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Find relative refractive index when Brewster's angle is given

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A ray of light is incident from a denser to a rarer medium. The critical angle for total internal reflection is $i_C$ and the Brewster's angle of incidence is $i_B$, such that $sini_C/sini_B =  = 1.28.$
The relative refractive index of the two media is

I know $tan i_B=\mu $(refractive index)
And in this case sin$i_C = \mu$
So putting these values in$sini_C/sini_B =  = 1.28.$
I got $\sqrt{1+\mu^2}=1.28$
From this I got approx $\mu =0.6$
But answer is 0.8

asked Mar 4, 2017 in Physics Problems by koolman (4,286 points)
arithmetic error : $1+\mu^2=1.28^2=1.6384, \mu^2=0.6384, \mu=\sqrt{0.6384}=0.8$.
Sorry , what a silly mistake I have done .

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