# 31 - Kapoor(mk9499 oldhomework 31 Turner(60230 This...

This preview shows pages 1–3. Sign up to view the full content.

Kapoor (mk9499) – oldhomework 31 – Turner – (60230) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points The index oF reFraction For violet light in silica ±int glass is 1 . 65, and that For red light is 1 . 61. R O Y G B V Visible light Deviation of yellow light Measure of dispersion Screen What is the angle oF deviation For the red ray passing through a prism oF apex angle 58 . 4 iF the angle oF incidence is 44 . 2 ? Correct answer: 46 . 3453 . Explanation: Given : n = 1 . 61 and θ 1 = 44 . 2 . Going From air to glass, sin θ 1 = n 2 sin θ 2 . θ 1 θ 4 θ 3 θ 2 φ α β Let the angles oF incidence and re±ection For the red light be θ 1 , θ r 2 , θ r 3 , and θ r 4 . At the frst surFace sin θ r 2 = sin θ 1 n θ r 2 = sin 1 p sin θ 1 n P = sin 1 p sin 44 . 2 1 . 61 P = 25 . 6595 . Considering the top triangle, α + β + φ = 180 (90 - θ 2 ) + (90 - θ 3 ) + φ = 180 θ 2 + θ 3 = φ θ r 3 = φ - θ r 2 = 58 . 4 - 25 . 6595 = 32 . 7405 . Going From glass to air, n sin θ r 3 = sin θ r 4 θ r 4 = sin 1 ( n sin θ r 3 ) = sin 1 (1 . 61 sin 32 . 7405 ) = 60 . 5453 . The angle oF deviation For the red ray is θ rd = θ r 4 + θ 1 - φ = 60 . 5453 + 44 . 2 - 58 . 4 = 46 . 3453 . 002 (part 2 oF 2) 10.0 points What is the angular dispersion oF visible light with the same angle oF incidence? Correct answer: 4 . 74481 . Explanation: Given : n = 1 . 65 . Let the angles oF incidence and reFraction For the violet light be θ 1 , θ v 2 , θ v 3 , and θ v 4 . θ v 2 = sin 1 p sin θ 1 n P = sin 1 p sin 44 . 2 1 . 65 P = 24 . 9941 . The incident angle at the second surFace For violent light is θ v 3 = φ - θ v 2 = 58 . 4 - 24 . 9941 = 33 . 4059 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Kapoor (mk9499) – oldhomework 31 – Turner – (60230) 2 The refraction angle at the second surface for violent light is θ v 4 = sin 1 ( n sin θ v 3 ) = sin 1 [(1 . 65) sin 33 . 4059 ] = 65 . 2901 .
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 06/28/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.

### Page1 / 5

31 - Kapoor(mk9499 oldhomework 31 Turner(60230 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online