the critical angle for the total internal reflection at the liquid air interface
ID: 1897932 • Letter: T
Question
the critical angle for the total internal reflection at the liquid air interface is 42.5 degrees. (a) If a ray of light traveling in the liquid has an angle of incidence of 35.0 degrees at the interface, what angle does the refracted ray in the air make with the normal? (b) If a ray of light traveling in air has an angle of incidence of 35.0 degree at the interface, what angle does the refracted ray in the liquid make with the normal?Explanation / Answer
critical angle = 42.5 degrees ==> when the ray of light travelling in liquid is incident at 42.5 deg with normal, the refracted ray just grazes the surface of liquid. ==> that the angle of refraction is 90 degrees. If the angle of incidence increases beyond 42.5 degrees, the ray will no longer be refracted, but it will be reflected as happens in a plane mirror. sin i / sin r = u sin 42.5/ sin 90 = u u= sin42.5 u of liqd w.r.t. air = 0.675 i = 35 deg sin 35 / sin r = u ==> sin r = sin 35/ u angle r = sin inverse of (sin 35 /sin 42.5) =sin inverse(0.849) =58.1 degrees Case 2 the ray travels from air to liquid So the refrective index of liquid wrt air = 1/u angle i = 35 angle r =? sin 35 / sin r = 1/u = 1/sin 42.5 so, sin r = sin 35 x sin 42.5 =0.387 r = arc sin (sin31 x sin 40) =22.77 degrees
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.