critical angle = the 42 degrees means when the ray of light drifting in liquid is occurrence at 42 deg with normal, the refracted ray just scratches the surface of liquid.
sin i / sin r = u (refractive index of air with respect to liquid)
so sin 42 / sin 90 = u since sin 90 =1 therefore, u of liquid w.r.t. air = sin 42 deg.
Case 1 since /_ i = 30.5 deg sin 30.5/ sin r = u (which is sin 42 deg) so sin r = sin 31 / u angle r = sin inverse of (sin 30.5 /sin 42) angle r = 49.33
Case 2 the ray travels from air to liquid
So the refractive index of liquid wrt air = 1/u
angle i = 30.5 angle r =? sin 30.5 / sin r = 1/u = 1/sin 42 so, sin r = sin 30.5 x sin 40 angle r = arc sin (sin 30.5 x sin 40)angle r = 19.04 degrees
