Correct Answer - `0.0513 m`
Here, `i_(1) = 60^(@), t = 0.1m, mu = 1.5`
As `(sin i_(1))/(sin r_(1)) = mu`
`:. Sin r_(1) = (sin i_(1))/(mu) = (sin 60^(@))/(1.5) = 0.5773`
`r_(1) = sin^(-1)(0.5773) = 35.3^(@)`
Lateral shift `= (t sin (t_(1) - r_(1)))/(cos r_(1))`
`= (0.1 sin (60^(@) - 35.3^(@)))/(cos 35.3^(@)) = (0.1 sin 24.7^(@))/(cos 35.3^(@))`
`= (0.1 xx 0.418)/(0.816) = 0.0513 m`