Ph12a_s9_2009

Ph12a_s9_2009 - on the moon: W L D (240000 miles) 632 . 8...

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

View Full Document Right Arrow Icon
Ph12a Solution Set 9 December 1, 2009 9.4 The geometry of this problem is illustrated in Fig. 1. The light entering the two slits will be reasonably coherent when the path difference to the two slits from an edge of the source is small compared to the wavelength of light. Let D 1 and D 2 be distance to the top and bottom slit from the top of the source, as shown. Then D 1 = r L 2 + ( a - d ) 2 4 L 0 ± 1 - ad 2 L 0 2 ² D 2 = r L 2 + ( a + d ) 2 4 L 0 ± 1 + ad 2 L 0 2 ² where L 0 2 = L 2 + a 2 + d 2 4 . We certainly require L ± a and L ± d , so we may take L = L 0 . Hence the path difference D 2 - D 1 = ad L . The light illuminating the slits is coherent if ad L ² λ , or L ± ad λ . 9.6 The actual angular width of Venus is roughly 8000 miles 93 × 10 6 miles 8 . 6 · 10 - 5 . Visible light has a wavelengh of roughly 550 nm. The diameter of the lens in the human eye may be approximately 0.5 cm. So, the smallest angle the human eye can resolve is roughly λ D 550 nm 0 . 5 cm 1 . 1 · 10 - 4 . Thus, you can not see the true size of Venus. The naked eye cannot resolve it. You must be seeing an effect due to the scattering from the Earth’s atmosphere. 9.16 Use formula (40) on page 475 of the book to find the width of the beam
Background image of page 1

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

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: on the moon: W L D (240000 miles) 632 . 8 nm 1 cm 15 miles. 1 Ph12a Solutions Problem Set 9 December 1, 2009 Figure 1: Geometry of problem 9.4 9.30 The circular wave a distance L from the source will appear to be planar in a circular region of radius R if its phase is approximately constant across this region. The maximum phase dierence will be between an edge and center of the region as shown in Fig. 2. The path dierence is given by p L 2 + R 2-L L (1 + R 2 2 L 2 )-L = R 2 2 L . Hence the phase dierence is = 2 path dierence = R 2 L . Hence the phase is approximately constant over the plane of the circle to the extent that its area is small compared to L . 2 Ph12a Solutions Problem Set 9 December 1, 2009 Figure 2: Geometry of problem 9.30 . 3...
View Full Document

This note was uploaded on 01/28/2011 for the course PH 2 taught by Professor Dudko during the Spring '09 term at UCSD.

Page1 / 3

Ph12a_s9_2009 - on the moon: W L D (240000 miles) 632 . 8...

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