A glass rod having square cross-section is bent into the shape as shown in the figure. The radius of the inner semi-circle is R and width of the rod is d. Find the minimum value of `d//R` so that the light that enters at A will emerge at B. Refractive index of glass is `mu=1.5`

Explanation

Minimum possible angle for internal reflection will be available for the ray coming along the inner edge as shown in figure.
For total internal reflection to take place at the point of incidence we can write the following condition :

`sin theta ge sin theta_(C )`
`rArr" "sin theta ge (1)/(1.5)`
`rArr" "(R )/(R+d)ge(1)/(1.5)`
`rArr" "1.5R ge R+d`
`rArr" "(d)/(R )le0.5`
Hence maximum possible value for `d//R` is 0.5.