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`

Answer

`((d)/(R ))_(max)=(1)/(2)`

Explanation

The path of incident parallel beam is shown in Fig.The beam entering face`A`will emerge from the face`B`,only if suffer total internal reflection at`Q`for ray`PQ`and at `F` for ray `EF`shown in fig.Obvisouly the angle of incidence is minimum for the light ray passing near inner wall.For total internal reflection angle of incidence must be greater than critical angle .For ray`EF`the angle of incident is given by
`sin i=(R+x)/(R+d)`
for minimum value ofi.x,=0
`:.sin i_(min)=(R)/(R+d)`
For total internal reflection
`i_(min)gtC`
`sin i_(min)gtsinC`
`(R)/(R+d)gtsin C`
As`SinC=(1)/(mu)=(1)/(1.5)=(2)/(3)`
`:.(R)/(R+d)gt(2)/(3)rArr3Rgt2R+2d rArrRgt2d`
`rArr (d)/(R)lt(1)/(2)`
`((d)/(R))_(max)=(1)/(2)`