Refraction and Prisms Light does the twist!

A prism (n = 1.45) with an apex angle of 60° has a beam of light strike it at 40° from the normal. Find the angle of deviation. 60° 40° n = 1.45

A prism (n = 1.45) with an apex angle of 60° has a beam of light strike it at 40° from the normal. Find the angle of deviation. 60° 40° n = 1.45 First, predict where the beam of light will go: 1 2 or 3? 1 2 3

A prism (n = 1.45) with an apex angle of 60° has a beam of light strike it at 40° from the normal. Find the angle of deviation. 60° 40° Step 1 nisini=nRsin R 1 sin 40°=1.45 sin R R= 26.3° 26.3° n = 1.45

A prism (n = 1.45) with an apex angle of 60° has a beam of light strike it at 40° from the normal. Find the angle of deviation. 60° 40° Step 1 nisini=nRsin R 1 sin 40°=1.45 sin R R= 26.3° Step 2 - big triangle A: 90°- 26.3°= 63.7° B: 180°-63.7° -60° = 56.3° A 26.3° B n = 1.45

A prism (n = 1.45) with an apex angle of 60° has a beam of light strike it at 40° from the normal. Find the angle of deviation. 60° 40° Step 1 nisini=nRsin R 1 sin 40°=1.45 sin R R= 26.3° Step 2 - big triangle A: 90°- 26.3°= 63.7° B: 180°-63.7° -60° = 56.3° Step 3 - Snell’s law again Find i : 90°- 56.3° = 33.7° nisini=nRsin R 1.45 sin 33.7° =1 sin R R= 53.6° A B 26.3° 33.7° n = 1.45

A prism (n = 1.45) with an apex angle of 60° has a beam of light strike it at 40° from the normal. Find the angle of deviation. 60° 40° Step 1 nisini=nRsin R 1 sin 40°=1.45 sin R R= 26.3° Step 2 - big triangle A: 90°- 26.3°= 63.7° B: 180°-63.7° -60° = 56.3° Step 3 - Snell’s law again Find i : 90°- 56.3° = 33.7° nisini=nRsin R 1.45 sin 33.7° =1 sin R R= 53.6° A B Step 4 - small triangle Extend entering and exiting rays until they cross & use ‘X’ patterns C: 40°- 26.3°= 13.7° D: 53.6° -33.7° = 19.9° C 26.3° D 33.7° n = 1.45

A prism (n = 1.45) with an apex angle of 60° has a beam of light strike it at 40° from the normal. Find the angle of deviation. 60° 40° Step 1 nisini=nRsin R 1 sin 40°=1.45 sin R R= 26.3° Step 2 - big triangle A: 90°- 26.3°= 63.7° B: 180°-63.7° -60° = 56.3° Step 3 - Snell’s law again Find i : 90°- 56.3° = 33.7° nisini=nRsin R 1.45 sin 33.7° =1 sin R R= 53.6° A B Step 4 - small triangle Extend entering and exiting rays until they cross & use ‘X’ patterns C: 40°- 26.3°= 13.7° D: 53.6° -33.7° = 19.9° E: 180°-13.7° -19.9° = 146.4° C E 26.3° D 33.7° n = 1.45

A prism (n = 1.45) with an apex angle of 60° has a beam of light strike it at 40° from the normal. Find the angle of deviation. 60° 40° Step 1 nisini=nRsin R 1 sin 40°=1.45 sin R R= 26.3° Step 2 - big triangle A: 90°- 26.3°= 63.7° B: 180°-63.7° -60° = 56.3° Step 3 - Snell’s law again Find i : 90°- 56.3° = 33.7° nisini=nRsin R 1.45 sin 33.7° =1 sin R R= 53.6° A B Step 4 - small triangle Extend entering and exiting rays until they cross & use ‘X’ patterns C: 40°- 26.3°= 13.7° D: 53.6° -33.7° = 19.9° E: 180°-13.7° -19.9° = 146.4° C E F 26.3° D 33.7° n = 1.45 Last step!!! deviation F: 180°-146.4° = 33.6°

© David Robinson -2003 Toronto District Christian High This file may be used by anyone for the purpose of showing others the beauty and order of the world around us. This file may not be sold or used in any commercial venture without my express permission.