ch36 - 1383 Chapter 36 1. (a) We use Eq. 36-3 to calculate...

Info iconThis preview shows pages 1–3. 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: 1383 Chapter 36 1. (a) We use Eq. 36-3 to calculate the separation between the first ( m 1 = 1) and fifth 2 ( 5) m minima: 2 1 sin . m D D y D D m m m a a a Solving for the slit width, we obtain a D m m y 2 1 6 400 550 10 5 1 0 35 2 5 b g b gc hb g mm mm mm mm . . . (b) For m = 1, sin . . . m a 1 550 10 2 5 2 2 10 6 4 bgc h mm mm The angle is = sin –1 (2.2 10 –4 ) = 2.2 10 –4 rad. 2. From Eq. 36-3, a l m sin 1 sin30.0 2.00. 3. (a) A plane wave is incident on the lens so it is brought to focus in the focal plane of the lens, a distance of 70 cm from the lens. (b) Waves leaving the lens at an angle to the forward direction interfere to produce an intensity minimum if a sin = m , where a is the slit width, is the wavelength, and m is an integer. The distance on the screen from the center of the pattern to the minimum is given by y = D tan , where D is the distance from the lens to the screen. For the conditions of this problem, sin . . . m a 1 590 10 0 40 10 1475 10 9 3 3 bgc h m m This means = 1.475 10 –3 rad and y = (0.70 m) tan(1.475 10 –3 rad) = 1.0 10 –3 m. 4. (a) Equations 36-3 and 36-12 imply smaller angles for diffraction for smaller wavelengths. This suggests that diffraction effects in general would decrease. CHAPTER 36 1384 (b) Using Eq. 36-3 with m = 1 and solving for 2 (the angular width of the central diffraction maximum), we find 2 2sin 1 l a 2sin 1 0.50m 6.0m 9.6 . (c) A similar calculation yields 0.23° for = 0.010 m. 5. (a) The condition for a minimum in a single-slit diffraction pattern is given by a sin = m , where a is the slit width, is the wavelength, and m is an integer. For = a and m = 1, the angle is the same as for = b and m = 2. Thus, a = 2 b = 2(350 nm) = 700 nm. (b) Let m a be the integer associated with a minimum in the pattern produced by light with wavelength a , and let m b be the integer associated with a minimum in the pattern produced by light with wavelength b . A minimum in one pattern coincides with a minimum in the other if they occur at the same angle. This means m a a = m b b . Since a = 2 b , the minima coincide if 2 m a = m b . Consequently, every other minimum of the b pattern coincides with a minimum of the a pattern. With m a = 2, we have m b = 4. (c) With m a = 3, we have m b = 6....
View Full Document

This document was uploaded on 04/22/2011.

Page1 / 35

ch36 - 1383 Chapter 36 1. (a) We use Eq. 36-3 to calculate...

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

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