Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 24 - The Wave Nature of Light - Problems - Page 708: 12

Answer

640 nm.

Work Step by Step

Equation 24–2a gives the condition for constructive interference, to be applied to the second-order blue light. $$d sin\theta=m\lambda_{blue}=2(480nm)$$ We use this with equation 24–2b because the unknown wavelength and unknown m’ result in an interference minimum. $$d sin\theta=(m’+\frac{1}{2})\lambda_x$$ Note that the slit separation and angle are the same in both situations. $$2(480nm)= (m’+\frac{1}{2})\lambda_x $$ Enumerate the possibilities and see which wavelength falls within the visible spectrum, 400-700 nm. For m’=0 $$2(480nm)= (0+\frac{1}{2})\lambda_x $$ $$\lambda_x =1920nm$$ For m’=1 $$2(480nm)= (1+\frac{1}{2})\lambda_x $$ $$\lambda_x =640nm$$ For m’=2 $$2(480nm)= (2+\frac{1}{2})\lambda_x $$ $$\lambda_x =384nm$$ The solutions will get smaller and smaller after this. The only solution in the visible region of 400-700 nm is 640 nm.
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