A beam of width t incident at `45^@` on an air-water boundary. The width of the beam is water is `________`

Explanation

`AB=(t)/(cos45^@)=(t^('))/(cosr)rArr t^(')=t(cosr)/(cos45^@)`
`mu=(sin45^@)/(sinr)rArr sin r =(1)/(sqrt(2)mu)`
Solving we get, `t^(')=(tsqrt(2mu^(2)-1))/(mu)`