The angle of incidence at which reflected light is totally polarized for reflection from air to glass (refractive index n), is

Question

The angle of incidence at which reflected light is totally polarized for reflection from air to glass (refractive index n), is (A)  sin -1(n)  (B)  sin -1(1/n)  (C)  tan -1(1/n)  (D)  tan -1(n)

Tardigrade Admin · Accepted Answer

The particular angle of incidence for which reflected light is totally polarized for reflection from air to glass, is called the angle of polarisation (ip) (Brewsters law). Accordingly, n= text tan text ip ⇒ ip= tan -1(n) where n= refractive index of glass.

Q. The angle of incidence at which reflected light is totally polarized for reflection from air to glass (refractive index n), is

$sin^{- 1} (n)$

$sin^{- 1} (1/ n)$

$tan^{- 1} (1/ n)$

$tan^{- 1} (n)$

The particular angle of incidence for which reflected light is totally polarized for reflection from air to glass, is called the angle of polarisation $(i_{p})$ (Brewsters law). Accordingly, $n = t an i_{p}$ $\Rightarrow$ $i_{p} = tan^{- 1} (n)$ where $n =$ refractive index of glass.

Q. The angle of incidence at which reflected light is totally polarized for reflection from air to glass (refractive index n), is

sin−1(n)

sin−1(1/n)

tan−1(1/n)

tan−1(n)

The particular angle of incidence for which reflected light is totally polarized for reflection from air to glass, is called the angle of polarisation (ip​) (Brewsters law). Accordingly, n=tanip​ ⇒ ip​=tan−1(n) where n= refractive index of glass.

$sin^{- 1} (n)$

$sin^{- 1} (1/ n)$

$tan^{- 1} (1/ n)$

$tan^{- 1} (n)$

The particular angle of incidence for which reflected light is totally polarized for reflection from air to glass, is called the angle of polarisation $(i_{p})$ (Brewsters law). Accordingly, $n = t an i_{p}$ $\Rightarrow$ $i_{p} = tan^{- 1} (n)$ where $n =$ refractive index of glass.