Correct Answer - `pi[(h)/(sqrt(mu_(w)^(2)-1))+(D)/(2)]^(2)`
For `T.I.R sin I =(1)/(mu_(w))`
`rArr tani =(1)/(sqrt(mu_(w)^(2)-1))=(x)/(h)`
`rArr x = (h)/(sqrt(mu_(w)^(2)-1))`
Hence required area `s=pi(x+(D)/(2))^(2)`
`=pi[(h)/(sqrt(mu_(w)^(2))-1)+(D)/(2)]^(2)`