In the figure shown for an angle of incidence i at the top of the surface, what is the minimum refractive index for total internal reflection at the vertical surface.

Explanation

The ray will total internally reflect at the vertical surface if `theta gt theta_(c)`
Now, `r=(90^(@)-theta)` and Snell.s law is `sin i=mu sin r`
`(sini)/(mu)=sin(90^(@)-theta) implies cos theta=(sin i)/(mu)`
or `sin theta=sqrt(1-cos^(2)theta)=sqrt(1-(sin^(2)i)/(mu^(2)))`

If `theta gt theta_(C)` , then `sin theta gt sin theta_(C)` ( As `sin theta` is an increasing function for `0 lt theta lt 90^(@))`
`sqrt(1-(sin^(2)i)/(mu^(2)))gt(1)/(mu)`
`1-(sin^(2)i)/(mu^(2))gt(1)/(mu^(2))`
`mu^(2)-sin^(2)i gt 1 "or" (mu^(2)-1)gtsin^(2)i`
If total internal reflection has to be larger for all value, the above inequality must be satisfied for all `(sin^(2)i)_("max")=1`
`impliesmu^(2)-1gt1 "or" mu gtsqrt(2)`
This total internal reflection phenomenon is used in fibre optics to bend light in a curved path.