Finding the refractive index using total internal reflection Total internal reflection gives us another method of finding the refractive index:
Refractive index n = 1
______ sin C
n = refractive index, C = critical angle (°) NOTE
You are free to use any or all methods of calculating refractive index in any question, unless specified otherwise:
n = sin i _____ sin r
AnB = 1 ____
BnA
n = Real depth Apparent depth
___ H c2
n = QUESTIONS AND ANSWERS
5. If the critical angle for a material is 45°, what is its refractive index?
Solution n = 1
______ sin C = 1
________ sin 45 = 1.4142
6. A coin is submerged in a beaker of fl uid whose critical angle is 42°. The coin appears to be 20 cm below the surface of the water. What is its real depth?
Solution n =
____________________ Apparent depth = 1
Real depth
so: Real depth =
sin C
______ sin C
Apparent depth ____________________
= 0.299 m You may now complete Exercise 14C (page 152). Applications of TIR
Total internal reflection is behind many natural phenomena and has important industrial applications.
Prisms
• If, as in Fig. 14.13a, a ray of light strikes the interior side of the prism at 45°, its angle of incidence i is greater than the critical angle. This means TIR occurs and i = r = 45°, which causes the light to turn 90° from its original path.
• If the prism is turned 90°, as in Fig. 14.13b, the ray of light undergoes TIR twice and ends up turning through 180° from its original path.
LEAVING CERTIFICATE PHYSICS 147 = 0.2
_________ sin 42°
H
7. By what factor has the speed of light decreased when it travels from a vacuum to a liquid of refractive index 1.25?
