Question

Show that Brewster's Law (where the incident angle( $\Theta_{i} = \Theta_{\rho }$ ) and Snell's Law together imply that $\Theta_{\rho } +\Theta _{2} = 90$ degrees.

Answer 1

Way A/c to Brewster's law mu = tan theta_p mu = sin theta_p/cos theta_p- (1) A/c to Snell's law mu = sin theta_i/sin theta_2- (2) given that theta_i = theta_p Put in (2) mu = sin theta_p/sin theta_r- (3) equate (1) & (3) sin theta_r = cos theta_p cos(90 degree- theta_2) = cos theta_p 90- theta_r = theta_p theta_p + theta_r = 90 degree \r\n Just for Understanding We know that reflected angle is equal to angle of incidence & given that theta_i = theta_p Also, reflected & refracted be ams are always at right angle to each other therefore theta_p + 90 degree + theta_r = 180 degree theta_p + theta_r = 90 degree (theta_r is angle of refraction)