ch36-p043

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

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

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
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