Answer
$\lambda=38pm$
Work Step by Step
We know that:
$2dsin\theta=m\lambda$
For $m=1$;
$2dsin\theta=(1)\lambda$
We plug in known values to obtain:
$2(0.94nm)sin(1.15^{\circ})=(1)\lambda$
$\lambda=0.038nm=38pm$
Fundamentals of Physics Extended (10th Edition)
by Halliday, David; Resnick, Robert; Walker, Jearl
Published by
Wiley
ISBN 10:
1-11823-072-8
ISBN 13:
978-1-11823-072-5
Chapter 36 - Diffraction - Problems - Page 1113: 67b
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