HW5 - PHYS 4D Solution to HW 5 February 5 2011 Problem...

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 4D Solution to HW 5 February 5, 2011 Problem Giancoli 32-43 (II) A light beam strikes a 2 . 0-cm-thick piece of plastic with a refractive index of 1 . 62 at a 45 ◦ angle. The plastic is on top of a 3 . 0-cm- thick piece of glass for which n = 1 . 47. What is the distance D in Fig. 32-48? Solution: The beam forms the hypotenuse of two right trangles as it passes through the plastic and then the glass. The upper angle of the triangle is the angle of refraction in that medium. Note that the sum of the opposite sides is equal to the displacement D . First, we calculate the angles of refraction in each medium using Snell’s Law (Eq. 32-5) sin 45 ◦ = n 1 sin θ 1 = n 2 sin θ 2 . θ 1 = sin − 1 ( sin 45 ◦ n 1 ) = sin − 1 ( sin 45 ◦ 1 . 62 ) = 25 . 88 ◦ , θ 2 = sin − 1 ( sin 45 ◦ n 2 ) = sin − 1 ( sin 45 ◦ 1 . 47 ) = 28 . 75 ◦ . We then use the trigonometric identity for tangent to calculate the two opposite sides, and sum to get the displace- ment. D = D 1 + D 2 = h 1 tan θ 1 + h 2 tan θ 2 = 2 . cm tan 25 . 88 ◦ + 3 . cm tan 28 . 75 ◦ = 2 . 6 cm. Problem Giancoli 32-45 (II) In searching the bottom of a pool at night, a watchman shines a narrow beam of light from his flashlight, 1 . 3 m above the water level, onto the surface of the water at a point 2 . 5 m from his foot at the edge of the pool (Fig. 32-50). Where does the spot of light hit the bottom of the pool, measured from the bottom of the wall beneath his foot, if the pool is 2 . 1 m deep? Solution: We find the angle of incidence from the distances tan θ 1 = l 1 h 1 = 2 . 5 m 1 . 3 m = 1 . 9231 ,θ 1 = 62 . 53 ◦ . For the refraction from air into water, we have n air sin θ 1 = n water sin θ 2 , sin θ 2 = 1 . 00 1 . 33 sin 62 . 53 ◦ ⇒ θ 2 = 41 . 84 ◦ We find the horizontal distance from the edge of the pool from l = l 1 + l 2 = l 1 + h 2 tan θ 2 = 2 . 5 m + 2 . 1 m tan 41 . 84 ◦ = 4 . 4 m Problem Giancoli 32-54 (II) A parallel beam of light containing two wavelengths, λ 1 = 465 nm and λ 2 = 652 nm , enters the silicate flint glass of an equilateral prism as shown in Fig. 32-54. At what angle does each beam leave the prism (give angle with normal to the face)? See Fig. 32-28.leave the prism (give angle with normal to the face)?...
View Full Document

This note was uploaded on 09/19/2011 for the course PHYS 4d taught by Professor Staff during the Winter '08 term at UCSD.

Page1 / 4

HW5 - PHYS 4D Solution to HW 5 February 5 2011 Problem...

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