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

Options

A. `sin^(-1) mu`
B. `"sin"^(-1)(1/mu)`
C. `"tan"^(-1)(1/mu)`
D. `tan^(-1)(mu)`

Answer

D

Explanation

Suppose `i_(P)` is the Brewster angle when unpolarised light is incident at an polarising angle the reflected component along OB and refracted component along OC in the following figure

Now, `angleBOY+angleYOC=90^(@)`
`(90-i_(P))+(90^(@)-r)=90^(@)`
Here, r is the angle of refractive
i.e., `90^(@)-i_(P)=r`
According to Shell's law `mu=(sini)/(sinr)`
When `i=i_(P)r=90^(@)-i_(p)`
Then `mu=(sin i_(p))/(sin(90^(@)-i_(p)))`
`=(sin i_(p))/(cos i_(p))=tani_(p)`
`i_(p)=tan^(-1)(mu)`
where, `i_(p)` is polarised angle