Phys 375 Sol HW 2

Phys 375 Sol HW 2 - Phys 375 HW 2 Fall 2009 Due 21 22...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Phys 375 HW 2 Fall 2009 Due 21 / 22 September, 2009 1. Pedrotti 3 , 3 rd edition, problem 2-7 (see Fig. 2-33). Solution: See FIGURE 2-33in the text P 3 From the geometry it is clear that h D C 4 / tan = θ , where h is the height of the slab and D is the diameter of the circle of light. From Snell’s law we know that the critical angle occurs when the angle of refraction is 2 π θ = R . Then applying Snell’s Law, 2 2 1 1 sin sin θ θ n n = we have: ( 29 55 . 1 4 / 4 / sin 2 / sin 2 2 = + = = h D D n n C air glass θ π Where I used 1 = air n . 2. Pedrotti 3 , 3 rd edition, problem 3-6. Solution: See FIGURE 3-9 of the text P 3 For the deviation angle, δ , we have from the figure and equation 3-9 and 3-11 : A- + = = ′- ′- + = 2 1 2 1 2 1 θ θ θ θ θ θ δ Combining this with equations 3-(7-10), we get: ( 29 A n A n- - + =-- 1 1 1 1 sin sin sin sin θ θ δ Here is a plot of the deviation angle vs. incident angle for n=1.52 and A=60 o ....
View Full Document

This note was uploaded on 12/29/2011 for the course PHYSICS 375 taught by Professor Eno during the Spring '11 term at Maryland.

Page1 / 3

Phys 375 Sol HW 2 - Phys 375 HW 2 Fall 2009 Due 21 22...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online