1018_Physics ProblemsTechnical Physics

# 1018_Physics ProblemsTechnical Physics - Chapter 36 P36.22...

This preview shows page 1. Sign up to view the full content.

Chapter 36 359 P36.22 When R →∞ , the equation describing image formation at a single refracting surface becomes qp n n =− F H G I K J 2 1 . We use this to locate the final images of the two surfaces of the glass plate. First, find the image the glass forms of the bottom of the plate. q B 1 133 166 800 641 F H G I K J . . .. cm cm a f This virtual image is 6.41 cm below the top surface of the glass of 18.41 cm below the water surface. Next, use this image as an object and locate the image the water forms of the bottom of the plate. q B 2 100 18 41 13 84 F H G I K J . . cm cm a f or 13.84 cm below the water surface. Now find image the water forms of the top surface of the glass. q 3 1 12 0 9 02 F H G I K J . cm cm a f or 9.02 cm below the water surface. Therefore, the apparent thickness of the glass is t = 13 84 9 02 4 82 . cm cm cm . P36.23 From Equation 36.8 n p n q nn R 122 1 += . Solve for q to find q nR p pn n = −− 2 21 1 bg . In this case, n 1 150 = ., n 2 = R 15 0 . cm and p = 10 0 cm . So q = 10 0 1 00 1 50 1 50 15 0 857
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

Ask a homework question - tutors are online