Answer
$\lambda_1=25pm$
Work Step by Step
We know that:
$2dsin\theta=m\lambda$
For $m=1$;
$2dsin\theta=(1)\lambda_1$
We plug in known values to obtain:
$2(0.94nm)sin(0.75^{\circ})=(1)\lambda_1$
$\lambda_1=0.025nm=25pm$
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: 67a
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