ch36-p043 - 43. We will make use of arctangents and sines...

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

View Full Document Right Arrow Icon
which suggests that the angle takes us to a point between the sixth minimum (which would have m = 6.5) and the seventh maximum (which corresponds to m = 7). (c) Similarly, consider Eq. 36-3 with “continuously variable” m (of course, m should be an integer for diffraction minima, but for the moment we will solve for it as if it could be any real number): ( ) 12.0 m sin9.93 sin 3.4 0.600 m a m µ θ ° == λ which suggests that the angle takes us to a point between the third diffraction minimum ( m = 3) and the fourth one ( m = 4). The maxima (in the smaller peaks of the diffraction pattern) are not exactly midway between the minima; their location would make use of mathematics not covered in the prerequisites of the usual sophomore-level physics course. 43. We will make use of arctangents and sines in our solution, even though they can be “shortcut” somewhat since the angles are [almost] small enough to justify the use of the
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

Ask a homework question - tutors are online