Light - 06 - Refraction - Critical Angle TIR

University:
High School
Course:
High School Physics
Academic year:

2024
Views:

204

Pages:

16
Author:

1 1

Refraction Critical Angle & Total Internal Reflection Please Write ☺ 7.3 Total Internal Reflection:  When light travels from one medium to another, refraction occurs, BUT also some reflection.  When light travels from higher n to lower n, light is refracted to a greater extent. Please Write ☺  Can increase θi until the angle of refraction θR reaches a maximum of 90o (i.e. it’s along the surface of the medium)  At this point, refraction ceases & all light is reflected back → Total Internal Reflection (TIR)  The angle of incidence at which TIR occurs is the critical angle, θc Refraction & reflection Air Glass R i i  c r Air Glass When calculating θc , angle of refraction θR is always 90o!!! R Refraction along medium surface & reflection i = c i r Total internal reflection Air Glass i r i  c Ex #1 Calculate the critical angle for the interface between water (n=1.33) & air (n=1) water air n1 sin1 = n2 sin 2 (1.33) sin 1 = (1) sin 90 o  c = 48.8 o Recall: critical angle will only exist if going from high n to low n Ex #2 Calculate the critical angle for the interface between crown glass (n=1.52) & air (n=1) glass air n1 sin1 = n2 sin 2 (1.52) sin 1 = (1) sin 90 o  c = 41.1 o Ex #3 A jeweler wants to display her diamonds in such a way that they capture the light & sparkle as much as possible. What range of incident angles must light strike inside the diamond to ensure TIR (n=2.42)? diamond air n1 sin1 = n2 sin 2 o (2.42) sin 1 = (1) sin 90  = 24.4 o Recall: i  c Glass n=1.5 Diamond n=2.42 Diamond cutters want TIR so must cut at correct angles & want to allow light to exit the top Glass has a high critical angle (42o), so light escapes Light undergoes TIR before exiting Ex #4 (1.33) sin( ) = (1) sin 90o  = 48.8 o i  c The angle of incidence of light in water & heading towards air is 50o. Calculate the angle of refraction. water air n1 sin1 = n2 sin 2 (1.33) sin(50) = (1) sin   = ??? The angle of incidence is greater than the critical angle, so TIR occurs, therefore no refraction Try this!  Please Write ☺  Substances with higher indices of refraction (n) refract the light to a greater degree & thus have lower critical angles Ex #5 Mark is lying by the side of a swimming pool and looking down at a light source located at the bottom of the pool, which is 1.35 m deep. Mark's eye n air = 1.00  n water= 1.33 1.35 m Light source d=? If the light ray is to enter Mark's eye, at what distance d from the side of the pool is the light source located? n1 sin1 = n2 sin 2 (1.33) sin( ) = (1) sin 90 o  c = 48.8 d tan 48.8 = 1.35 d = 1.54m If further away, angle of incidence increases beyond critical angle & TIR occurs! Quantum HW: pg 86 #3, 4, 5, 11 & 13 pg 88 #1, 2, 3, 5 & 6