore ` sinphi =cos r`
`rArr" " sinphige(1)/(mu)`
`rArr" " cos r ge(1)/(mu)`
or `" " 1-cos^(2)rle1-(1)/(mu^(2))`
i.e., `" " sin^(2)rle(1)/(mu^(2))`
i.,e, `" " sin^(2)rle1-(1)/(mu^(2))`
since, `" " sini-musinr`
`" " (1)/(mu_(2))sin^(2)ile1-(1)/(mu^(2))or sin^(2)ilemu^(2)-1`
when `" " i=(pi)/(2)`
Then, we have smallest angle `phi`
If that is greater than the critical angle, then all other angles of inidence shall be more than the ciritical angle
Thus `" " 1lemu^(2)-1`
or `" " mu^(2)ge2`
`rArr " " mulesqrt(2)`
This is the required result.