In the figure shown , for an angle of incidence `45^(@)`, at the top surface , what is the minimum refractive index needed for the internal reflection at vertical face ?

Options

A. `(sqrt(2)+1)/(2)`
B. `sqrt((3)/(2))`
C. `sqrt((1)/(2))`
D. `sqrt(2)+1`

Answer

B

Explanation

At point `A` , by Snell's law
`mu=(sini 45)/(sin r)implies sin r =(1)/(sqrt(2))` …(i)
At point `B`, for total internal reflection `sin i_(1)=(1)/(mu)` From figure ,`i_(1)=90-r :. sin (90-r)=(1)/(mu)`
`implies cos r=(1)/(mu)` …(ii)
Now `cos r =sqrt(1-sin^(2)r)=sqrt(1-(1)/(2mu^(2)))=sqrt((2mu^(2)-1)/(2mu^(2)))` ...(iii)
From equation (ii) and (iii) `(1)/(mu)=sqrt((2mu^(2)-1)/(2mu^(2)))`
Squaring both side and then solving we get `(1)/(mu)=sqrt((3)/(2))`