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 = (sin 45^@)/(sin r) implies sin r = (sin 45^@)/(mu)`
`therefore sin r = 1/(mu sqrt(2))`
At point B for total internal reflection
`sin i = 1/(mu)`
`i= 90 - r `
` therefore sin (90^@ - r) = 1/mu = cos r = 1/mu " " ...(1)`
Now `cos r = sqrt(1 - sin^2 r) = sqrt(1 - 1/(2 mu^2)) = sqrt((2mu^2 - 1)/(2mu)) " " ....(2)`
Putting the value of cos r in equation (2) and squaring both sides
`1/mu = sqrt((2mu^2 - 1)/(2 mu^2)) implies 1/(mu^2) = (2 mu^2 - 1)/(2 mu^2)`
`2 = 2 mu^2 - 1 implies 3/2 = mu^2 implies mu = sqrt(3/2)`